Everything you need to know about vectors. Books download
During the execution of the computer of the current program inside the machine and in the environment associated with it (for example, in a technological process controlled by a computer), events may occur that require an immediate response to them from the machine.
The reaction is that the machine interrupts the processing of the current program and proceeds to the execution of some other program specifically designed for this event. Upon completion of this program, the computer returns to the execution of the interrupted program.
The process in question is called program interruption. It is fundamentally important that the moments of occurrence of events that require program interruption are not known in advance and therefore cannot be taken into account when programming.
Each event that requires an interrupt is accompanied by a signal notifying the computer - interrupt requests. The program requested by the interrupt request is called the interrupt program, as opposed to the interrupted program that was being executed by the machine before the request occurred.
The ability to interrupt programs is an important architectural property of a computer that allows efficient use of processor performance in the presence of several processes running in parallel in time that require control and maintenance by the processor at arbitrary times. First of all, this refers to the organization of parallel operation of the processor and peripheral devices of the machine, as well as to the use of a computer for real-time control of technological processes.
In order for the computer to be able, without requiring much effort from the programmer, to implement program interruptions with high speed, the machine must be given appropriate hardware and software, the totality of which is called the program interrupt system.
The main functions of the interrupt system are:
storing the state of the interrupted program and making a transition to the interrupted program
restoring the state of the interrupted program and returning to it.
The interrupt vector is the "initial state of the interrupting program" vector. The interrupt vector contains all the necessary information to jump to the interrupt program, including its start address. Each interrupt request (number) has its own interrupt vector, capable of initiating the execution of the corresponding interrupt program. Interrupt vectors are located in specially allocated fixed memory cells - the interrupt vector table.
The main place in the procedure for switching to an interrupting program is occupied by the procedure for transferring from the corresponding register (registers) of the processor to memory (in particular, to the stack) to save the current state vector of the interrupted program (so that you can return to its execution) and load the interrupt vector of the interrupting program into the register (registers) of the processor, to which control of the processor passes.
Interrupt classification
Interrupt requests can occur within the computer itself and in its external environment. The former include, for example, requests when such events occur in the computer as the appearance of an error in the operation of its equipment, an overflow of the bit grid, an attempt to divide by 0, an exit from the memory area established for a given program, a request by a peripheral device for an I / O operation, completion of an I / O operation by a peripheral device or the occurrence of an exception during this operation, etc. Although some of these events are generated by the program itself, the moments of their occurrence, as a rule, cannot be foreseen. Requests in the external environment may arise from other computers, from emergency and some other process sensors, etc.
The Intel 80x86 microprocessor family supports 256 priority interrupt levels triggered by three types of events:
internal hardware interrupts
external hardware interrupts
software interrupts
Internal hardware interrupts, sometimes called failures, are generated by certain events that occur during program execution, such as an attempt to divide by 0. Assigning certain interrupt numbers to such events is hardwired into the processor and cannot be changed.
External hardware interrupts initiated by peripheral equipment controllers or coprocessors (for example, 8087/80287). Interrupt sources are connected to either the processor's non-maskable interrupt (NMI) pin or the maskable interrupt pin (INTR). The NMI line is usually reserved for interrupts caused by catastrophic events such as memory parity errors or a power failure.
Software interrupts. Any program can initiate a synchronous software interrupt by executing the instruction int. MS-DOS uses interrupts from 20H to 3FH to interact with its modules and application programs (for example, the MS-DOS function manager is accessed by executing the command int 21h). BIOS programs stored in ROM and IBM PC applications use other interrupts, higher or lower. This distribution of interrupt numbers is conditional and is not fixed in hardware in any way.
Interrupt vector table
In order to associate the address of the interrupt handler with the interrupt number, the interrupt vector table is used, which occupies the first kilobyte of RAM. This table is in the address range from 0000:0000 to 0000:03FFh and consists of 256 entries - far addresses of interrupt handlers.
The entries in the interrupt vector table are called interrupt vectors. The first word of the table entry contains the offset component, and the second word contains the segment component of the interrupt handler address.
Interrupt vector 0 is at address 0000:0000, interrupt vector 1 is at 0000:0004, and so on. In general, the address of the interrupt vector is found by multiplying the interrupt number by 4.
Table initialization is performed partly by the BIOS basic input/output system after testing the hardware and before the operating system starts booting, partly when MS-DOS boots. The MS-DOS operating system can change some of the interrupt vectors set by the BIOS.
Interrupt vector table
Number |
Description |
division error. Called automatically after executing the DIV or IDIV commands if an overflow occurs as a result of division (for example, when dividing by 0). Typically, when this interrupt is handled, MS-DOS displays an error message and stops the program from running. In this case, for the i8086 processor, the return address points to the instruction following the division instruction, and for the i80286 processor and later models, to the first byte of the instruction that caused the interrupt |
|
Interruption of the step mode. Issued after each machine instruction is executed if the step trace bit TF is set in the flags word. Used for debugging programs. This interrupt is not generated after transferring data to segment registers with MOV and POP instructions. |
|
Hardware NMI. This interrupt can be used differently on different machines. It is usually generated when a parity error occurs in RAM and when an interrupt is requested from the coprocessor. |
|
Trace interrupt. Generated when executing a single-byte machine instruction with code CCh and is commonly used by debuggers to set a breakpoint |
|
Overflow. Generated by the INTO machine instruction if the OF overflow flag is set. If the flag is not set, the INTO command is executed as NOP. This interrupt is used to handle errors when performing arithmetic operations. |
|
Print a copy of the screen. Fired when the user has pressed a key |
|
Undefined opcode or instruction length greater than 10 bytes |
|
Special case of no arithmetic coprocessor |
|
IRQ0 - interval timer interrupt, occurs 18.2 times per second |
|
IRQ1 - keyboard interrupt. Emitted when the user presses and releases keys. Used to read data from the keyboard |
|
IRQ2 - used to cascade hardware interrupts |
|
IRQ3 - interruption of the asynchronous port COM2 |
|
IRQ4 - interrupt asynchronous port COM1 |
|
IRQ5 - hard disk controller interrupt (IBM PC/XT computers only) |
|
IRQ6 - an interrupt is generated by the floppy disk controller after the completion of the I / O operation |
|
IRQ7 - interrupt from parallel adapter. Generated when the printer connected to the adapter is ready to perform the next operation. Usually not used |
|
Video adapter maintenance |
|
Determining the configuration of devices in the system |
|
Determining the size of RAM |
|
Disk system maintenance |
|
Working with an asynchronous serial adapter |
|
Advanced Service |
|
Keyboard Maintenance |
|
Printer Maintenance |
|
Run BASIC in ROM if available |
|
Watch service |
|
Interrupt handler that occurs when the user presses a key combination |
|
Software interrupt, called 18.2 times per second by the timer hardware interrupt handler |
|
Video table address for 6845 video adapter controller |
|
Pointer to floppy parameter table |
|
Pointer to graphics table for ASCII characters 128-255 |
|
Used by MS-DOS or reserved for MS-DOS |
|
Interrupts reserved for user programs |
|
Not used |
|
IRQ8 - real time clock interrupt |
|
IRQ9 - interrupt from the EGA controller |
|
IRQ10 - reserved |
|
IRQ11 - reserved |
|
IRQ12 - reserved |
|
IRQ13 - interrupt from the arithmetic coprocessor |
|
IRQ14 - interrupt from the hard disk controller |
|
IRQ15 - reserved |
|
Not used |
|
Reserved for BASIC |
|
Used by the BASIC interpreter |
|
Not used |
Interrupts designated as IRQ0 - IRQ15 are external hardware interrupts.
Interrupt Service Order
The CPU, upon detecting an interrupt signal, pushes the program status word (defining various CPU flags), program segment register (CS), and instruction pointer (IP) onto the machine stack and disables the interrupt system. The CPU then uses the 8-bit number (interrupt number) set on the system bus by the interrupting process to retrieve the handler address from the vector table and resumes execution at that address.
If there are several sources of interrupt requests, a certain order (discipline) in servicing incoming requests must be established. In other words, priority relationships must be established between requests (and the corresponding interrupting programs), determining which of several incoming requests is to be processed first, and establishing whether this request (interrupting program) has the right or not to interrupt this or that program. If the highest priority of the exposed interrupt requests does not exceed the priority level of the program being executed by the processor, then the interrupt request is ignored or its service is deferred until the execution of the current program is completed. Each interrupt corresponds to a specific number, which determines the priority. A request with a lower number is considered to be of higher priority, i.e. Interrupt request number 0 has the highest priority, and interrupt request number 255 has the lowest priority.
The state of the system at the moment control is transferred to the interrupt handler is completely independent of whether the interrupt was initiated by an external device or was the result of the program executing the INT instruction. This circumstance is convenient to use when writing and testing external interrupt handlers, which can be debugged almost completely by exciting them with simple software tools.
Arguments are passed to interrupt handlers via registers or the stack.
Program model of x86 microprocessor. Classification, list and purpose of user registers.
The software model of the microprocessor is understood as that part of it that is left visible and accessible for programming. We will consider the program model on the example of the i80486 processor, which contains 32 registers, to some extent available for use by the programmer. These registers can be divided into two large groups:
16 user registers that the user can freely use in their programs to implement the task;
16 system registers registers designed to support various modes of operation, service functions.
Registers called high-speed memory areas located inside the processor in close proximity to its execution core. Access to them is incomparably faster than to RAM cells. Accordingly, machine instructions with operands in registers are executed as quickly as possible.
Custom registers include:
eight 32-bit registers that can be used by programmers to store data and addresses. They are called general purpose registers (RON):
six segment registers:
status and control registers:
flag register EFlags/Flags;
EIP/IP command pointer register.
Rice. 1.3i486 microprocessor user registers
Many of the register names are given with a slash separator. It should be noted that these are not different registers - they are parts of one large 32-bit register. They can be used in the program as separate objects. This was done to ensure the operability of programs written for the younger 16-bit microprocessor models from Intel, starting with the i8086. The i486 and Pentium microprocessors have mostly 32-bit registers. Their number, with the exception of segment registers, is the same as that of the i8086, but the dimension is larger, which is reflected in their designations - they have the prefix E (Extended).
Consider the composition and purpose of user registers.
General purpose registers
All registers of this group allow you to access their "lower" parts (see Fig. 1.3). Note that only the lower 16 and 8 bit parts of these registers can be used as independent objects. The upper 16 bits of these registers are not available as independent objects. This is done, as noted above, for compatibility with the younger 16-bit microprocessor models from Intel.
Let us list in more detail the registers belonging to the group of general purpose registers. Since these registers are physically located in the microprocessor inside the arithmetic logic unit (ALU), they are often called ALU registers:
EAX/AX/AH/AL (Accumulator register) – battery. Used to store intermediate data. In some commands, the use of this register is mandatory;
EBX/BX/BH/BL (Base register) – base register. Used to store the base address of some object in memory;
ECX/CX/CH/CL (Counter register) – register- counter. It is used in commands that perform some repetitive actions. Its use is often implicit and hidden in the algorithm of the corresponding command. For example, the loop organization command, in addition to transferring control to a command located at a certain address, decreases by one and analyzes the value of the ECX / CX register;
EDX / DX / DH / DL (Data register) - register data. Just like the EAX/AX/AH/AL register, it stores intermediate data. Some commands require its use; for some commands this happens implicitly (for example, multiplication and division).
The following two registers are used to support the so-called chain operations, that is, operations that sequentially process chains of elements, each of which can be 32, 16, or 8 bits long:
ESI/SI (Source Index register) - index source. This register in chain operations contains the current address of the element in the source chain;
EDI/DI (Destination Index register) – index receiver(recipient). This register in chain operations contains the current address in the destination chain.
The microprocessor architecture at the hardware and software level supports such a data structure as stack .
Stack is an area of memory specially allocated for temporary storage of program data. The microprocessor organizes work with the stack according to the following principle: the last item entered in this area is retrieved first.
To work with the stack in the microprocessor instruction system there are special commands, and in the microprocessor software model there are special registers for this:
ESP / SP (Stack Pointer register) - register pointer stack. Contains a pointer to the top of the stack in the current stack segment.
EBP/BP (Base Pointer register) - register stack frame base pointer. Designed to organize random access to data inside the stack.
In more detail, the features of using the stack are discussed in module No. 4 "Instructions for the i80486 microprocessor", section "Instructions for working with the stack".
In fact, the functionality of the ALU registers is not rigid. Most of the registers can be used in programming to store operands in almost any combination. But, as noted above, some commands use fixed registers to perform their actions.
segment registers
There are six segment registers in the microprocessor software model: CS, SS, D.S., ES, FS, GS. Their existence is due to the specifics of the organization and use of RAM by Intel microprocessors. It lies in the fact that the microprocessor hardware supports the structural organization of the program in the form of three parts, called segments. Accordingly, such an organization of memory is called segment.
In order to indicate the segments to which the program has access at a particular moment in time, and are intended segment registers. In fact, with a slight correction, as we will see later, these registers contain the memory addresses from which the corresponding segments begin. The logic of processing a machine instruction is constructed in such a way that addresses in well-defined segment registers are implicitly used when fetching an instruction, accessing program data or accessing the stack. The types of segments and their corresponding registers are discussed in more detail in Section 3 "Memory Segment Organization" of this module.
Status and control registers
The microprocessor includes two registers that constantly contain information about the state of both the microprocessor itself and the program whose commands are currently loaded onto the pipeline:
register flags EFlags/Flags;
register instruction pointer EIP/IP.
Using these registers, you can get information about the results of command execution and influence the state of the microprocessor itself. Let us consider in more detail the purpose and contents of these registers:
EFlags/ Flags(Flag register). Individual bits of this register have a specific functional purpose and are called flags. The lower part of this register is exactly the same as the Flags register for the i8086. On fig. 1.4 shows the contents of the EFlags register.
Based on usage patterns, the EFlags/Flags register flags can be divided into three groups:
8 status flags. These flags may change after machine instructions have been executed. Status flags register EFlags reflect the features of the result of the execution of arithmetic or logical operations. This makes it possible to analyze the state of the computational process and respond to it using conditional jump commands and subroutine calls. In table. 1.1 shows the main status flags and their purpose;
1 control flag. Designated DF (Direction Flag). It is located in bit 10 of the EFlags register and is used by chained instructions. The value of the DF flag determines the direction of element-by-element processing in these operations: from the beginning of the string to the end (DF = 0) or vice versa, from the end of the string to its beginning (DF = 1). There are special commands for working with the DF flag: cld(remove DF flag) and std(set the DF flag). The use of these commands allows you to bring the DF flag in line with the algorithm and provide automatic increment or decrement of counters when performing operations with strings;
5 system flags, which control I/O, maskable interrupts, debugging, task switching, and 8086 virtual mode. It is not recommended for application programs to modify these flags unnecessarily, as this will cause the program to terminate in most cases. In table. 1.2 lists the system flags and their purpose.
Rice. 1.4Register contents EFlags
Table 1.1
Basic Status Flags
Flag mnemonic |
Flag |
Bit number in EFlags | |
Carry Flag |
1 - the arithmetic operation carried over from the high bit of the result. The most significant bit is 7, 15, or 31, depending on the size of the operand; 0 - there was no transfer |
||
Parity Flag |
1 - 8 least significant bits (this flag is only for the 8 least significant bits of an operand of any size) of the result contains even number units; 0 - 8 least significant digits of the result contain an odd number of ones |
||
Zero Flag |
1 - the result is zero; 0 - non-zero result |
||
sign flag |
Reflects the state of the high bit of the result (bits 7, 15, or 31 for 8, 16, or 32-bit operands, respectively): 1 – the high bit of the result is 1; 0 - the most significant bit of the result is 0 |
||
Overflow Flag |
The flag of is used to fix the fact of the loss of a significant bit during arithmetic operations: 1 - as a result of the operation, a transfer (borrow) to (from) the most significant, sign bit of the result (bits 7, 15 or 31 for 8, 16 or 32-bit operands, respectively); 0 - as a result of the operation, there is no transfer (loan) to (from) the high, sign bit of the result |
Definition
Scalar- a value that can be characterized by a number. For example, length, area, mass, temperature, etc.
