EntranceNormal plane vector, coordinates of the normal plane vector. Direct normal vector (normal vector) How to find a normal vector
What is normal? In simple words, normal is perpendicular. That is, the normal vector of a line is perpendicular to a given line. Obviously, any straight line has an infinite number of them (as well as direction vectors), and all the normal vectors of the straight line will be collinear (codirectional or not, it makes no difference).
Dealing with them will be even easier than with guide vectors:
If a line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this line.
If the coordinates of the direction vector have to be carefully “pulled out” from the equation, then the coordinates of the normal vector can be simply “removed”.
The normal vector is always orthogonal to the direction vector of the line. Let us verify the orthogonality of these vectors using the scalar product:
I will give examples with the same equations as for the direction vector: 
Is it possible to construct an equation of a straight line given one point and a normal vector? If the normal vector is known, then the direction of the straight line itself is clearly defined - this is a “rigid structure” with an angle of 90 degrees.
How to write an equation of a straight line given a point and a normal vector?
If a certain point belonging to a line and the normal vector of this line are known, then the equation of this line is expressed by the formula:
Write an equation of a straight line given a point and a normal vector. Find the direction vector of the line.
Solution: Use the formula: 
The general equation of the straight line has been obtained, let’s check:
1) “Remove” the coordinates of the normal vector from the equation: 
– yes, indeed, the original vector was obtained from the condition (or a collinear vector should be obtained).
2) Let's check whether the point satisfies the equation: 
True equality.
After we are convinced that the equation is composed correctly, we will complete the second, easier part of the task. We take out the directing vector of the straight line: 
Answer: 
In the drawing the situation looks like this: 
For training purposes, a similar task for independent decision:
Write an equation of a straight line given a point and a normal vector. Find the direction vector of the line.
The final section of the lesson will be devoted to less common, but also important species equations of a straight line on a plane
Equation of a straight line in segments.
Equation of a line in parametric form
The equation of a straight line in segments has the form , where are nonzero constants. Some types of equations cannot be represented in this form, for example, direct proportionality (since the free term is equal to zero and there is no way to get one on the right side).
This is, figuratively speaking, a “technical” type of equation. A common task is to represent the general equation of a line as an equation of a line in segments. How is it convenient? The equation of a line in segments allows you to quickly find the points of intersection of the line with the coordinate axes, which can be very important in some problems of higher mathematics.
Let's find the point of intersection of the line with the axis. We reset the “y” to zero, and the equation takes the form . Desired point it turns out automatically: .
Same with the axis 
– the point at which the straight line intersects the ordinate axis.
The actions I have just explained in detail are performed verbally.
Given a straight line. Write an equation of a straight line in segments and determine the points of intersection of the graph with the coordinate axes.
Solution: Let's reduce the equation to the form . First we move the free term to the right side:
To get one on the right, divide each term in the equation by –11:
Making fractions three-story:
The points of intersection of the straight line with the coordinate axes surfaced: 
Answer: 
All that remains is to attach a ruler and draw a straight line.
It is easy to see that this line is uniquely determined by the red and green segments, hence the name - “equation of a line in segments.”
Of course, the points are not so difficult to find from the equation, but the task is still useful. The considered algorithm will be required to find the points of intersection of the plane with the coordinate axes, to reduce the equation of a second-order line to canonical form, and in some other problems. Therefore, a couple of straight lines for an independent solution:
Draw up an equation of a straight line in segments and determine the points of its intersection with the coordinate axes.
Solutions and answers at the end. Don't forget that you can draw everything if you want.
How to write parametric equations for a straight line?
Parametric equations of a straight line are more relevant for straight lines in space, but without them our abstract will be orphaned.
If a certain point belonging to a line and the direction vector of this line are known, then the parametric equations of this line are given by the system:
Compose parametric equations of a straight line using a point and a direction vector
The solution ended before it even began: 
The “te” parameter can take any value from “minus infinity” to “plus infinity”, and each parameter value corresponds to a specific point on the plane. For example, if , then we get the point 
.
Inverse problem: how to check whether a condition point will belong to a given line?
Let's substitute the coordinates of the point into the resulting parametric equations:
From both equations it follows that, that is, the system is consistent and has a unique solution.
Let's consider more meaningful tasks:
Write parametric equations of a straight line
Solution: By condition, the line is given in general view. In order to compose parametric equations of a line, you need to know its direction vector and some point belonging to this line.