Vector a directed segment is called $\overline(A B)$; point $A$ - the beginning, point $B$ - the end of the vector (Fig. 1).
A vector is denoted by either two capital letters- by its beginning and end: $\overline(A B)$ or by one small letter: $\overline(a)$.
Definition
If the beginning and end of a vector are the same, then such a vector is called zero. Most often, the null vector is denoted as $\overline(0)$.
The vectors are called collinear, if they lie either on the same line or on parallel lines (Fig. 2).
Definition
Two collinear vectors $\overline(a)$ and $\overline(b)$ are called co-directional, if their directions are the same: $\overline(a) \uparrow \uparrow \overline(b)$ (Fig. 3, a). Two collinear vectors $\overline(a)$ and $\overline(b)$ are called opposite directions, if their directions are opposite: $\overline(a) \uparrow \downarrow \overline(b)$ (Fig. 3b).
Definition
The vectors are called coplanar if they are parallel to the same plane or lie in the same plane (Fig. 4).
Two vectors are always coplanar.
Definition
Length (module) vector $\overline(A B)$ is the distance between its start and end: $|\overline(A B)|$
A detailed theory about the length of a vector is at the link.
The length of the null vector is zero.
Definition
A vector whose length is equal to one is called unit vector or ortom.
The vectors are called equal if they lie on one or parallel lines; their directions coincide and lengths are equal.
In other words, two vectors equal, if they are collinear, co-directed and have equal lengths:
$\overline(a)=\overline(b)$ if $\overline(a) \uparrow \uparrow \overline(b),|\overline(a)|=|\overline(b)|$
At an arbitrary point $M$ in space, one can construct a single vector $\overline(M N)$ equal to the given vector $\overline(A B)$.
In this article, you and I will begin a discussion of one "magic wand" that will allow you to reduce many problems in geometry to simple arithmetic. This “wand” can make your life much easier, especially when you feel insecure in building spatial figures, sections, etc. All this requires a certain imagination and practical skills. The method, which we will begin to consider here, will allow you to abstract almost completely from all kinds of geometric constructions and reasoning. The method is called "coordinate method". In this article, we will consider the following questions:
- Coordinate plane
- Points and vectors on the plane
- Building a vector from two points
- Vector length (distance between two points)
- Midpoint coordinates
- Dot product of vectors
- Angle between two vectors
I think you already guessed why the coordinate method is called that? It is true that it received such a name, since it operates not with geometric objects, but with their numerical characteristics(coordinates). And the transformation itself, which makes it possible to move from geometry to algebra, consists in introducing a coordinate system. If the original figure was flat, then the coordinates are two-dimensional, and if the figure is three-dimensional, then the coordinates are three-dimensional. In this article, we will consider only the two-dimensional case. And the main purpose of the article is to teach you how to use some basic techniques of the coordinate method (they sometimes turn out to be useful when solving problems in planimetry in part B of the Unified State Examination). The following two sections on this topic are devoted to the discussion of methods for solving problems C2 (the problem of stereometry).
Where would it be logical to start discussing the coordinate method? Probably with the concept of a coordinate system. Remember when you first met her. It seems to me that in the 7th grade, when you learned about the existence of a linear function, for example. Let me remind you that you built it point by point. Do you remember? You chose an arbitrary number, substituted it into the formula and calculated in this way. For example, if, then, if, then, etc. What did you get as a result? And you received points with coordinates: and. Next, you drew a “cross” (coordinate system), chose a scale on it (how many cells you will have single segment) and marked on it the points you received, which you then connected with a straight line, the resulting line is the graph of the function.
There are a few things that need to be explained to you in a little more detail:
1. You choose a single segment for reasons of convenience, so that everything fits nicely and compactly in the picture
2. It is assumed that the axis goes from left to right, and the axis goes from bottom to top
3. They intersect at a right angle, and the point of their intersection is called the origin. It is marked with a letter.
4. In the record of the coordinate of a point, for example, on the left in brackets is the coordinate of the point along the axis, and on the right, along the axis. In particular, simply means that the point
5. In order to set any point on the coordinate axis, you need to specify its coordinates (2 numbers)
6. For any point lying on the axis,
7. For any point lying on the axis,
8. The axis is called the x-axis
9. The axis is called the y-axis
Now let's take the next step with you: mark two points. Connect these two points with a line. And we will put the arrow as if we are drawing a segment from point to point: that is, we will make our segment directed!
Remember what another name for a directed segment is? That's right, it's called a vector!
Thus, if we connect a dot to a dot, and the beginning will be point A, and the end will be point B, then we get a vector. You also did this construction in the 8th grade, remember?
It turns out that vectors, like points, can be denoted by two numbers: these numbers are called the coordinates of the vector. Question: do you think it is enough for us to know the coordinates of the beginning and end of the vector to find its coordinates? It turns out that yes! And it's very easy to do:
Thus, since in the vector the point is the beginning, and the end, the vector has the following coordinates:
For example, if, then the coordinates of the vector
Now let's do the opposite, find the coordinates of the vector. What do we need to change for this? Yes, you need to swap the beginning and end: now the beginning of the vector will be at a point, and the end at a point. Then:
Look closely, what is the difference between vectors and? Their only difference is the signs in the coordinates. They are opposite. This fact is written like this:
Sometimes, if it is not specifically stated which point is the beginning of the vector, and which is the end, then the vectors are denoted not by two capital letters, but one lowercase, for example: , etc.
Now a little practice and find the coordinates of the following vectors:
Examination:
Now solve the problem a little more difficult:
A vector torus with on-cha-scrap at a point has co-or-di-on-you. Find-di-te abs-cis-su points.
All the same is quite prosaic: Let be the coordinates of the point. Then
I compiled the system by determining what the coordinates of a vector are. Then the point has coordinates. We are interested in the abscissa. Then
Answer:
What else can you do with vectors? Yes, almost everything is the same as with ordinary numbers (except that you cannot divide, but you can multiply in two ways, one of which we will discuss here a little later)
- Vectors can be stacked with each other
- Vectors can be subtracted from each other
- Vectors can be multiplied (or divided) by an arbitrary non-zero number
- Vectors can be multiplied with each other
All these operations have a quite visual geometric representation. For example, the triangle (or parallelogram) rule for addition and subtraction:
A vector stretches or shrinks or changes direction when multiplied or divided by a number:
However, here we will be interested in the question of what happens to the coordinates.
1. When adding (subtracting) two vectors, we add (subtract) their coordinates element by element. That is:
2. When multiplying (dividing) a vector by a number, all its coordinates are multiplied (divided) by this number:
For example:
· Find-di-the sum of ko-or-di-nat century-to-ra.
Let's first find the coordinates of each of the vectors. Both of them have the same origin - the origin point. Their ends are different. Then, . Now we calculate the coordinates of the vector Then the sum of the coordinates of the resulting vector is equal to.
Answer:
Now solve the following problem yourself:
· Find the sum of the coordinates of the vector
We check:
Let's now consider the following problem: we have two points on the coordinate plane. How to find the distance between them? Let the first point be, and the second. Let's denote the distance between them as . Let's make the following drawing for clarity:
What I've done? I first connected points and, a also drew a line parallel to the axis from the point, and drew a line parallel to the axis from the point. Did they intersect at a point, forming a wonderful figure? Why is she wonderful? Yes, you and I almost know everything about right triangle. Well, the Pythagorean theorem, for sure. The desired segment is the hypotenuse of this triangle, and the segments are the legs. What are the coordinates of the point? Yes, they are easy to find from the picture: Since the segments are parallel to the axes and, respectively, their lengths are easy to find: if we denote the lengths of the segments, respectively, through, then
Now let's use the Pythagorean theorem. We know the lengths of the legs, we will find the hypotenuse:
Thus, the distance between two points is the root sum of the squared differences from the coordinates. Or - the distance between two points is the length of the segment connecting them. It is easy to see that the distance between the points does not depend on the direction. Then:
From this we draw three conclusions:
Let's practice a bit on calculating the distance between two points:
For example, if, then the distance between and is
Or let's go differently: find the coordinates of the vector
And find the length of the vector:
As you can see, it's the same!
Now practice a little on your own:
Task: find the distance between the given points:
We check:
Here are a couple more problems for the same formula, though they sound a little different:
1. Find-di-te the square of the length of the eyelid-to-ra.
2. Nai-di-te square of eyelid length-to-ra
I'm guessing you can handle them easily? We check:
1. And this is for attentiveness) We have already found the coordinates of the vectors before: . Then the vector has coordinates. The square of its length will be:
2. Find the coordinates of the vector
Then the square of its length is
Nothing complicated, right? Simple arithmetic, nothing more.
The following tasks cannot be unambiguously classified, they are rather general erudition and the ability to draw simple pictures.
1. Find-di-those sine of the angle on-clo-on-from-cut, connect-one-n-th-th point, with the abscissa axis.
And
How are we going to do it here? You need to find the sine of the angle between and the axis. And where can we look for the sine? That's right, in a right triangle. So what do we need to do? Build this triangle!
Since the coordinates of the point and, then the segment is equal, and the segment. We need to find the sine of the angle. Let me remind you that the sine is the ratio of the opposite leg to the hypotenuse, then
What are we left to do? Find the hypotenuse. You can do it in two ways: using the Pythagorean theorem (the legs are known!) or using the formula for the distance between two points (actually the same as the first method!). I will go the second way:
Answer:
The next task will seem even easier to you. She - on the coordinates of the point.
Task 2. From the point, the per-pen-di-ku-lar is lowered onto the abs-ciss axis. Nai-di-te abs-cis-su os-no-va-niya per-pen-di-ku-la-ra.
Let's make a drawing:
The base of the perpendicular is the point at which it intersects the x-axis (axis) for me this is a point. The figure shows that it has coordinates: . We are interested in the abscissa - that is, the "X" component. She is equal.
Answer: .
Task 3. Under the conditions of the previous problem, find the sum of the distances from the point to the coordinate axes.
The task is generally elementary if you know what the distance from a point to the axes is. You know? I hope, but still I remind you:
So, in my drawing, located a little higher, I have already depicted one such perpendicular? What axis is it? to the axis. And what is its length then? She is equal. Now draw a perpendicular to the axis yourself and find its length. It will be equal, right? Then their sum is equal.
Answer: .
Task 4. In the conditions of problem 2, find the ordinate of the point symmetrical to the point about the x-axis.
I think you intuitively understand what symmetry is? Very many objects have it: many buildings, tables, planes, many geometric figures: ball, cylinder, square, rhombus, etc. Roughly speaking, symmetry can be understood as follows: a figure consists of two (or more) identical halves. This symmetry is called axial. What then is an axis? This is exactly the line along which the figure can, relatively speaking, be “cut” into identical halves (in this picture, the axis of symmetry is straight):
Now let's get back to our task. We know that we are looking for a point that is symmetric about the axis. Then this axis is the axis of symmetry. So, we need to mark a point so that the axis cuts the segment into two equal parts. Try to mark such a point yourself. Now compare with my solution:
Did you do the same? Fine! At the found point, we are interested in the ordinate. She is equal
Answer:
Now tell me, after thinking for a second, what will be the abscissa of the point symmetrical to point A about the y-axis? What is your answer? Correct answer: .
In general, the rule can be written like this:
A point symmetrical to a point about the x-axis has the coordinates:
A point symmetrical to a point about the y-axis has coordinates:
Well, now it's really scary. task: Find the coordinates of a point that is symmetrical to a point, relative to the origin. You first think for yourself, and then look at my drawing!
Answer:
Now parallelogram problem:
Task 5: The points are ver-shi-na-mi-pa-ral-le-lo-gram-ma. Find-dee-te or-dee-on-tu points.
You can solve this problem in two ways: logic and the coordinate method. I will first apply the coordinate method, and then I will tell you how you can decide otherwise.
It is quite clear that the abscissa of the point is equal. (it lies on the perpendicular drawn from the point to the x-axis). We need to find the ordinate. Let's take advantage of the fact that our figure is a parallelogram, which means that. Find the length of the segment using the formula for the distance between two points:
We lower the perpendicular connecting the point with the axis. The point of intersection is denoted by a letter.
The length of the segment is equal. (find the problem yourself, where we discussed this moment), then we will find the length of the segment using the Pythagorean theorem:
The length of the segment is exactly the same as its ordinate.
Answer: .
Another solution (I'll just provide a picture that illustrates it)
Solution progress:
1. Spend
2. Find point coordinates and length
3. Prove that.
Another one cut length problem:
The points are-la-yut-xia top-shi-on-mi tri-angle-no-ka. Find the length of his midline, par-ral-lel-noy.
Do you remember what the middle line of a triangle is? Then for you this task is elementary. If you don’t remember, then I’ll remind you: the middle line of a triangle is the line that connects the midpoints opposite sides. It is parallel to the base and equal to half of it.
The base is a segment. We had to look for its length earlier, it is equal. Then the length of the midline is half as long and equal.
Answer: .
Comment: This problem can be solved in another way, which we will turn to a little later.
In the meantime, here are a few tasks for you, practice on them, they are quite simple, but they help to “fill your hand” using the coordinate method!
1. The points appear-la-yut-xia top-shi-on-mi tra-pe-tion. Find the length of its midline.
2. Points and yav-la-yut-xia ver-shi-na-mi pa-ral-le-lo-gram-ma. Find-dee-te or-dee-on-tu points.
3. Find the length from the cut, connect the second point and
4. Find-di-te the area for-the-red-shen-noy fi-gu-ry on the ko-or-di-nat-noy plane.
5. A circle centered at na-cha-le ko-or-di-nat passes through a point. Find-de-te her ra-di-mustache.
6. Find-di-te ra-di-us of the circle, describe-san-noy near the right-angle-no-ka, the tops of someone-ro-go have co-or-di-on-you co-from-reply-stven-but
Solutions:
1. It is known that the midline of a trapezoid is equal to half the sum of its bases. The base is equal, but the base. Then
Answer:
2. The easiest way to solve this problem is to notice that (parallelogram rule). Calculate the coordinates of the vectors and is not difficult: . When adding vectors, the coordinates are added. Then has coordinates. The point has the same coordinates, since the beginning of the vector is a point with coordinates. We are interested in the ordinate. She is equal.
Answer:
3. We act immediately according to the formula for the distance between two points:
Answer:
4. Look at the picture and say, between which two figures is the shaded area “squeezed”? It is sandwiched between two squares. Then the area of the desired figure is equal to the area of the large square minus the area of the small one. Side small square is a line segment that connects points and its length is
Then the area of the small square is
We do the same with a large square: its side is a segment connecting the points and its length is equal to
Then the area of the large square is
The area of the desired figure is found by the formula:
Answer:
5. If the circle has the origin as its center and passes through a point, then its radius will be exactly equal to the length of the segment (make a drawing and you will understand why this is obvious). Find the length of this segment:
Answer:
6. It is known that the radius of a circle circumscribed about a rectangle is equal to half of its diagonal. Let's find the length of any of the two diagonals (after all, in a rectangle they are equal!)
Answer:
Well, did you manage everything? It wasn't that hard to figure it out, was it? There is only one rule here - to be able to make a visual picture and simply “read” all the data from it.
We have very little left. There are literally two more points that I would like to discuss.
Let's try to solve this simple problem. Let two points and be given. Find the coordinates of the middle of the segment. The solution to this problem is as follows: let the point be the desired middle, then it has coordinates:
That is: coordinates of the middle of the segment = arithmetic mean of the corresponding coordinates of the ends of the segment.
This rule is very simple and usually does not cause difficulties for students. Let's see in what problems and how it is used:
1. Find-di-te or-di-na-tu se-re-di-us from-cut, connect-nya-yu-th-th point and
2. The points are yav-la-yut-xia ver-shi-na-mi-che-you-reh-coal-no-ka. Find-di-te or-di-na-tu points of re-re-se-che-niya of his dia-go-on-lei.
3. Find-di-te abs-cis-su of the center of the circle, describe-san-noy near the right-angle-no-ka, the tops of someone-ro-go have co-or-di-on-you co-from-reply-but.
Solutions:
1. The first task is just a classic. We act immediately by determining the midpoint of the segment. She has coordinates. The ordinate is equal.
Answer:
2. It is easy to see that the given quadrilateral is a parallelogram (even a rhombus!). You can prove it yourself by calculating the lengths of the sides and comparing them with each other. What do I know about a parallelogram? Its diagonals are bisected by the intersection point! Aha! So the point of intersection of the diagonals is what? This is the middle of any of the diagonals! I will choose, in particular, the diagonal. Then the point has coordinates. The ordinate of the point is equal to.
Answer:
3. What is the center of the circle circumscribed about the rectangle? It coincides with the point of intersection of its diagonals. What do you know about the diagonals of a rectangle? They are equal and the intersection point is divided in half. The task has been reduced to the previous one. Take, for example, the diagonal. Then if is the center of the circumscribed circle, then is the middle. I'm looking for coordinates: The abscissa is equal.
Answer:
Now practice a little on your own, I will only give the answers to each problem so that you can check yourself.
1. Find-di-te ra-di-us of the circle, describe-san-noy near the triangle-no-ka, the tops of someone-ro-go have co-or-di-on-you
2. Find-di-te or-di-na-that center of the circle-no-sti, describe-san-noy near a triangle-no-ka, the tops of someone-ro-go have co-or-di-na-you
3. What kind of ra-di-y-sa should there be a circle with a center at a point so that it touches the abs-ciss axis?
4. Find-di-te or-di-on-that point of re-re-se-che-ing of the axis and from-cut, connect-nya-yu-th-th point and
Answers:
Did everything work out? I really hope for it! Now - the last push. Now be especially careful. The material that I will now explain is directly related not only to simple tasks to the coordinate method from part B, but also occurs everywhere in problem C2.
Which of my promises have I not yet kept? Remember what operations on vectors I promised to introduce and which ones I eventually introduced? Am I sure I haven't forgotten anything? Forgot! I forgot to explain what multiplication of vectors means.
There are two ways to multiply a vector by a vector. Depending on the chosen method, we will get objects of a different nature:
The vector product is quite tricky. How to do it and why it is needed, we will discuss with you in the next article. And in this we will focus on the scalar product.
There are already two ways that allow us to calculate it:
As you guessed, the result should be the same! So let's look at the first way first:
Dot product through coordinates
Find: - common notation for dot product
The formula for the calculation is as follows:
That is, the dot product = the sum of the products of the coordinates of the vectors!
Example:
Find-dee-te
Solution:
Find the coordinates of each of the vectors:
We calculate the scalar product by the formula:
Answer:
You see, absolutely nothing complicated!
Well, now try it yourself:
Find-di-te scalar-noe pro-from-ve-de-nie century-to-ditch and
Did you manage? Maybe he noticed a little trick? Let's check:
Vector coordinates, as in the previous task! Answer: .
In addition to the coordinate, there is another way to calculate the scalar product, namely, through the lengths of the vectors and the cosine of the angle between them:
Denotes the angle between the vectors and.
That is, the scalar product is equal to the product of the lengths of the vectors and the cosine of the angle between them.
Why do we need this second formula, if we have the first one, which is much simpler, at least there are no cosines in it. And we need it so that from the first and second formulas we can deduce how to find the angle between vectors!
Let Then remember the formula for the length of a vector!
Then if I plug this data into the dot product formula, I get:
But in other way:
So what have we got? We now have a formula to calculate the angle between two vectors! Sometimes, for brevity, it is also written like this:
That is, the algorithm for calculating the angle between vectors is as follows:
- We calculate the scalar product through the coordinates
- Find the lengths of vectors and multiply them
- Divide the result of point 1 by the result of point 2
Let's practice with examples:
1. Find the angle between the eyelids-to-ra-mi and. Give your answer in degrees.
2. Under the conditions of the previous problem, find the cosine between the vectors
Let's do this: I'll help you solve the first problem, and try to do the second one yourself! Agree? Then let's start!
1. These vectors are our old friends. We have already considered their scalar product and it was equal. Their coordinates are: , . Then we find their lengths:
Then we are looking for the cosine between the vectors:
What is the cosine of the angle? This is the corner.
Answer:
Well, now solve the second problem yourself, and then compare! I'll just give a very short solution:
2. has coordinates, has coordinates.
Let be the angle between the vectors and, then
Answer:
It should be noted that the tasks directly on the vectors and the method of coordinates in part B examination work quite rare. However, the vast majority of C2 problems can be easily solved by introducing a coordinate system. So you can consider this article as a foundation, on the basis of which we will make quite tricky constructions that we will need to solve complex problems.
COORDINATES AND VECTORS. INTERMEDIATE LEVEL
You and I continue to study the method of coordinates. In the last part, we derived a number of important formulas that allow:
- Find vector coordinates
- Find the length of a vector (alternatively: the distance between two points)
- Add, subtract vectors. Multiply them by a real number
- Find the midpoint of a segment
- Calculate dot product of vectors
- Find the angle between vectors
Of course, the entire coordinate method does not fit into these 6 points. It underlies such a science as analytical geometry, which you will get acquainted with at the university. I just want to build a foundation that will allow you to solve problems in a single state. exam. We figured out the tasks of part B in Now it's time to move to a qualitatively new level! This article will be devoted to a method for solving those C2 problems in which it would be reasonable to switch to the coordinate method. This reasonableness is determined by what needs to be found in the problem, and what figure is given. So, I would use the coordinate method if the questions are:
- Find the angle between two planes
- Find the angle between a line and a plane
- Find the angle between two lines
- Find the distance from a point to a plane
- Find the distance from a point to a line
- Find the distance from a straight line to a plane
- Find the distance between two lines
If the figure given in the condition of the problem is a body of revolution (ball, cylinder, cone ...)
Suitable figures for the coordinate method are:
- cuboid
- Pyramid (triangular, quadrangular, hexagonal)
Also in my experience it is inappropriate to use the coordinate method for:
- Finding the areas of sections
- Calculations of volumes of bodies
However, it should be immediately noted that three “unfavorable” situations for the coordinate method are quite rare in practice. In most tasks, it can become your savior, especially if you are not very strong in three-dimensional constructions (which are sometimes quite intricate).
What are all the figures I have listed above? They are no longer flat, such as a square, triangle, circle, but voluminous! Accordingly, we need to consider not a two-dimensional, but a three-dimensional coordinate system. It is built quite easily: just in addition to the abscissa and ordinates, we will introduce another axis, the applicate axis. The figure schematically shows their relative position:
All of them are mutually perpendicular, intersect at one point, which we will call the origin. The abscissa axis, as before, will be denoted, the ordinate axis - , and the introduced applicate axis - .
If earlier each point on the plane was characterized by two numbers - the abscissa and the ordinate, then each point in space is already described by three numbers - the abscissa, the ordinate, the applicate. For example:
Accordingly, the abscissa of the point is equal, the ordinate is , and the applicate is .
Sometimes the abscissa of a point is also called the projection of the point on the abscissa axis, the ordinate is the projection of the point on the ordinate axis, and the applicate is the projection of the point on the applicate axis. Accordingly, if a point is given then, a point with coordinates:
called the projection of a point onto a plane
called the projection of a point onto a plane
A natural question arises: are all the formulas derived for the two-dimensional case valid in space? The answer is yes, they are just and have the same appearance. For a small detail. I think you already guessed which one. In all formulas, we will have to add one more term responsible for the applicate axis. Namely.
1. If two points are given: , then:
- Vector coordinates:
- Distance between two points (or vector length)
- The middle of the segment has coordinates
2. If two vectors are given: and, then:
- Their dot product is:
- The cosine of the angle between the vectors is:
However, space is not so simple. As you understand, the addition of one more coordinate introduces a significant variety in the spectrum of figures "living" in this space. And for further narration, I need to introduce some, roughly speaking, "generalization" of the straight line. This "generalization" will be a plane. What do you know about plane? Try to answer the question, what is a plane? It's very difficult to say. However, we all intuitively imagine what it looks like:
Roughly speaking, this is a kind of endless “leaf” thrust into space. "Infinity" should be understood that the plane extends in all directions, that is, its area is equal to infinity. However, this explanation "on the fingers" does not give the slightest idea about the structure of the plane. And we will be interested in it.
Let's remember one of the basic axioms of geometry:
- A straight line passes through two different points on a plane, moreover, only one:
Or its analog in space:
Of course, you remember how to derive the equation of a straight line from two given points, this is not at all difficult: if the first point has coordinates: and the second, then the equation of the straight line will be as follows:
You went through this in 7th grade. In space, the equation of a straight line looks like this: let us have two points with coordinates: , then the equation of a straight line passing through them has the form:
For example, a line passes through points:
How should this be understood? This should be understood as follows: a point lies on a line if its coordinates satisfy the following system:
We will not be very interested in the equation of a straight line, but we need to pay attention to the very important concept of the directing vector of a straight line. - any non-zero vector lying on a given line or parallel to it.
For example, both vectors are direction vectors of a straight line. Let be a point lying on a straight line, and be its directing vector. Then the equation of a straight line can be written in the following form:
Once again, I will not be very interested in the equation of a straight line, but I really need you to remember what a direction vector is! Again: it is ANY non-zero vector lying on a line, or parallel to it.
Withdraw three-point equation of a plane is no longer so trivial, and usually this issue is not considered in the course high school. But in vain! This technique is vital when we resort to the coordinate method to solve complex problems. However, I assume that you are full of desire to learn something new? Moreover, you will be able to impress your teacher at the university when it turns out that you already know how to use the technique that is usually studied in the course of analytic geometry. So let's get started.
The equation of a plane is not too different from the equation of a straight line on a plane, namely, it has the form:
some numbers (not all equal to zero), but variables, for example: etc. As you can see, the equation of a plane is not very different from the equation of a straight line (linear function). However, remember what we argued with you? We said that if we have three points that do not lie on one straight line, then the equation of the plane is uniquely restored from them. But how? I'll try to explain to you.
Since the plane equation is:
And the points belong to this plane, then when substituting the coordinates of each point into the equation of the plane, we should get the correct identity:
Thus, there is a need to solve three equations already with unknowns! Dilemma! However, we can always assume that (for this we need to divide by). Thus, we get three equations with three unknowns:
However, we will not solve such a system, but write out the cryptic expression that follows from it:
Equation of a plane passing through three given points
\[\left| (\begin(array)(*(20)(c))(x - (x_0))&((x_1) - (x_0))&((x_2) - (x_0))\\(y - (y_0))&((y_1) - (y_0))&((y_2) - (y_0))\\(z - (z_0))&((z_1) - (z_0))&((z_2) - (z_0 ))\end(array)) \right| = 0\]
Stop! What else is this? Some very unusual module! However, the object that you see in front of you has nothing to do with the module. This object is called a third-order determinant. From now on, when you deal with the method of coordinates on a plane, you will often come across these very determinants. What is a third order determinant? Oddly enough, it's just a number. It remains to understand what specific number we will compare with the determinant.
Let's first write the third-order determinant in more general view:
Where are some numbers. Moreover, by the first index we mean the row number, and by the index - the column number. For example, means that given number stands at the intersection of the second row and third column. Let's pose the following question: how exactly are we going to calculate such a determinant? That is, what specific number will we compare it with? For the determinant of precisely the third order, there is a heuristic (visual) triangle rule, it looks like this:
- The product of the elements of the main diagonal (from top left to bottom right) the product of the elements that form the first triangle "perpendicular" to the main diagonal the product of the elements that form the second triangle "perpendicular" to the main diagonal
- The product of the elements of the secondary diagonal (from the upper right corner to the lower left) the product of the elements that form the first triangle "perpendicular" of the secondary diagonal the product of the elements that form the second triangle "perpendicular" of the secondary diagonal
- Then the determinant is equal to the difference values obtained at the step and
If we write all this in numbers, then we get the following expression:
However, you don’t need to memorize the calculation method in this form, it’s enough to just keep the triangles in your head and the very idea of \u200b\u200bwhat is added to what and what is then subtracted from what).
Let's illustrate the triangle method with an example:
1. Calculate the determinant:
Let's figure out what we add and what we subtract:
Terms that come with a "plus":
This is the main diagonal: the product of the elements is
The first triangle, "perpendicular to the main diagonal: the product of the elements is
The second triangle, "perpendicular to the main diagonal: the product of the elements is
We add three numbers:
Terms that come with a "minus"
This is a side diagonal: the product of the elements is
The first triangle, "perpendicular to the secondary diagonal: the product of the elements is
The second triangle, "perpendicular to the secondary diagonal: the product of the elements is
We add three numbers:
All that remains to be done is to subtract from the sum of the plus terms the sum of the minus terms:
Thus,
As you can see, there is nothing complicated and supernatural in the calculation of third-order determinants. It is simply important to remember about triangles and not to make arithmetic mistakes. Now try to calculate yourself:
We check:
- The first triangle perpendicular to the main diagonal:
- The second triangle perpendicular to the main diagonal:
- The sum of the plus terms:
- First triangle perpendicular to the side diagonal:
- The second triangle, perpendicular to the side diagonal:
- The sum of terms with a minus:
- Sum of plus terms minus sum of minus terms:
Here's a couple more determinants for you, calculate their values yourself and compare with the answers:
Answers:
Well, did everything match? Great, then you can move on! If there are difficulties, then my advice is this: on the Internet there are a bunch of programs for calculating the determinant online. All you need is to come up with your own determinant, calculate it yourself, and then compare it with what the program calculates. And so on until the results start to match. I'm sure this moment will not be long in coming!
Now let's return to the determinant that I wrote out when I talked about the equation of a plane passing through three given points:
All you have to do is calculate its value directly (using the triangle method) and set the result equal to zero. Naturally, since they are variables, you will get some expression that depends on them. It is this expression that will be the equation of a plane passing through three given points that do not lie on one straight line!
Let's illustrate this with a simple example:
1. Construct the equation of the plane passing through the points
We compose a determinant for these three points:
Simplifying:
Now we calculate it directly according to the rule of triangles:
\[(\left| (\begin(array)(*(20)(c))(x + 3)&2&6\\(y - 2)&0&1\\(z + 1)&5&0\end(array)) \right| = \left((x + 3) \right) \cdot 0 \cdot 0 + 2 \cdot 1 \cdot \left((z + 1) \right) + \le ft((y - 2) \right) \cdot 5 \cdot 6 - )\]
Thus, the equation of the plane passing through the points is:
Now try to solve one problem yourself, and then we will discuss it:
2. Find the equation of the plane passing through the points
Well, let's discuss the solution now:
We make a determinant:
And calculate its value:
Then the equation of the plane has the form:
Or, reducing by, we get:
Now two tasks for self-control:
- Construct the equation of a plane passing through three points:
Answers:
Did everything match? Again, if there are certain difficulties, then my advice is this: take three points from your head (with a high degree of probability they will not lie on one straight line), build a plane on them. And then check yourself online. For example, on the site:
However, with the help of determinants, we will construct not only the equation of the plane. Remember, I told you that for vectors, not only the dot product is defined. There is also a vector, as well as a mixed product. And if the scalar product of two vectors will be a number, then the vector product of two vectors will be a vector, and this vector will be perpendicular to the given ones:
And its module will be equal to area parallelogram built on vectors and. We will need this vector to calculate the distance from a point to a line. How can we count vector product vectors and if their coordinates are given? The determinant of the third order again comes to our aid. However, before I move on to the algorithm for calculating the cross product, I have to make a small lyrical digression.