Let's find the direction vector:
Now you need to find some point belonging to the line (any one will do); for these purposes, it is convenient to rewrite the general equation in the form of an equation with an angular coefficient:
This suggests, of course, the point
Let's compose the parametric equations of the straight line: 
And finally, a small creative task for you to solve on your own.
Compose parametric equations of a line if the point belonging to it and the normal vector are known
There is more than one way to formulate a task. One version of the solution and the answer at the end.
Solutions and answers:
Example 2: Solution: Let's find the slope: 
Let's compose the equation of a straight line using a point and an angular coefficient:
Answer:
Example 4: Solution: Let's compose the equation of a straight line using the formula: 
Answer:
Example 6: Solution: Use the formula: 
Answer: (y-axis)
Example 8: Solution: Let’s create an equation of a straight line using two points:
Multiply both sides by –4:
And divide by 5:
Answer:
Example 10: Solution: We use the formula:
Reduce by -2:
Direct vector: 
Answer:
Example 12:
A) Solution: Let's transform the equation:
Thus:
Answer:
b) Solution: Let's transform the equation:
Thus:
Answer:
Example 15: Solution: First, let's create a general equation of a straight line at a point and the normal vector :
Multiply by 12:
We multiply by 2 more to get rid of the fraction after opening the second bracket:
Direct vector:
Let's compose parametric equations of a straight line from a point and direction vector :
Answer:
The simplest problems with a straight line on a plane.
The relative position of the lines. Angle between straight lines
We continue to consider these endless, endless straight lines.
How to find the distance from a point to a line?
How to find the distance between two parallel lines?
How to find the angle between two straight lines?
The relative position of two straight lines
Let's consider two straight lines defined by equations in general form: 
This is the case when the audience sings along in chorus. Two lines can:
1) match;
2) be parallel: ;
3) or intersect at a single point: .
Please remember the mathematical intersection sign, it will appear very often. The notation means that the line intersects with the line at point .
How to determine the relative position of two lines?
Let's start with the first case:
Two lines coincide if and only if their corresponding coefficients are proportional, that is, there is a number “lambda” such that the equalities hold
Let's consider the straight lines and create three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.
Indeed, if all the coefficients of the equation 
multiply by –1 (change signs), and all coefficients of the equation 
cut by 2, you get the same equation: .
The second case, when the lines are parallel:
Two lines are parallel if and only if their coefficients of the variables are proportional: 
, But .
As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables: 
However, it is quite obvious that.
And the third case, when the lines intersect:
Two lines intersect if and only if their coefficients for the variables are NOT proportional, that is, there is NO such “lambda” value that the equalities hold
So, for straight lines we will create a system: 
From the first equation it follows that , and from the second equation: , which means that the system is inconsistent (there are no solutions). Thus, the coefficients of the variables are not proportional.
Conclusion: lines intersect
In practical problems, you can use the solution scheme just discussed. By the way, it is very reminiscent of the algorithm for checking vectors for collinearity. But there is a more civilized packaging:
Find out the relative position of the lines: 
The solution is based on the study of directing vectors of straight lines:
a) From the equations we find the direction vectors of the lines: 
.
, which means that the vectors are not collinear and the lines intersect.
b) Find the direction vectors of the lines: 
The lines have the same direction vector, which means they are either parallel or coincident. There is no need to count the determinant here.
It is obvious that the coefficients of the unknowns are proportional, and .
Let's find out whether the equality is true: 
Thus,
c) Find the direction vectors of the lines: 
Let's calculate the determinant made up of the coordinates of these vectors: 
, therefore, the direction vectors are collinear. The lines are either parallel or coincident.
The proportionality coefficient “lambda” can be found directly from the relationship of collinear direction vectors. However, it is also possible through the coefficients of the equations themselves: .
Now let's find out whether the equality is true. Both free terms are zero, so:
The resulting value satisfies this equation (any number in general satisfies it).
Thus, the lines coincide.
How to construct a line parallel to a given one?
The straight line is given by the equation. Write an equation for a parallel line that passes through the point.
Solution: Let's denote the unknown line by the letter . What does the condition say about her? The straight line passes through the point. And if the lines are parallel, then it is obvious that the direction vector of the straight line “tse” is also suitable for constructing the straight line “de”.
We take the direction vector out of the equation:
The example geometry looks simple:
Analytical testing consists of the following steps:
1) We check that the lines have the same direction vector (if the equation of the line is not simplified properly, then the vectors will be collinear).
2) Check whether the point satisfies the resulting equation.
In most cases, analytical testing can be easily performed orally. Look at the two equations, and many of you will quickly determine the parallelism of the lines without any drawing.