This digression concerns the basis vectors.
Schematically they are shown in the figure:
Why do you think they are called basic? The fact is that :
Or in the picture:
The validity of this formula is obvious, because:
vector product
Now I can start introducing the cross product:
The vector product of two vectors is a vector that is calculated according to the following rule:
Now let's give some examples of calculating the cross product:
Example 1: Find the cross product of vectors:
Solution: I make a determinant:
And I calculate it:
Now, from writing through basis vectors, I will return to the usual vector notation:
Thus:
Now try.
Ready? We check:
And traditionally two tasks to control:
- Find the cross product of the following vectors:
- Find the cross product of the following vectors:
Answers:
Mixed product of three vectors
The last construction I need is the mixed product of three vectors. It, like a scalar, is a number. There are two ways to calculate it. - through the determinant, - through the mixed product.
Namely, let's say we have three vectors:
Then the mixed product of three vectors, denoted by can be calculated as:
1. - that is, the mixed product is the scalar product of a vector and the vector product of two other vectors
For example, the mixed product of three vectors is:
Try to calculate it yourself using the vector product and make sure that the results match!
And again - two examples for an independent solution:
Answers:
Choice of coordinate system
Well, now we have all the necessary foundation of knowledge to solve complex stereometric problems in geometry. However, before proceeding directly to the examples and algorithms for solving them, I believe that it will be useful to dwell on the following question: how exactly choose a coordinate system for a particular figure. After all, it is the choice of the relative position of the coordinate system and the figure in space that will ultimately determine how cumbersome the calculations will be.
I remind you that in this section we are considering the following figures:
- cuboid
- Straight prism (triangular, hexagonal…)
- Pyramid (triangular, quadrangular)
- Tetrahedron (same as triangular pyramid)
For a cuboid or cube, I recommend the following construction:
That is, I will place the figure “in the corner”. The cube and the box are very good figures. For them, you can always easily find the coordinates of its vertices. For example, if (as shown in the picture)
then the vertex coordinates are:
Of course, you don’t need to remember this, but remember how best to position the cube or cuboid- desirable.
straight prism
Prism is a more harmful figure. You can arrange it in space in different ways. However, I think the following is the best option:
Triangular prism:
That is, we put one of the sides of the triangle entirely on the axis, and one of the vertices coincides with the origin.
Hexagonal prism:
That is, one of the vertices coincides with the origin, and one of the sides lies on the axis.
Quadrangular and hexagonal pyramid:
A situation similar to a cube: we combine two sides of the base with the coordinate axes, we combine one of the vertices with the origin. The only small difficulty will be to calculate the coordinates of the point.
For a hexagonal pyramid - the same as for a hexagonal prism. The main task will again be in finding the coordinates of the vertex.
Tetrahedron (triangular pyramid)
The situation is very similar to the one I gave for the triangular prism: one vertex coincides with the origin, one side lies on the coordinate axis.
Well, now you and I are finally close to starting to solve problems. From what I said at the very beginning of the article, you could draw the following conclusion: most C2 problems fall into 2 categories: problems for the angle and problems for the distance. First, we will consider problems for finding an angle. They, in turn, are divided into the following categories (as the complexity increases):
Problems for finding corners
- Finding the angle between two lines
- Finding the angle between two planes
Let's consider these problems sequentially: let's start by finding the angle between two straight lines. Come on, remember, have you and I solved similar examples before? You remember, because we already had something similar ... We were looking for an angle between two vectors. I remind you, if two vectors are given: and, then the angle between them is found from the relation:
Now we have a goal - finding the angle between two straight lines. Let's turn to the "flat picture":
How many angles do we get when two lines intersect? Already things. True, only two of them are not equal, while others are vertical to them (and therefore coincide with them). So what angle should we consider the angle between two straight lines: or? Here the rule is: the angle between two straight lines is always no more than degrees. That is, from two angles, we will always choose the angle with the smallest degree measure. That is, in this picture, the angle between the two lines is equal. In order not to bother with finding the smallest of the two angles every time, cunning mathematicians suggested using the module. Thus, the angle between two straight lines is determined by the formula:
You, as an attentive reader, should have had a question: where, in fact, do we get these very numbers that we need to calculate the cosine of an angle? Answer: we will take them from the direction vectors of the lines! Thus, the algorithm for finding the angle between two lines is as follows:
- We apply formula 1.
Or in more detail:
- We are looking for the coordinates of the direction vector of the first straight line
- We are looking for the coordinates of the direction vector of the second line
- Calculate the modulus of their scalar product
- We are looking for the length of the first vector
- We are looking for the length of the second vector
- Multiply the results of point 4 by the results of point 5
- We divide the result of point 3 by the result of point 6. We get the cosine of the angle between the lines
- If given result allows you to accurately calculate the angle, we are looking for it
- Otherwise, we write through the arccosine
Well, now it's time to move on to the tasks: I will demonstrate the solution of the first two in detail, I will present the solution of another one in summary, and for the last two problems I will only give answers, you must carry out all the calculations for them yourself.
Tasks:
1. In the right tet-ra-ed-re, find-di-te the angle between you-so-that tet-ra-ed-ra and the me-di-a-noy bo-ko-how side.
2. In the right-forward six-coal-pi-ra-mi-de, the hundred-ro-ns of the os-no-va-niya are somehow equal, and the side ribs are equal, find the angle between the straight lines and.
3. The lengths of all edges of the right-handed four-you-rech-coal-noy pi-ra-mi-dy are equal to each other. Find-di-the angle between the straight lines and if from-re-zok - you-so-that given pi-ra-mi-dy, the point - se-re-di-on her bo-ko-go edge
4. On the edge of the cube from-me-che-to a point so that Find-di-te the angle between the straight lines and
5. Point - se-re-di-on the edges of the cube Nai-di-te the angle between the straight lines and.
It is no coincidence that I placed the tasks in this order. While you have not yet had time to begin to navigate the coordinate method, I myself will analyze the most “problematic” figures, and I will leave you to deal with the simplest cube! Gradually you have to learn how to work with all the figures, I will increase the complexity of the tasks from topic to topic.
Let's start solving problems:
1. Draw a tetrahedron, place it in the coordinate system as I suggested earlier. Since the tetrahedron is regular, then all its faces (including the base) are regular triangles. Since we are not given the length of the side, I can take it equal. I think you understand that the angle will not really depend on how much our tetrahedron will be "stretched" ?. I will also draw the height and median in the tetrahedron. Along the way, I will draw its base (it will also come in handy for us).
I need to find the angle between and. What do we know? We only know the coordinate of the point. So, we need to find more coordinates of the points. Now we think: a point is a point of intersection of heights (or bisectors or medians) of a triangle. A dot is an elevated point. The point is the midpoint of the segment. Then finally we need to find: the coordinates of the points: .
Let's start with the simplest: point coordinates. Look at the figure: It is clear that the applicate of a point is equal to zero (the point lies on a plane). Its ordinate is equal (because it is the median). It is more difficult to find its abscissa. However, this is easily done on the basis of the Pythagorean theorem: Consider a triangle. Its hypotenuse is equal, and one of the legs is equal Then:
Finally we have:
Now let's find the coordinates of the point. It is clear that its applicate is again equal to zero, and its ordinate is the same as that of a point, that is. Let's find its abscissa. This is done rather trivially if one remembers that heights equilateral triangle the intersection point is divided in proportion counting from the top. Since: , then the desired abscissa of the point, equal to the length segment is equal to: . Thus, the coordinates of the point are:
Let's find the coordinates of the point. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. And the applique is equal to the length of the segment. - this is one of the legs of the triangle. The hypotenuse of a triangle is a segment - a leg. It is searched for the reasons that I highlighted in bold:
The point is the midpoint of the segment. Then we need to remember the formula for the coordinates of the middle of the segment:
That's it, now we can look for the coordinates of the direction vectors:
Well, everything is ready: we substitute all the data into the formula:
Thus,
Answer:
You should not be afraid of such "terrible" answers: for problems C2 this is a common practice. I would rather be surprised by the "beautiful" answer in this part. Also, as you noted, I practically did not resort to anything other than the Pythagorean theorem and the property of the heights of an equilateral triangle. That is, to solve the stereometric problem, I used the very minimum of stereometry. The gain in this is partially "extinguished" by rather cumbersome calculations. But they are quite algorithmic!
2. Draw the correct hexagonal pyramid together with the coordinate system, as well as its base:
We need to find the angle between the lines and. Thus, our task is reduced to finding the coordinates of points: . We will find the coordinates of the last three from the small drawing, and we will find the coordinate of the vertex through the coordinate of the point. Lots of work, but gotta get started!
a) Coordinate: it is clear that its applicate and ordinate are zero. Let's find the abscissa. To do this, consider a right triangle. Alas, in it we only know the hypotenuse, which is equal to. We will try to find the leg (because it is clear that twice the length of the leg will give us the abscissa of the point). How can we look for it? Let's remember what kind of figure we have at the base of the pyramid? This is a regular hexagon. What does it mean? This means that all sides and all angles are equal. We need to find one such corner. Any ideas? There are a lot of ideas, but there is a formula:
The sum of the angles of a regular n-gon is .
So the sum of the angles regular hexagon equals degrees. Then each of the angles is equal to:
Let's look at the picture again. It is clear that the segment is the bisector of the angle. Then the angle is degrees. Then:
Then where.
So it has coordinates
b) Now we can easily find the coordinate of the point: .
c) Find the coordinates of the point. Since its abscissa coincides with the length of the segment, it is equal. Finding the ordinate is also not very difficult: if we connect the points and and denote the point of intersection of the line, say for. (do it yourself simple construction). Then Thus, the ordinate of point B is equal to the sum of the lengths of the segments. Let's look at the triangle again. Then
Then since Then the point has coordinates
d) Now find the coordinates of the point. Consider a rectangle and prove that Thus, the coordinates of the point are:
e) It remains to find the coordinates of the vertex. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. Let's find an app. Since then. Consider a right triangle. By the condition of the problem, the lateral edge. This is the hypotenuse of my triangle. Then the height of the pyramid is the leg.
Then the point has coordinates:
That's it, I have the coordinates of all points of interest to me. I am looking for the coordinates of the directing vectors of the straight lines:
We are looking for the angle between these vectors:
Answer:
Again, when solving this problem, I did not use any sophisticated tricks, except for the formula for the sum of the angles of a regular n-gon, as well as the definition of the cosine and sine of a right triangle.
3. Since we are again not given the lengths of the edges in the pyramid, I will consider them equal to one. Thus, since ALL edges, and not just the side ones, are equal to each other, then at the base of the pyramid and me lies a square, and the side faces are regular triangles. Let's depict such a pyramid, as well as its base on a plane, marking all the data given in the text of the problem:
We are looking for the angle between and. I will make very brief calculations when I am looking for the coordinates of points. You will need to "decrypt" them:
b) - the middle of the segment. Her coordinates:
c) I will find the length of the segment using the Pythagorean theorem in a triangle. I will find by the Pythagorean theorem in a triangle.
Coordinates:
d) - the middle of the segment. Its coordinates are
e) Vector coordinates
f) Vector coordinates
g) Looking for an angle:
Cube - simplest figure. I'm sure you can figure it out on your own. The answers to problems 4 and 5 are as follows:
Finding the angle between a line and a plane
Well, the time for simple puzzles is over! Now the examples will be even more difficult. To find the angle between a line and a plane, we will proceed as follows:
- Using three points, we build the equation of the plane
,
using a third order determinant. - By two points we are looking for the coordinates of the directing vector of the straight line:
- We apply the formula to calculate the angle between a straight line and a plane:
As you can see, this formula is very similar to the one we used to find the angles between two lines. The structure of the right side is just the same, and on the left we are now looking for a sine, and not a cosine, as before. Well, one nasty action was added - the search for the equation of the plane.
Let's not shelve solving examples:
1. Os-no-va-ni-em straight-my prize-we are-la-et-xia equal-but-poor-ren-ny triangle-nick you-with-that prize-we are equal. Find the angle between the straight line and the plane
2. In a rectangular pa-ral-le-le-pi-pe-de from the West Nai-di-te the angle between the straight line and the plane
3. In the right-handed six-coal prism, all edges are equal. Find the angle between the straight line and the plane.
4. In the right triangular pi-ra-mi-de with the os-no-va-ni-em from the west of the rib Nai-di-te angle, ob-ra-zo-van-ny plane of the os-no-va-nia and my straight line, passing through the se-re-di-na of the ribs and
5. The lengths of all edges of the right quadrangular pi-ra-mi-dy with the top are equal to each other. Find the angle between the straight line and the plane, if the point is se-re-di-on the bo-ko-in-th edge of the pi-ra-mi-dy.
Again, I will solve the first two problems in detail, the third - briefly, and I leave the last two for you to solve on your own. In addition, you already had to deal with triangular and quadrangular pyramids, but with prisms - not yet.
Solutions:
1. Draw a prism, as well as its base. Let's combine it with the coordinate system and mark all the data that are given in the problem statement:
I apologize for some non-observance of proportions, but for solving the problem this, in fact, is not so important. The plane is just the "back wall" of my prism. It is enough to simply guess that the equation of such a plane has the form:
However, this can also be shown directly:
We choose arbitrary three points on this plane: for example, .
Let's make the equation of the plane:
Exercise for you: calculate this determinant yourself. Did you succeed? Then the equation of the plane has the form:
Or simply
Thus,
To solve the example, I need to find the coordinates of the directing vector of the straight line. Since the point coincided with the origin, the coordinates of the vector will simply coincide with the coordinates of the point. To do this, we first find the coordinates of the point.
To do this, consider a triangle. Let's draw a height (it is also a median and a bisector) from the top. Since, then the ordinate of the point is equal. In order to find the abscissa of this point, we need to calculate the length of the segment. By the Pythagorean theorem we have:
Then the point has coordinates:
A dot is a "raised" on a dot:
Then the coordinates of the vector:
Answer:
As you can see, there is nothing fundamentally difficult in solving such problems. In fact, the “straightness” of a figure such as a prism simplifies the process a little more. Now let's move on to the next example:
2. We draw a parallelepiped, draw a plane and a straight line in it, and also separately draw its lower base:
First, we find the equation of the plane: The coordinates of the three points lying in it:
(the first two coordinates are obtained in an obvious way, and you can easily find the last coordinate from the picture from the point). Then we compose the equation of the plane:
We calculate:
We are looking for the coordinates of the direction vector: It is clear that its coordinates coincide with the coordinates of the point, isn't it? How to find coordinates? These are the coordinates of the point, raised along the applicate axis by one! . Then we are looking for the desired angle:
Answer:
3. Draw a regular hexagonal pyramid, and then draw a plane and a straight line in it.
Here it is even problematic to draw a plane, not to mention the solution of this problem, but the coordinate method does not care! It is in its versatility that its main advantage lies!
The plane passes through three points: . We are looking for their coordinates:
1) . Display the coordinates for the last two points yourself. You will need to solve the problem with a hexagonal pyramid for this!
2) We build the equation of the plane:
We are looking for the coordinates of the vector: . (See triangular pyramid problem again!)
3) We are looking for an angle:
Answer:
As you can see, there is nothing supernaturally difficult in these tasks. You just need to be very careful with the roots. To the last two problems, I will give only answers:
As you can see, the technique for solving problems is the same everywhere: the main task is to find the coordinates of the vertices and substitute them into some formulas. It remains for us to consider one more class of problems for calculating angles, namely:
Calculating angles between two planes
The solution algorithm will be as follows:
- For three points we are looking for the equation of the first plane:
- For the other three points, we are looking for the equation of the second plane:
- We apply the formula:
As you can see, the formula is very similar to the previous two, with the help of which we were looking for angles between straight lines and between a straight line and a plane. So remembering this one will not be difficult for you. Let's jump right into the problem:
1. A hundred-ro-on the basis of the right triangular prism is equal, and the dia-go-nal of the side face is equal. Find the angle between the plane and the plane of the base of the prize.
2. In the right four-you-rech-coal-noy pi-ra-mi-de, all the edges of a swarm are equal, find the sine of the angle between the plane and the plane, passing through the point of pen-di-ku-lyar-but straight.
3. In a regular four-coal prism, the sides of the os-no-va-nia are equal, and the side edges are equal. On the edge from-me-che-to the point so that. Find the angle between the planes and
4. In the right quadrangular prism, the sides of the bases are equal, and the side edges are equal. On the edge from-me-che-to a point so that Find the angle between the planes and.
5. In the cube, find the co-si-nus of the angle between the planes and
Problem solutions:
1. I draw the correct one (at the base is an equilateral triangle) triangular prism and I mark on it the planes that appear in the condition of the problem:
We need to find the equations of two planes: The base equation is obtained trivially: you can make the corresponding determinant for three points, but I will make the equation right away:
Now let's find the equation The point has coordinates The point - Since - the median and the height of the triangle, it is easy to find by the Pythagorean theorem in a triangle. Then the point has coordinates: Find the applicate of the point To do this, consider a right triangle
Then we get the following coordinates: We compose the equation of the plane.
We calculate the angle between the planes:
Answer:
2. Making a drawing:
The most difficult thing is to understand what kind of mysterious plane it is, passing through a point perpendicularly. Well, the main thing is what is it? The main thing is attentiveness! Indeed, the line is perpendicular. The line is also perpendicular. Then the plane passing through these two lines will be perpendicular to the line, and, by the way, will pass through the point. This plane also passes through the top of the pyramid. Then the desired plane - And the plane is already given to us. We are looking for coordinates of points.
We find the coordinate of the point through the point. From small drawing it is easy to deduce that the coordinates of the point will be as follows: What is now left to find in order to find the coordinates of the top of the pyramid? Still need to calculate its height. This is done using the same Pythagorean theorem: first, prove that (trivially from small triangles forming a square at the base). Since by condition, we have:
Now everything is ready: vertex coordinates:
We compose the equation of the plane:
You are already an expert in calculating determinants. Easily you will receive:
Or otherwise (if we multiply both parts by the root of two)
Now let's find the equation of the plane:
(You didn’t forget how we get the equation of the plane, right? If you don’t understand where this minus one came from, then go back to the definition of the equation of the plane! It just always turned out that my plane belonged to the origin!)
We calculate the determinant:
(You may notice that the equation of the plane coincided with the equation of the straight line passing through the points and! Think why!)
Now we calculate the angle:
We need to find the sine:
Answer:
3. A tricky question: what is a rectangular prism, what do you think? It's just a well-known parallelepiped to you! Drawing right away! You can even not separately depict the base, there is little use from it here:
The plane, as we noted earlier, is written as an equation:
Now we make a plane
We immediately compose the equation of the plane:
Looking for an angle
Now the answers to the last two problems:
Well, now is the time to take a break, because you and I are great and have done a great job!
Coordinates and vectors. Advanced level
In this article, we will discuss with you another class of problems that can be solved using the coordinate method: distance problems. Namely, we will consider the following cases:
- Calculating the distance between skew lines.
I have ordered the given tasks as their complexity increases. The easiest is to find point to plane distance and the hardest part is finding distance between intersecting lines. Although, of course, nothing is impossible! Let's not procrastinate and immediately proceed to the consideration of the first class of problems:
Calculating the distance from a point to a plane
What do we need to solve this problem?
1. Point coordinates
So, as soon as we get all the necessary data, we apply the formula:
You should already know how we build the equation of the plane from the previous problems that I analyzed in the last part. Let's get down to business right away. The scheme is as follows: 1, 2 - I help you decide, and in some detail, 3, 4 - only the answer, you make the decision yourself and compare. Started!
Tasks:
1. Given a cube. The edge length of the cube is Find-di-te distance from se-re-di-ny from cut to flat
2. Given the right-vil-naya four-you-rekh-coal-naya pi-ra-mi-da Bo-ko-voe edge hundred-ro-on the os-no-va-nia is equal. Find-di-those distances from a point to a plane where - se-re-di-on the edges.
3. In the right triangular pi-ra-mi-de with os-but-va-ni-em, the other edge is equal, and the hundred-ro-on os-no-va-nia is equal. Find-di-those distances from the top to the plane.
4. In the right-handed six-coal prism, all edges are equal. Find-di-those distances from a point to a plane.
Solutions:
1. Draw a cube with single edges, build a segment and a plane, denote the middle of the segment by the letter
.
First, let's start with an easy one: find the coordinates of a point. Since then (remember the coordinates of the middle of the segment!)
Now we compose the equation of the plane on three points
\[\left| (\begin(array)(*(20)(c))x&0&1\\y&1&0\\z&1&1\end(array)) \right| = 0\]
Now I can start finding the distance:
2. We start again with a drawing, on which we mark all the data!
For a pyramid, it would be useful to draw its base separately.
Even the fact that I draw like a chicken paw will not prevent us from easily solving this problem!
Now it's easy to find the coordinates of a point
Since the coordinates of the point
2. Since the coordinates of the point a are the middle of the segment, then
We can easily find the coordinates of two more points on the plane. We compose the equation of the plane and simplify it:
\[\left| (\left| (\begin(array)(*(20)(c))x&1&(\frac(3)(2))\\y&0&(\frac(3)(2))\\z&0&(\frac((\sqrt 3 ))(2))\end(array)) \right|) \right| = 0\]
Since the point has coordinates: , then we calculate the distance:
Answer (very rare!):
Well, did you understand? It seems to me that everything here is just as technical as in the examples that we considered with you in the previous part. So I am sure that if you have mastered that material, then it will not be difficult for you to solve the remaining two problems. I'll just give you the answers:
Calculating the Distance from a Line to a Plane
In fact, there is nothing new here. How can a line and a plane be located relative to each other? They have all the possibilities: to intersect, or a straight line is parallel to the plane. What do you think is the distance from the line to the plane with which the given line intersects? It seems to me that it is clear that such a distance is equal to zero. Uninteresting case.
The second case is trickier: here the distance is already non-zero. However, since the line is parallel to the plane, then each point of the line is equidistant from this plane:
Thus:
And this means that my task has been reduced to the previous one: we are looking for the coordinates of any point on the line, we are looking for the equation of the plane, we calculate the distance from the point to the plane. In fact, such tasks in the exam are extremely rare. I managed to find only one problem, and the data in it was such that the coordinate method was not very applicable to it!
Now let's move on to another, much more important class of problems:
Calculating the Distance of a Point to a Line
What will we need?
1. The coordinates of the point from which we are looking for the distance:
2. Coordinates of any point lying on a straight line
3. Direction vector coordinates of the straight line
What formula do we use?
What does the denominator of this fraction mean to you and so it should be clear: this is the length of the directing vector of the straight line. Here is a very tricky numerator! The expression means the module (length) of the vector product of vectors and How to calculate the vector product, we studied in the previous part of the work. Refresh your knowledge, it will be very useful to us now!
Thus, the algorithm for solving problems will be as follows:
1. We are looking for the coordinates of the point from which we are looking for the distance:
2. We are looking for the coordinates of any point on the line to which we are looking for the distance:
3. Building a vector
4. We build the direction vector of the straight line
5. Calculate the cross product
6. We are looking for the length of the resulting vector:
7. Calculate the distance:
We have a lot of work, and the examples will be quite complex! So now focus all your attention!
1. Dana is a right-handed triangular pi-ra-mi-da with a vertex. One hundred-ro-on the os-no-va-niya pi-ra-mi-dy is equal, you-so-ta is equal. Find-di-those distances from the se-re-di-na of the b-to-the-th edge to the straight line, where the points and are the se-re-di-na of the ribs and, accordingly.
2. The lengths of the ribs and the right-angle-no-para-ral-le-le-pi-pe-yes are equal, respectively, and Find-di-te distances from the top of the bus to my straight
3. In the right six-coal prism, all the edges of a swarm are equal find-di-those distance from a point to a straight line
Solutions:
1. We make a neat drawing, on which we mark all the data:
We have a lot of work for you! I would first like to describe in words what we will look for and in what order:
1. Coordinates of points and
2. Point coordinates
3. Coordinates of points and
4. Coordinates of vectors and
5. Their cross product
6. Vector length
7. The length of the vector product
8. Distance from to
Well, we have a lot of work to do! Let's roll up our sleeves!
1. To find the coordinates of the height of the pyramid, we need to know the coordinates of the point. Its applicate is zero, and the ordinate is equal to its abscissa. Finally, we got the coordinates:
Point coordinates
2. - middle of the segment
3. - the middle of the segment
midpoint
4.Coordinates
Vector coordinates
5. Calculate the vector product:
6. The length of the vector: the easiest way is to replace that the segment is the middle line of the triangle, which means it is equal to half the base. So.
7. We consider the length of the vector product:
8. Finally, find the distance:
Phew, that's all! Honestly, I'll tell you: solving this problem by traditional methods (through constructions) would be much faster. But here I reduced everything to a ready-made algorithm! I think that the solution algorithm is clear to you? Therefore, I will ask you to solve the remaining two problems on your own. Compare answers?
Again, I repeat: it is easier (faster) to solve these problems through constructions, rather than resorting to the coordinate method. I demonstrated this way of solving only to show you a universal method that allows you to “do not finish anything”.
Finally, consider the last class of problems:
Calculating the distance between skew lines
Here the algorithm for solving problems will be similar to the previous one. What we have:
3. Any vector connecting the points of the first and second lines:
How do we find the distance between lines?
The formula is:
The numerator is the module of the mixed product (we introduced it in the previous part), and the denominator is the same as in the previous formula (the module of the vector product of the directing vectors of the lines, the distance between which we are looking for).
I will remind you that
Then the distance formula can be rewritten as:
Divide this determinant by the determinant! Although, to be honest, I'm not in the mood for jokes here! This formula, in fact, is very cumbersome and leads to rather complicated calculations. If I were you, I would only use it as a last resort!
Let's try to solve a few problems using the above method:
1. In the right triangular prism, all the edges are somehow equal, find the distance between the straight lines and.
2. Given the right triangular prism, all the edges of the os-no-va-niya are somehow equal to Se-che-thing, passing through the bo-ko-th edge and se-re-di-well, the edges are yav-la-et-sya square-ra-tom. Find-di-te dis-sto-I-nie between straight-we-mi and
I decide the first, and based on it, you decide the second!
1. I draw a prism and mark the lines and
Point C coordinates: then
Point coordinates
Vector coordinates
Point coordinates
Vector coordinates
Vector coordinates
\[\left((B,\overrightarrow (A(A_1)) \overrightarrow (B(C_1)) ) \right) = \left| (\begin(array)(*(20)(l))(\begin(array)(*(20)(c))0&1&0\end(array))\\(\begin(array)(*(20)(c))0&0&1\end(array))\\(\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2))&( - \frac(1)(2 ))&1\end(array))\end(array)) \right| = \frac((\sqrt 3 ))(2)\]
We consider the cross product between the vectors and
\[\overrightarrow (A(A_1)) \cdot \overrightarrow (B(C_1)) = \left| \begin(array)(l)\begin(array)(*(20)(c))(\overrightarrow i )&(\overrightarrow j )&(\overrightarrow k )\end(array)\\\begin(array)(*(20)(c))0&0&1\end(array)\\\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2)) &( - \frac(1)(2))&1\end(array)\end(array) \right| - \frac((\sqrt 3 ))(2)\overrightarrow k + \frac(1)(2)\overrightarrow i \]
Now we consider its length:
Answer:
Now try to carefully complete the second task. The answer to it will be:.
Coordinates and vectors. Brief description and basic formulas
A vector is a directed segment. - the beginning of the vector, - the end of the vector.
The vector is denoted by or.
Absolute value vector - the length of the segment representing the vector. Designated as.
Vector coordinates:
,
where are the ends of the vector \displaystyle a .
Sum of vectors: .
The product of vectors:
Dot product of vectors:
The scalar product of vectors is equal to the product of their absolute values and the cosine of the angle between them:
Well, the topic is over. If you are reading these lines, then you are very cool.
Because only 5% of people are able to master something on their own. And if you have read to the end, then you are in the 5%!
Now the most important thing.
You've figured out the theory on this topic. And, I repeat, it's ... it's just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough ...
For what?
For successful passing the exam, for admission to the institute on the budget and, MOST IMPORTANTLY, for life.
I will not convince you of anything, I will just say one thing ...
People who received a good education, earn much more than those who did not receive it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because much more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the exam and be ultimately ... happier?
FILL YOUR HAND, SOLVING PROBLEMS ON THIS TOPIC.
On the exam, you will not be asked theory.
You will need solve problems on time.
And, if you haven’t solved them (LOTS!), you will definitely make a stupid mistake somewhere or simply won’t make it in time.
It's like in sports - you need to repeat many times to win for sure.
Find a collection anywhere you want necessarily with solutions detailed analysis and decide, decide, decide!
You can use our tasks (not necessary) and we certainly recommend them.
In order to get a hand with the help of our tasks, you need to help extend the life of the YouClever textbook that you are currently reading.
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In conclusion...
If you don't like our tasks, find others. Just don't stop with theory.
“Understood” and “I know how to solve” are completely different skills. You need both.
Find problems and solve!
Finally, I got my hands on an extensive and long-awaited topic analytical geometry. First, a little about this section higher mathematics…. Surely you now remembered the school geometry course with numerous theorems, their proofs, drawings, etc. What to hide, an unloved and often obscure subject for a significant proportion of students. Analytic geometry, oddly enough, may seem more interesting and accessible. What does the adjective "analytical" mean? Two stamped mathematical turns immediately come to mind: “graphic method of solution” and “analytical method of solution”. Graphic method , of course, is associated with the construction of graphs, drawings. Analytical same method involves problem solving predominantly through algebraic operations. In this regard, the algorithm for solving almost all problems of analytical geometry is simple and transparent, often it is enough to accurately apply the necessary formulas - and the answer is ready! No, of course, it will not do without drawings at all, besides, for a better understanding of the material, I will try to bring them in excess of the need.
The open course of lessons in geometry does not claim to be theoretical completeness, it is focused on solving practical problems. I will include in my lectures only what, from my point of view, is important in in practical terms. If you need a more complete reference on any subsection, I recommend the following quite accessible literature:
1) A thing that, no joke, is familiar to several generations: School textbook on geometry, authors - L.S. Atanasyan and Company. This hanger of the school locker room has already withstood 20 (!) reissues, which, of course, is not the limit.
2) Geometry in 2 volumes. Authors L.S. Atanasyan, Bazylev V.T.. This is literature for high school, you will need first volume. Rarely occurring tasks may fall out of my field of vision, and tutorial will provide invaluable assistance.
Both books are free to download online. Also, you can use my archive with ready-made solutions, which can be found on the page Download higher mathematics examples.
From tools I offer again my own development - software package on analytical geometry, which will greatly simplify life and save a lot of time.
It is assumed that the reader is familiar with basic geometric concepts and figures: point, line, plane, triangle, parallelogram, parallelepiped, cube, etc. It is advisable to remember some theorems, at least the Pythagorean theorem, hello repeaters)
And now we will sequentially consider: the concept of a vector, actions with vectors, vector coordinates. Further I recommend reading the most important article Dot product of vectors, as well as Vector and mixed product of vectors. The local task will not be superfluous - Division of the segment in this regard. Based on the above information, you can equation of a straight line in a plane With the simplest examples of solutions, which will allow learn how to solve problems in geometry. The following articles are also helpful: Equation of a plane in space, Equations of a straight line in space, Basic problems on the line and plane , other sections of analytic geometry. Naturally, standard tasks will be considered along the way.
The concept of a vector. free vector
First, let's repeat the school definition of a vector. Vector called directed a segment for which its beginning and end are indicated:
In this case, the beginning of the segment is the point , the end of the segment is the point . The vector itself is denoted by . Direction is essential, if you rearrange the arrow to the other end of the segment, you get a vector, and this is already completely different vector. It is convenient to identify the concept of a vector with the movement of a physical body: you must admit that entering the doors of an institute or leaving the doors of an institute are completely different things.
It is convenient to consider individual points of a plane, space as the so-called zero vector. Such a vector has the same end and beginning.
!!! Note: Here and below, you can assume that the vectors lie in the same plane or you can assume that they are located in space - the essence of the material presented is valid for both the plane and space.
Designations: Many immediately drew attention to a stick without an arrow in the designation and said that they also put an arrow at the top! That's right, you can write with an arrow: , but admissible and record that I will use later. Why? Apparently, such a habit has developed from practical considerations, my shooters at school and university turned out to be too diverse and shaggy. IN educational literature sometimes they don’t bother with cuneiform at all, but highlight the letters in bold: , thereby implying that this is a vector.
That was the style, and now about the ways of writing vectors:
1) Vectors can be written in two capital Latin letters:
and so on. While the first letter Necessarily denotes the start point of the vector, and the second letter denotes the end point of the vector.
2) Vectors are also written in small Latin letters:
In particular, our vector can be redesignated for brevity by a small Latin letter .
Length or module non-zero vector is called the length of the segment. The length of the null vector is zero. Logically.
The length of a vector is denoted by the modulo sign: ,
How to find the length of a vector, we will learn (or repeat, for someone how) a little later.
That was elementary information about the vector, familiar to all schoolchildren. In analytic geometry, the so-called free vector.
If it's quite simple - vector can be drawn from any point:
We are accustomed to call such vectors equal (definition equal vectors will be given below), but from a purely mathematical point of view, this is the SAME VECTOR or free vector. Why free? Because in the course of solving problems, you can “attach” one or another vector to ANY point of the plane or space you need. This is a very cool property! Imagine a vector of arbitrary length and direction - it can be "cloned" an infinite number of times and at any point in space, in fact, it exists EVERYWHERE. There is such a student's proverb: Each lecturer in f ** u in the vector. After all, not just a witty rhyme, everything is mathematically correct - a vector can be attached there too. But do not rush to rejoice, students themselves suffer more often =)
So, free vector- This a bunch of identical directional segments. The school definition of a vector, given at the beginning of the paragraph: “A directed segment is called a vector ...”, implies specific a directed segment taken from a given set, which is attached to a certain point in the plane or space.
It should be noted that from the point of view of physics, the concept of a free vector is generally incorrect, and the point of application of the vector matters. Indeed, a direct blow of the same force on the nose or on the forehead is enough to develop my stupid example entails different consequences. However, not free vectors are also found in the course of vyshmat (do not go there :)).
Actions with vectors. Collinearity of vectors
IN school course geometry considers a number of actions and rules with vectors: addition according to the triangle rule, addition according to the parallelogram rule, the rule of the difference of vectors, multiplication of a vector by a number, the scalar product of vectors, etc. As a seed, we repeat two rules that are especially relevant for solving problems of analytical geometry.
Rule of addition of vectors according to the rule of triangles
Consider two arbitrary non-zero vectors and :
It is required to find the sum of these vectors. Due to the fact that all vectors are considered free, we postpone the vector from end vector :
The sum of vectors is the vector . For a better understanding of the rule, it is advisable to put a physical meaning into it: let some body make a path along the vector , and then along the vector . Then the sum of the vectors is the vector of the resulting path starting at the point of departure and ending at the point of arrival. A similar rule is formulated for the sum of any number of vectors. As they say, the body can go its way strongly zigzag, or maybe on autopilot - along the resulting sum vector.
By the way, if the vector is postponed from start vector , then we get the equivalent parallelogram rule addition of vectors.
First, about the collinearity of vectors. The two vectors are called collinear if they lie on the same line or on parallel lines. Roughly speaking, it is parallel vectors. But in relation to them, the adjective "collinear" is always used.
Imagine two collinear vectors. If the arrows of these vectors are directed in the same direction, then such vectors are called co-directional. If the arrows look in different directions, then the vectors will be oppositely directed.
Designations: collinearity of vectors is written with the usual parallelism icon: , while detailing is possible: (vectors are co-directed) or (vectors are directed oppositely).
work of a nonzero vector by a number is a vector whose length is equal to , and the vectors and are co-directed at and oppositely directed at .
The rule for multiplying a vector by a number is easier to understand with a picture:
We understand in more detail:
1) Direction. If the multiplier is negative, then the vector changes direction to the opposite.
2) Length. If the factor is contained within or , then the length of the vector decreases. So, the length of the vector is twice less than the length of the vector . If the modulo multiplier is greater than one, then the length of the vector increases in time.
3) Please note that all vectors are collinear, while one vector is expressed through another, for example, . The reverse is also true: if one vector can be expressed in terms of another, then such vectors are necessarily collinear. Thus: if we multiply a vector by a number, we get collinear(relative to original) vector.
4) The vectors are codirectional. The vectors and are also codirectional. Any vector of the first group is directed opposite to any vector of the second group.
What vectors are equal?
Two vectors are equal if they are codirectional and have the same length. Note that co-direction implies that the vectors are collinear. The definition will be inaccurate (redundant) if you say: "Two vectors are equal if they are collinear, co-directed and have the same length."
From the point of view of the concept of a free vector, equal vectors are the same vector, which was already discussed in the previous paragraph.
Vector coordinates on the plane and in space
The first point is to consider vectors on a plane. Draw a Cartesian rectangular coordinate system and set aside from the origin single vectors and :
Vectors and orthogonal. Orthogonal = Perpendicular. I recommend slowly getting used to the terms: instead of parallelism and perpendicularity, we use the words respectively collinearity And orthogonality.
Designation: orthogonality of vectors is written with the usual perpendicular sign, for example: .
The considered vectors are called coordinate vectors or orts. These vectors form basis on surface. What is the basis, I think, is intuitively clear to many, more detailed information can be found in the article Linear (non) dependence of vectors. Vector basis.In simple words, the basis and the origin of coordinates define the entire system - this is a kind of foundation on which a full and rich geometric life boils.
Sometimes the constructed basis is called orthonormal basis of the plane: "ortho" - because the coordinate vectors are orthogonal, the adjective "normalized" means unit, i.e. the lengths of the basis vectors are equal to one.
Designation: the basis is usually written in parentheses, inside which in strict order basis vectors are listed, for example: . Coordinate vectors it is forbidden swap places.
Any plane vector the only way expressed as:
, Where - numbers, which are called vector coordinates in this basis. But the expression itself called vector decompositionbasis .
Dinner served:
Let's start with the first letter of the alphabet: . The drawing clearly shows that when decomposing the vector in terms of the basis, the ones just considered are used:
1) the rule of multiplication of a vector by a number: and ;
2) addition of vectors according to the triangle rule: .
Now mentally set aside the vector from any other point on the plane. It is quite obvious that his corruption will "relentlessly follow him." Here it is, the freedom of the vector - the vector "carries everything with you." This property, of course, is true for any vector. It's funny that the basis (free) vectors themselves do not have to be set aside from the origin, one can be drawn, for example, at the bottom left, and the other at the top right, and nothing will change from this! True, you don’t need to do this, because the teacher will also show originality and draw you a “pass” in an unexpected place.
Vectors , illustrate exactly the rule for multiplying a vector by a number, the vector is co-directed with the basis vector , the vector is directed opposite to the basis vector . For these vectors, one of the coordinates is equal to zero, it can be meticulously written as follows:
And the basis vectors, by the way, are like this: (in fact, they are expressed through themselves).
And finally: , . By the way, what is vector subtraction, and why didn't I tell you about the subtraction rule? Somewhere in linear algebra, I don’t remember where, I noted that subtraction is special case addition. So, the expansions of the vectors "de" and "e" are calmly written as a sum: . Rearrange the terms in places and follow the drawing how clearly the good old addition of vectors according to the triangle rule works in these situations.
Considered decomposition of the form sometimes called a vector decomposition in the system ort(i.e. in the system of unit vectors). But this is not the only way to write a vector, the following option is common:
Or with an equals sign:
The basis vectors themselves are written as follows: and
That is, the coordinates of the vector are indicated in parentheses. In practical tasks, all three recording options are used.
I doubted whether to speak, but still I will say: vector coordinates cannot be rearranged. Strictly in first place write down the coordinate that corresponds to the unit vector , strictly in second place write down the coordinate that corresponds to the unit vector . Indeed, and are two different vectors.
We figured out the coordinates on the plane. Now consider vectors in three-dimensional space, everything is almost the same here! Only one more coordinate will be added. It is difficult to perform three-dimensional drawings, so I will limit myself to one vector, which for simplicity I will postpone from the origin:
Any 3d space vector the only way
expand in an orthonormal basis:
, where are the coordinates of the vector (number) in the given basis.
Example from the picture: . Let's see how the vector action rules work here. First, multiplying a vector by a number: (red arrow), (green arrow) and (magenta arrow). Secondly, here is an example of adding several, in this case three, vectors: . The sum vector starts at the starting point of departure (the beginning of the vector ) and ends up at the final point of arrival (the end of the vector ).
All vectors of three-dimensional space, of course, are also free, try to mentally postpone the vector from any other point, and you will understand that its expansion "remains with it."
Similarly to the plane case, in addition to writing versions with brackets are widely used: either .
If one (or two) coordinate vectors are missing in the expansion, then zeros are put instead. Examples:
vector (meticulously ) – write down ;
vector (meticulously ) – write down ;
vector (meticulously ) – write down .
Basis vectors are written as follows:
Here, perhaps, is all the minimum theoretical knowledge necessary for solving problems of analytical geometry. Perhaps there are too many terms and definitions, so I recommend dummies to re-read and comprehend this information again. And it will be useful for any reader to refer to the basic lesson from time to time for better assimilation of the material. Collinearity, orthogonality, orthonormal basis, vector decomposition - these and other concepts will be often used in what follows. I note that the materials of the site are not enough to pass a theoretical test, a colloquium in geometry, since I carefully encrypt all theorems (and without proofs) - to the detriment of scientific style presentation, but a plus to your understanding of the subject. For detailed theoretical information, I ask you to bow to Professor Atanasyan.
Now let's move on to the practical part:
The simplest problems of analytic geometry.
Actions with vectors in coordinates
The tasks that will be considered, it is highly desirable to learn how to solve them fully automatically, and the formulas memorize, don’t even remember it on purpose, they will remember it themselves =) This is very important, because on the simplest elementary examples other problems of analytic geometry are based, and it would be annoying to spend Extra time to eat pawns. You do not need to fasten the top buttons on your shirt, many things are familiar to you from school.
The presentation of the material will follow a parallel course - both for the plane and for space. For the reason that all the formulas ... you will see for yourself.
How to find a vector given two points?
If two points of the plane and are given, then the vector has the following coordinates:
If two points in space and are given, then the vector has the following coordinates:
That is, from the coordinates of the end of the vector you need to subtract the corresponding coordinates vector start.
Exercise: For the same points, write down the formulas for finding the coordinates of the vector. Formulas at the end of the lesson.
Example 1
Given two points in the plane and . Find vector coordinates
Solution: according to the corresponding formula:
Alternatively, the following notation could be used:
Aesthetes will decide like this:
Personally, I'm used to the first version of the record.
Answer:
According to the condition, it was not required to build a drawing (which is typical for problems of analytical geometry), but in order to explain some points to dummies, I will not be too lazy:
Must be understood difference between point coordinates and vector coordinates:
Point coordinates are the usual coordinates in a rectangular coordinate system. I think everyone knows how to plot points on the coordinate plane since grade 5-6. Each point has a strict place on the plane, and they cannot be moved anywhere.
The coordinates of the same vector is its expansion with respect to the basis , in this case . Any vector is free, therefore, if necessary, we can easily postpone it from some other point in the plane. Interestingly, for vectors, you can not build axes at all, a rectangular coordinate system, you only need a basis, in this case, an orthonormal basis of the plane.
The records of point coordinates and vector coordinates seem to be similar: , and sense of coordinates absolutely different, and you should be well aware of this difference. This difference, of course, is also true for space.
Ladies and gentlemen, we fill our hands:
Example 2
a) Given points and . Find vectors and .
b) Points are given And . Find vectors and .
c) Given points and . Find vectors and .
d) Points are given. Find Vectors .
Perhaps enough. These are examples for an independent decision, try not to neglect them, it will pay off ;-). Drawings are not required. Solutions and answers at the end of the lesson.
What is important in solving problems of analytical geometry? It is important to be EXTREMELY CAREFUL in order to avoid the masterful “two plus two equals zero” error. I apologize in advance if I made a mistake =)
How to find the length of a segment?
The length, as already noted, is indicated by the modulus sign.
If two points of the plane and are given, then the length of the segment can be calculated by the formula
If two points in space and are given, then the length of the segment can be calculated by the formula
Note: The formulas will remain correct if the corresponding coordinates are swapped: and , but the first option is more standard
Example 3
Solution: according to the corresponding formula:
Answer:
For clarity, I will make a drawing
Line segment - it's not a vector, and you can't move it anywhere, of course. In addition, if you complete the drawing to scale: 1 unit. \u003d 1 cm (two tetrad cells), then the answer can be checked with a regular ruler by directly measuring the length of the segment.
Yes, the solution is short, but there are a couple of important points in it that I would like to clarify:
First, in the answer we set the dimension: “units”. The condition does not say WHAT it is, millimeters, centimeters, meters or kilometers. Therefore, the general formulation will be a mathematically competent solution: “units” - abbreviated as “units”.
Second, let's repeat school material, which is useful not only for the considered problem:
pay attention to important technical trick – taking the multiplier out from under the root. As a result of the calculations, we got the result and good mathematical style involves taking the multiplier out from under the root (if possible). The process looks like this in more detail: . Of course, leaving the answer in the form will not be a mistake - but it is definitely a flaw and a weighty argument for nitpicking on the part of the teacher.
Here are other common cases:
Often under the root it turns out enough big number, For example . How to be in such cases? On the calculator, we check if the number is divisible by 4:. Yes, split completely, thus: . Or maybe the number can be divided by 4 again? . Thus: . The last digit of the number is odd, so dividing by 4 for the third time is clearly not possible. Trying to divide by nine: . As a result:
Ready.
Conclusion: if under the root we get a completely non-extractable number, then we try to take out the factor from under the root - on the calculator we check whether the number is divisible by: 4, 9, 16, 25, 36, 49, etc.
In the course of solving various problems, roots are often found, always try to extract factors from under the root in order to avoid a lower score and unnecessary troubles with finalizing your solutions according to the teacher's remark.
Let's repeat the squaring of the roots and other powers at the same time:
Rules for actions with powers in general form can be found in school textbook in algebra, but, I think, from the examples given, everything or almost everything is already clear.
Task for an independent solution with a segment in space:
Example 4
Given points and . Find the length of the segment.
Solution and answer at the end of the lesson.
How to find the length of a vector?
If a plane vector is given, then its length is calculated by the formula.
If a space vector is given, then its length is calculated by the formula .
2018 Olshevsky Andrey Georgievich
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Vectors on the plane and in space, ways to solve problems, examples, formulas
1 Vectors in space
Vectors in space include geometry 10, class 11 and analytic geometry. Vectors allow you to effectively solve the geometric problems of the second part of the exam and analytical geometry in space. Vectors in space are given in the same way as vectors in the plane, but the third coordinate z is taken into account. Exclusion from vectors in the space of the third dimension gives vectors on the plane, which explains the geometry of 8, 9 class.
1.1 Vector on the plane and in space
A vector is a directed segment with a beginning and an end, indicated by an arrow in the figure. An arbitrary point in space can be considered a null vector. The zero vector has no specific direction, since the beginning and end are the same, so it can be given any direction.
Vector translated from English means vector, direction, course, guidance, direction setting, aircraft heading.
The length (modulus) of a non-zero vector is the length of the segment AB, which is denoted
. Vector length denoted . Zero vector has length equal to zero
= 0.
Collinear vectors are non-zero vectors that lie on the same line or on parallel lines.
The zero vector is collinear to any vector.
Codirectional are called collinear non-zero vectors that have one direction. Codirectional vectors are denoted by . For example, if the vector is codirectional with the vector , then the notation is used.
The zero vector is codirectional with any vector.
Oppositely directed are two collinear non-zero vectors that have the opposite direction. Oppositely directed vectors are denoted by ↓. For example, if the vector is opposite to the vector , then the notation ↓ is used.
Codirectional vectors of equal length are called equal.
Many physical quantities are vector quantities: force, speed, electric field.
If the point of application (beginning) of the vector is not set, then it is chosen arbitrarily.
If the beginning of the vector is placed at the point O, then it is considered that the vector is postponed from the point O. From any point, a single vector equal to the given vector can be plotted.
1.2 Sum of vectors
When adding vectors according to the triangle rule, vector 1 is drawn, from the end of which vector 2 is drawn and the sum of these two vectors is vector 3, drawn from the beginning of vector 1 to the end of vector 2:
For arbitrary points A , B and C, you can write the sum of vectors:
+
=
If two vectors start from the same point
then it is better to add them according to the parallelogram rule.
When two vectors are added according to the parallelogram rule, the added vectors are laid off from one point, a parallelogram is completed from the ends of these vectors by applying the beginning of another to the end of one vector. The vector formed by the diagonal of the parallelogram, originating from the start point of the added vectors, will be the sum of the vectors
The parallelogram rule contains a different order of addition of vectors according to the triangle rule.
Vector addition laws:
1. The commutative law + = + .
2. Associative law ( + ) + = + ( + ).
If it is necessary to add several vectors, then the vectors are added in pairs or according to the polygon rule: vector 2 is drawn from the end of vector 1, vector 3 is drawn from the end of vector 2, vector 4 is drawn from the end of vector 3, vector 5 is drawn from the end of vector 4, etc. A vector that is the sum of several vectors is drawn from the beginning of vector 1 to the end of the last vector.
According to the laws of vector addition, the order of vector addition does not affect the resulting vector, which is the sum of several vectors.
Opposite are two non-zero oppositely directed vectors of equal length. Vector - is the opposite of a vector
These vectors are oppositely directed and equal in absolute value.
1.3 Vector difference
The difference of vectors can be written as the sum of vectors
- = + (-),
where "-" is the vector opposite to the vector .
Vectors and - can be added according to the rule of a triangle or a parallelogram.
Let vectors and
To find the difference of vectors - we build a vector -
We add the vectors and - according to the triangle rule, applying the beginning of the vector - to the end of the vector, we got the vector + (-) = -
We add the vectors and - according to the parallelogram rule, postponing the beginnings of the vectors and - from one point
If the vectors and originate from the same point
,
then the difference of vectors - gives a vector connecting their ends and the arrow at the end of the resulting vector is placed in the direction of the vector from which the second vector is subtracted
The figure below shows the addition and difference of vectors
The figure below shows the addition and difference of vectors in different ways.
Task. Given vectors and .
Draw the sum and difference of vectors by all possible ways in all possible combinations of vectors.
1.4 Collinear vector lemma
= k
1.5 Multiplication of a vector by a number
The product of a non-zero vector by a number k gives a vector = k , collinear to the vector . Vector length :
| | = |k |·| |
If k > 0, then the vectors and are codirectional.
If k = 0, then the vector is zero.
If k< 0, то векторы и противоположно направленные.
If | k | = 1, then the vectors and are of equal length.
If k = 1, then and equal vectors.
If k = -1, then opposite vectors.
If | k | > 1, then the length of the vector is greater than the length of the vector .
If k > 1, then the vectors and are codirectional and the length is greater than the length of the vector .
If k< -1, то векторы и противоположно направленные и длина больше длины вектора .
If | k |< 1, то длина вектора меньше длины вектора .
If 0< k< 1, то векторы и сонаправленные и длина меньше длины вектора .
If -1< k< 0, то векторы и противоположно направленные и длина меньше длины вектора .
The product of a zero vector by a number gives a zero vector.
Task. Given a vector .
Construct vectors 2 , -3 , 0.5 , -1.5 .
Task. Given vectors and .
Construct vectors 3 + 2 , 2 - 2 , -2 - .
Laws describing the multiplication of a vector by a number
1. Combination law (kn) = k (n)
2. The first distributive law k ( + ) = k + k .
3. The second distributive law (k + n) = k + n.
For collinear vectors and , if ≠ 0, there is a single number k that allows expressing the vector in terms of:
= k
1.6 Coplanar vectors
Coplanar vectors are those that lie in the same plane or in parallel planes. If you draw vectors equal to given coplanar vectors from one point, then they will lie in the same plane. Therefore, we can say that vectors are called coplanar if there are equal vectors lying in the same plane.
Two arbitrary vectors are always coplanar. The three vectors may or may not be coplanar. Three vectors, of which at least two are collinear, are coplanar. Collinear vectors are always coplanar.
1.7 Decomposition of a vector in two non-collinear vectors
Any vector uniquely decomposes on the plane in two noncollinear nonzero vectors And with only expansion coefficients x and y :
= x+y
Any vector coplanar to non-zero vectors and is uniquely decomposed in two non-collinear vectors and with unique expansion coefficients x and y :
= x+y
Let us expand the given vector on the plane according to the given noncollinear vectors and :
Draw from one point the given coplanar vectors
From the end of the vector we draw lines parallel to the vectors and to the intersection with the lines drawn through the vectors and . Get a parallelogram
The lengths of the sides of the parallelogram are obtained by multiplying the lengths of the vectors and by the numbers x and y, which are determined by dividing the lengths of the sides of the parallelogram by the lengths of the corresponding vectors and. We get the decomposition of the vector in given non-collinear vectors and :
= x+y
In the problem being solved, x ≈ 1.3, y ≈ 1.9, so the expansion of the vector in given noncollinear vectors and can be written as
1,3 + 1,9 .
In the problem being solved, x ≈ 1.3, y ≈ -1.9, so the expansion of the vector in given non-collinear vectors and can be written as
1,3 - 1,9 .
1.8 Box rule
A parallelepiped is a three-dimensional figure whose opposite faces consist of two equal parallelograms lying in parallel planes.
The parallelepiped rule allows you to add three non-coplanar vectors that are drawn from one point and construct a parallelepiped so that the summed vectors form its edges, and the remaining edges of the parallelepiped are respectively parallel and equal to the lengths of the edges formed by the summed vectors. The diagonal of the parallelepiped forms a vector that is the sum of the given three vectors, which starts from the start point of the added vectors.
1.9 Decomposition of a vector in three non-coplanar vectors
Any vector expands in three given non-coplanar vectors , and with single expansion coefficients x, y, z:
= x + y + z .
1.10 Rectangular coordinate system in space
In three-dimensional space, the rectangular coordinate system Oxyz is defined by the origin O and the mutually perpendicular coordinate axes Ox , Oy and Oz intersecting in it with selected positive directions indicated by arrows and the unit of measurement of the segments. If the scale of the segments is the same along all three axes, then such a system is called a Cartesian coordinate system.
Coordinate x is called the abscissa, y is the ordinate, z is the applicate. Point M coordinates are written in brackets M (x ; y ; z ).
1.11 Vector coordinates in space
In space, let's set a rectangular coordinate system Oxyz . From the origin in the positive directions of the axes Ox , Oy , Oz we draw the corresponding unit vectors , , , which are called coordinate vectors and are non-coplanar. Therefore, any vector can be decomposed into three given non-coplanar coordinate vectors , and with the only expansion coefficients x , y , z :
= x + y + z .
The expansion coefficients x , y , z are the coordinates of the vector in a given rectangular coordinate system, which are written in brackets (x ; y ; z ). Zero vector has coordinates equal to zero (0; 0; 0). For equal vectors, the corresponding coordinates are equal.
Rules for finding the coordinates of the resulting vector:
1. When summing two or more vectors, each coordinate of the resulting vector is equal to the sum of the corresponding coordinates of the given vectors. If two vectors are given (x 1 ; y 1 ; z 1) and (x 1 ; y 1 ; z 1), then the sum of vectors + gives a vector with coordinates (x 1 + x 1 ; y 1 + y 1 ; z 1 + z 1)
+ = (x 1 + x 1 ; y 1 + y 1 ; z1 + z1)
2. The difference is a kind of sum, so the difference of the corresponding coordinates gives each coordinate of the vector obtained by subtracting the two given vectors. If two vectors are given (x a ; y a ; z a ) and (x b ; y b ; z b ), then the difference of the vectors - gives a vector with coordinates (x a - x b ; y a - y b ; z a - z b )
- = (x a - x b ; y a - y b ; z a - z b )
3. When multiplying a vector by a number, each coordinate of the resulting vector is equal to the product of this number by the corresponding coordinate of the given vector. Given a number k and a vector (x ; y ; z ), then multiplying the vector by the number k gives a vector k with coordinates
k = (kx ; ky ; kz ).
Task. Find the coordinates of the vector = 2 - 3 + 4 if the coordinates of the vectors are (1; -2; -1), (-2; 3; -4), (-1; -3; 2).
Solution
2 + (-3) + 4
2 = (2 1; 2 (-2); 2 (-1)) = (2; -4; -2);
3 = (-3 (-2); -3 3; -3 (-4)) = (6; -9; 12);
4 = (4 (-1); 4 (-3); 4 2) = (-4; -12; 8).
= (2 + 6 - 4; -4 - 9 -12; -2 + 12 + 8) = (4; -25; 18).
1.12 Vector, radius vector and point coordinates
The vector coordinates are the coordinates of the end of the vector, if the beginning of the vector is placed at the origin.
A radius vector is a vector drawn from the origin to a given point, the coordinates of the radius vector and the point are equal.
If the vector
given by points M 1 (x 1; y 1; z 1) and M 2 (x 2; y 2; z 2), then each of its coordinates is equal to the difference between the corresponding coordinates of the end and beginning of the vector
For collinear vectors = (x 1 ; y 1 ; z 1) and = (x 2 ; y 2 ; z 2), if ≠ 0, there is a single number k that allows expressing the vector in terms of:
= k
Then the coordinates of the vector are expressed in terms of the coordinates of the vector
= (kx 1 ; ky1; kz 1)
The ratio of the corresponding coordinates of collinear vectors is equal to the single number k
1.13 Vector length and distance between two points
The length of the vector (x; y; z) is equal to the square root of the sum of the squares of its coordinates
Vector length , given by points the beginning of M 1 (x 1; y 1; z 1) and the end of M 2 (x 2; y 2; z 2) is equal to the square root of the sum of the squares of the difference between the corresponding coordinates of the end of the vector and the beginning
Distance d between two points M 1 (x 1 ; y 1 ; z 1) and M 2 (x 2 ; y 2 ; z 2) is equal to the length of the vector
There is no z coordinate on the plane
Distance between points M 1 (x 1; y 1) and M 2 (x 2; y 2)
1.14 Coordinates of the middle of the segment
If point C is the midpoint of the segment AB , then the radius vector of point C in an arbitrary coordinate system with origin at point O is equal to half the sum of the radius vectors of points A and B
If the coordinates of the vectors
(x ; y ; z ),
(x 1 ; y 1 ; z 1),
(x 2; y 2; z 2), then each vector coordinate is equal to half the sum of the corresponding coordinates of the vectors and
,
,
=
(x, y, z) =
Each of the coordinates of the middle of the segment is equal to half the sum of the corresponding coordinates of the ends of the segment.
1.15 Angle between vectors
The angle between vectors is equal to the angle between the rays drawn from one point and co-directed with these vectors. The angle between vectors can be from 0 0 to 180 0 inclusive. The angle between codirectional vectors is equal to 0 0 . If one vector or both are zero, then the angle between the vectors, at least one of which is zero, is equal to 0 0 . The angle between perpendicular vectors is 90 0 . The angle between oppositely directed vectors is 180 0 .
1.16 Vector projection
1.17 Dot product of vectors
The scalar product of two vectors is a number (scalar) equal to the product of the lengths of the vectors and the cosine of the angle between the vectors
If = 0 0 , then the vectors are codirectional
And
= cos 0 0 = 1, therefore, the scalar product of codirectional vectors is equal to the product of their lengths (modules)
.
If the angle between vectors is 0<
< 90 0 , то косинус угла между такими векторами больше
нуля
, hence the scalar product is greater than zero
.
If non-zero vectors are perpendicular, then their scalar product is zero
, since cos 90 0 = 0. The scalar product of perpendicular vectors is equal to zero.
If
, then the cosine of the angle between such vectors is less than zero
, so the scalar product is less than zero
.
As the angle between vectors increases, the cosine of the angle between them
decreases and reaches a minimum value at = 180 0 when the vectors are oppositely directed
. Since cos 180 0 = -1, then
. The scalar product of oppositely directed vectors is equal to the negative product of their lengths (modules).
The scalar square of a vector is equal to the modulus of the vector squared
The scalar product of vectors, at least one of which is zero, is equal to zero.
1.18 The physical meaning of the scalar product of vectors
From the course of physics it is known that the work A of the force while moving the body is equal to the product of the lengths of the force and displacement vectors and the cosine of the angle between them, that is, it is equal to the scalar product of the force and displacement vectors
If the force vector is co-directed with the movement of the body, then the angle between the vectors
= 0 0 , therefore, the work of the force on displacement is maximum and is equal to A =
.
If 0< < 90 0 , то работа силы на перемещении положительна A > 0.
If = 90 0 , then the work of the force on displacement is equal to zero A = 0.
If 90 0< < 180 0 , то работа силы на перемещении отрицательна A < 0.
If the force vector is opposite to the movement of the body, then the angle between the vectors = 180 0, therefore, the work of the force on the movement is negative and equal to A = -.
Task. Determine the work of gravity when lifting a passenger car weighing 1 ton along a 1 km long track with an inclination angle of 30 0 to the horizon. How many liters of water at a temperature of 20 0 can be boiled using this energy?
Solution
Job A gravity when moving the body, it is equal to the product of the lengths of the vectors and the cosine of the angle between them, that is, it is equal to the scalar product of the vectors of gravity and displacement
Gravity
G \u003d mg \u003d 1000 kg 10 m / s 2 \u003d 10,000 N.
= 1000 m.
Angle between vectors = 1200. Then
cos 120 0 \u003d cos (90 0 + 30 0) \u003d - sin 30 0 \u003d - 0.5.
Substitute
A \u003d 10,000 N 1000 m (-0.5) \u003d - 5,000,000 J \u003d - 5 MJ.
1.19 Dot product of vectors in coordinates
Dot product of two vectors = (x 1 ; y 1 ; z 1) and \u003d (x 2; y 2; z 2) in a rectangular coordinate system is equal to the sum of the products of the coordinates of the same name
= x 1 x 2 + y 1 y 2 + z 1 z 2 .
1.20 The condition of perpendicularity of vectors
If non-zero vectors \u003d (x 1; y 1; z 1) and \u003d (x 2; y 2; z 2) are perpendicular, then their scalar product is zero
If one non-zero vector = (x 1; y 1; z 1) is given, then the coordinates of the vector perpendicular (normal) to it = (x 2; y 2; z 2) must satisfy the equality
x 1 x 2 + y 1 y 2 + z 1 z 2 = 0.
There are an infinite number of such vectors.
If one non-zero vector = (x 1; y 1) is set on the plane, then the coordinates of the vector perpendicular (normal) to it = (x 2; y 2) must satisfy the equality
x 1 x 2 + y 1 y 2 = 0.
If a non-zero vector = (x 1 ; y 1) is set on the plane, then it is sufficient to set arbitrarily one of the coordinates of the vector perpendicular (normal) to it = (x 2 ; y 2) and from the condition of perpendicularity of the vectors
x 1 x 2 + y 1 y 2 = 0
express the second coordinate of the vector .
For example, if we substitute an arbitrary x 2 coordinate, then
y 1 y 2 = - x 1 x 2 .
The second coordinate of the vector
If you give x 2 \u003d y 1, then the second coordinate of the vector
If a non-zero vector = (x 1; y 1) is given on the plane, then the vector perpendicular (normal) to it = (y 1; -x 1).
If one of the coordinates of a non-zero vector is equal to zero, then the vector has the same coordinate not equal to zero, and the second coordinate is equal to zero. Such vectors lie on the coordinate axes, therefore they are perpendicular.
Let's define the second vector, perpendicular to the vector = (x 1 ; y 1), but opposite to the vector , that is, the vector - . Then it suffices to change the signs of the coordinates of the vector
- = (-y1; x1)
1 = (y1; -x1)
2 = (-y1; x1).
Task.
Solution
Coordinates of two vectors perpendicular to the vector = (x 1; y 1) on the plane
1 = (y1; -x1)
2 = (-y1; x1).
We substitute the coordinates of the vector = (3; -5)
1 = (-5; -3),
2 = (-(-5); 3) = (5; 3).
x 1 x 2 + y 1 y 2 = 0
3 (-5) + (-5) (-3) = -15 + 15 = 0
right!
3 5 + (-5) 3 = 15 - 15 = 0
right!
Answer: 1 = (-5; -3), 2 = (5; 3).
If we assign x 2 = 1, substitute
x 1 + y 1 y 2 = 0.
y 1 y 2 = -x 1
Get the y 2 coordinate of a vector perpendicular to the vector = (x 1; y 1)
To obtain a second vector perpendicular to the vector = (x 1; y 1), but opposite to the vector . Let
Then it is enough to change the signs of the coordinates of the vector .
Coordinates of two vectors perpendicular to the vector = (x 1; y 1) on the plane
Task. Given a vector = (3; -5). Find two normal vectors with different orientation.
Solution
Coordinates of two vectors perpendicular to the vector = (x 1; y 1) on the plane
Single vector coordinates
Second vector coordinates
To check the perpendicularity of vectors, we substitute their coordinates into the condition of perpendicularity of vectors
x 1 x 2 + y 1 y 2 = 0
3 1 + (-5) 0.6 = 3 - 3 = 0
right!
3 (-1) + (-5) (-0.6) = -3 + 3 = 0
right!
Answer: and.
If you assign x 2 \u003d - x 1, substitute
x 1 (-x 1) + y 1 y 2 = 0.
-x 1 2 + y 1 y 2 = 0.
y 1 y 2 = x 1 2
Get the coordinate of the vector perpendicular to the vector
If you assign x 2 \u003d x 1, substitute
x 1 x 1 + y 1 y 2 = 0.
x 1 2 + y 1 y 2 = 0.
y 1 y 2 = -x 1 2
Get the y coordinate of the second vector perpendicular to the vector
Coordinates of one vector perpendicular to the vector in the plane = (x 1; y 1)
Coordinates of the second vector, perpendicular to the vector in the plane = (x 1; y 1)
Coordinates of two vectors perpendicular to the vector = (x 1; y 1) on the plane
1.21 Cosine of the angle between vectors
The cosine of the angle between two non-zero vectors \u003d (x 1; y 1; z 1) and \u003d (x 2; y 2; z 2) is equal to the scalar product of vectors divided by the product of the lengths of these vectors
If
= 1, then the angle between the vectors is equal to 0 0 , the vectors are codirectional.
If 0< < 1, то 0 0 < < 90 0 .
If = 0, then the angle between the vectors is equal to 90 0 , the vectors are perpendicular.
If -1< < 0, то 90 0 < < 180 0 .
If = -1, then the angle between the vectors is 180 0 , the vectors are oppositely directed.
If some vector is given by the coordinates of the beginning and end, then subtracting the coordinates of the beginning from the corresponding coordinates of the end of the vector, we obtain the coordinates of this vector.
Task. Find the angle between the vectors (0; -2; 0), (-2; 0; -4).
Solution
Dot product of vectors
= 0 (-2) + (-2) 0 + 0 (-4) = 0,
hence the angle between the vectors is = 90 0 .
1.22 Properties of the Dot Product of Vectors
The properties of the scalar product are valid for any , , ,k :
1.
, If
, That
, If
=, That
=
0.
2. Displacement law
3. Distributive law
4. Combination law
.
1.23 Direction vector direct
The directing vector of a line is a non-zero vector lying on a line or on a line parallel to the given line.
If the line is given by two points M 1 (x 1; y 1; z 1) and M 2 (x 2; y 2; z 2), then the vector is the guide
or its opposite vector
= - , whose coordinates
It is desirable to set the coordinate system so that the line passes through the origin, then the coordinates of the only point on the line will be the coordinates of the direction vector.
Task. Determine the coordinates of the directing vector of the straight line passing through the points M 1 (1; 0; 0), M 2 (0; 1; 0).
Solution
The direction vector of the straight line passing through the points M 1 (1; 0; 0), M 2 (0; 1; 0) is denoted
. Each of its coordinates is equal to the difference between the corresponding coordinates of the end and beginning of the vector
= (0 - 1; 1 - 0; 0 - 0) = (-1; 1; 0)
Let's depict the directing vector of the straight line in the coordinate system with the beginning at the point M 1, with the end at the point M 2 and the vector equal to it
from origin with end at point M (-1; 1; 0)
1.24 Angle between two straight lines
Possible options for the relative position of 2 lines on the plane and the angle between such lines:
1. The lines intersect at a single point, forming 4 angles, 2 pairs of vertical angles are equal in pairs. The angle φ between two intersecting lines is the angle not exceeding the other three angles between these lines. Therefore, the angle between the lines φ ≤ 90 0 .
Intersecting lines can be, in particular, perpendicular φ = 90 0 .
Possible options for the relative position of 2 lines in space and the angle between such lines:
1. The lines intersect at a single point, forming 4 angles, 2 pairs of vertical angles are equal in pairs. The angle φ between two intersecting lines is the angle not exceeding the other three angles between these lines.
2. The lines are parallel, that is, they do not coincide and do not intersect, φ=0 0 .
3. The lines coincide, φ = 0 0 .
4. The lines intersect, that is, they do not intersect in space and are not parallel. The angle φ between intersecting lines is the angle between lines drawn parallel to these lines so that they intersect. Therefore, the angle between the lines φ ≤ 90 0 .
The angle between 2 lines is equal to the angle between the lines drawn parallel to these lines in the same plane. Therefore, the angle between the lines is 0 0 ≤ φ ≤ 90 0 .
Angle θ (theta) between vectors and 0 0 ≤ θ ≤ 180 0 .
If the angle φ between the lines α and β is equal to the angle θ between the direction vectors of these lines φ = θ, then
cos φ = cos θ.
If the angle between the lines φ = 180 0 - θ, then
cos φ \u003d cos (180 0 - θ) \u003d - cos θ.
cos φ = - cos θ.
Therefore, the cosine of the angle between the lines is equal to the modulus of the cosine of the angle between the vectors
cos φ = |cos θ|.
If the coordinates of non-zero vectors = (x 1 ; y 1 ; z 1) and = (x 2 ; y 2 ; z 2) are given, then the cosine of the angle θ between them
The cosine of the angle between the lines is equal to the modulus of the cosine of the angle between the direction vectors of these lines
cos φ = |cos θ| =
The lines are the same geometric objects, therefore the same trigonometric functions cos are present in the formula.
If each of the two lines is given by two points, then the direction vectors of these lines and the cosine of the angle between the lines can be determined.
If cos φ \u003d 1, then the angle φ between the lines is 0 0, one of the directing vectors of these lines can be taken for these lines, the lines are parallel or coincide. If the lines do not coincide, then they are parallel. If the lines coincide, then any point of one line belongs to the other line.
If 0< cos φ ≤ 1, then the angle between the lines is 0 0< φ ≤ 90 0 , прямые пересекаются или скрещиваются. Если прямые не пересекаются, то они скрещиваются. Если прямые пересекаются, то они имеют общую точку.
If cos φ \u003d 0, then the angle φ between the lines is 90 0 (the lines are perpendicular), the lines intersect or intersect.
Task. Determine the angle between the lines M 1 M 3 and M 2 M 3 with the coordinates of the points M 1 (1; 0; 0), M 2 (0; 1; 0) and M 3 (0; 0; 1).
Solution
Let's construct the given points and straight lines in the Oxyz coordinate system.
We direct the directing vectors of the lines so that the angle θ between the vectors coincides with the angle φ between the given lines. Draw the vectors =
and =
, as well as the angles θ and φ:
Let us determine the coordinates of the vectors and
= = (1 - 0; 0 - 0; 0 - 1) = (1; 0; -1);
= = (0 - 0; 1 - 0; 0 - 1) = (0; 1; -1). d = 0 and ax + by + cz = 0;
The plane is parallel to that coordinate axis, the designation of which is absent in the equation of the plane and, therefore, the corresponding coefficient is equal to zero, for example, at c = 0, the plane is parallel to the Oz axis and does not contain z in the equation ax + by + d = 0;
The plane contains the axis of coordinates, the designation of which is missing, therefore, the corresponding coefficient is zero and d = 0, for example, at c = d = 0, the plane is parallel to the Oz axis and does not contain z in the equation ax + by = 0;
The plane is parallel to the coordinate plane, the notation of which is absent in the equation of the plane and, therefore, the corresponding coefficients are equal to zero, for example, for b = c = 0, the plane is parallel to the coordinate plane Oyz and does not contain y, z in the equation ax + d = 0.
If the plane coincides with coordinate plane, then the equation of such a plane is the equality to zero of the designation of the coordinate axis perpendicular to the given coordinate plane, for example, for x = 0, the given plane is the coordinate plane Oyz .
Task. The normal vector is given by the equation
Represent the equation of the plane in normal form.
Solution
Normal vector coordinates
A ; b; c ), then we can substitute the coordinates of the point M 0 (x 0 ; y 0 ; z 0) and the coordinates a , b , c of the normal vector in general equation plane
ax + by + cz + d = 0 (1)
We get an equation with one unknown d
ax 0 + by 0 + cz 0 + d = 0
From here
d = -(ax 0 + by 0 + cz 0 )
Plane equation (1) after substitution d
ax + by + cz - (ax 0 + by 0 + cz 0) = 0
We obtain the equation of a plane passing through the point M 0 (x 0 ; y 0 ; z 0) perpendicular to a non-zero vector (a ; b ; c )
a (x - x 0) + b (y - y 0) + c (z - z 0) = 0
Let's open the brackets
ax - ax 0 + by - by 0 + cz - cz 0 = 0
ax + by + cz - ax 0 - by 0 - cz 0 = 0
Denote
d = - ax 0 - by 0 - cz 0
We obtain the general equation of the plane
ax + by + cz + d = 0.
1.29 Equation of a plane passing through two points and the origin
ax + by + cz + d = 0.
It is desirable to set the coordinate system so that the plane passes through the origin of this coordinate system. Points M 1 (x 1 ; y 1 ; z 1) and M 2 (x 2 ; y 2 ; z 2) lying in this plane must be set so that the straight line connecting these points does not pass through the origin.
The plane will pass through the origin, so d = 0. Then the general equation of the plane becomes
ax + by + cz = 0.
Unknown 3 coefficients a , b , c . Substituting the coordinates of two points into the general equation of the plane gives a system of 2 equations. If we take some coefficient in the general equation of the plane equal to one, then the system of 2 equations will allow us to determine 2 unknown coefficients.
If one of the coordinates of the point is zero, then the coefficient corresponding to this coordinate is taken as one.
If some point has two zero coordinates, then the coefficient corresponding to one of these zero coordinates is taken as unity.
If a = 1 is accepted, then a system of 2 equations will allow us to determine 2 unknown coefficients b and c:
It is easier to solve the system of these equations by multiplying some equation by such a number that the coefficients for some unknown steel are equal. Then the difference of the equations will allow us to exclude this unknown, to determine another unknown. Substituting the found unknown into any equation will allow us to determine the second unknown.
1.30 Equation of a plane passing through three points
Let us define the coefficients of the general equation of the plane
ax + by + cz + d = 0,
passing through points M 1 (x 1 ; y 1 ; z 1), M 2 (x 2 ; y 2 ; z 2) and M 3 (x 3 ; y 3 ; z 3). Points must not have two identical coordinates.
Unknown 4 coefficients a , b , c and d . Substituting the coordinates of three points into the general equation of the plane gives a system of 3 equations. Take some coefficient in the general equation of the plane equal to one, then the system of 3 equations will allow you to determine 3 unknown coefficients. Usually accepted a = 1, then the system of 3 equations will allow you to determine 3 unknown coefficients b, c and d:
The system of equations is best solved by the elimination of unknowns (Gauss method). You can rearrange the equations in the system. Any equation can be multiplied or divided by any non-zero factor. Any two equations can be added, and the resulting equation can be written instead of either of these two added equations. The unknowns are excluded from the equations by obtaining a zero coefficient in front of them. In one equation, usually the lowest one is left with one variable that is defined. The found variable is substituted into the second equation from the bottom, in which 2 unknowns usually remain. The equations are solved from the bottom up and all unknown coefficients are determined.
The coefficients are placed in front of the unknowns, and the terms free from unknowns are transferred to the right side of the equations
The top row usually contains an equation that has a factor of 1 before the first or any unknown, or the entire first equation is divided by the factor before the first unknown. In this system of equations, we divide the first equation by y 1
Before the first unknown we got a coefficient of 1:
To reset the coefficient in front of the first variable of the second equation, we multiply the first equation by -y 2 , add it to the second equation, and write the resulting equation instead of the second equation. The first unknown in the second equation will be eliminated because
y 2 b - y 2 b = 0.
Similarly, we exclude the first unknown in the third equation by multiplying the first equation by -y 3 , adding it to the third equation and writing the resulting equation instead of the third equation. The first unknown in the third equation will also be eliminated because
y 3 b - y 3 b = 0.
Similarly, we exclude the second unknown in the third equation. We solve the system from the bottom up.
Task.
ax + by + cz + d = 0,
passing through the points M 1 (0; 0; 0), M 2 (0; 1; 0) and y+ 0 z + 0 = 0
x = 0.
The given plane is the coordinate plane Oyz .
Task. Determine the general equation of the plane
ax + by + cz + d = 0,
passing through the points M 1 (1; 0; 0), M 2 (0; 1; 0) and M 3 (0; 0; 1). Find the distance from this plane to the point M 0 (10; -3; -7).
Solution
Let's build the given points in the Oxyz coordinate system.
Accept a= 1. Substituting the coordinates of three points into the general equation of the plane gives a system of 3 equations
ℓ =
Web pages: 1 2 Vectors in the plane and in space (continued)
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