Examples for independent solutions today will be creative.
Write an equation for a line passing through a point parallel to the line if
The shortest path is at the end.
How to find the point of intersection of two lines?
If straight intersect at point , then its coordinates are a solution to the system linear equations 
How to find the point of intersection of lines? Solve the system.
Here you go geometric meaning systems of two linear equations with two unknowns are two intersecting (most often) lines in a plane.
Find the point of intersection of lines
Solution: There are two ways to solve it - graphical and analytical.
The graphical method is to simply draw the given lines and find out the intersection point directly from the drawing: 
Here is our point: . To check, you should substitute its coordinates into each equation of the line, they should fit both there and there. In other words, the coordinates of a point are a solution to the system. Essentially, we looked at a graphical way to solve a system of linear equations with two equations, two unknowns.
The graphical method is, of course, not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to create a correct and ACCURATE drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.
Therefore, it is more expedient to look for the intersection point analytical method. Let's solve the system: 
To solve the system, the method of term-by-term addition of equations was used.
The check is trivial - the coordinates of the intersection point must satisfy each equation of the system.
Find the point of intersection of the lines if they intersect.
This is an example for you to solve on your own. It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Write down the equation of the straight line.
2) Write down the equation of the straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.
Development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.
Complete solution and the answer at the end:
Perpendicular lines. Distance from a point to a line.
Angle between straight lines
How to construct a line perpendicular to a given one?
The straight line is given by the equation. Write an equation perpendicular to the line passing through the point.
Solution: By condition it is known that . It would be nice to find the directing vector of the line. Since the lines are perpendicular, the trick is simple:
From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.
Let's compose the equation of a straight line using a point and a direction vector:
Answer: 
Let's expand the geometric sketch:
Analytical verification of the solution:
1) We take out the direction vectors from the equations 
and using the scalar product of vectors we come to the conclusion that the lines are indeed perpendicular: .
By the way, you can use normal vectors, it's even easier.
2) Check whether the point satisfies the resulting equation .
The test, again, is easy to perform orally.
Find the point of intersection of perpendicular lines if the equation is known 
and period.
This is an example for you to solve on your own. There are several actions in the problem, so it is convenient to formulate the solution point by point.
Distance from point to line
Distance in geometry is traditionally denoted by the Greek letter “p”, for example: – the distance from point “m” to straight line “d”.
Distance from point to line 
expressed by the formula 
Find the distance from a point to a line 
Solution: all you need to do is carefully substitute the numbers into the formula and carry out the calculations:
Answer: 
Let's make the drawing: 
The found distance from the point to the line is exactly the length of the red segment. If you draw up a drawing on checkered paper on a scale of 1 unit. = 1 cm (2 cells), then the distance can be measured with an ordinary ruler.
Let's consider another task based on the same drawing:
How to construct a point that is symmetrical about a straight line?
The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line 
. I suggest performing the steps yourself, but I will outline the solution algorithm with intermediate results:
1) Find a line that is perpendicular to the line.
2) Find the point of intersection of the lines: 
.
In geometry, the angle between two straight lines is taken to be the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered the angle between intersecting lines. And its “green” neighbor or the oppositely oriented “raspberry” corner is considered as such.
If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.
How are the angles different? Orientation. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .
Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing, for a negative angle, be sure to indicate its orientation with an arrow (clockwise).
Based on the above, it is convenient to formalize the solution in two steps:
1) Let's calculate the scalar product of the direction vectors of the lines:
, which means the lines are not perpendicular.
2) Find the angle between straight lines using the formula:
By using inverse function It's easy to find the corner itself. In this case, we use the oddness of the arctangent: 
Answer: 
In your answer, we indicate the exact value, as well as an approximate value (preferably in both degrees and radians), calculated using a calculator.
Well, minus, minus, no big deal. Here is a geometric illustration: 
It is not surprising that the angle turned out to be of a negative orientation, because in the problem statement the first number is a straight line and the “unscrewing” of the angle began precisely with it.
There is a third solution. The idea is to calculate the angle between the direction vectors of the lines:
Here we are no longer talking about an oriented angle, but “just about an angle,” that is, the result will certainly be positive. The catch is that you may end up with an obtuse angle (not the one you need). In this case, you will have to make a reservation that the angle between straight lines is a smaller angle, and subtract the resulting arc cosine from “pi” radians (180 degrees).
Find the angle between the lines.
This is an example for you to solve on your own. Try to solve it in two ways.
Solutions and answers:
Example 3: Solution: Find the directing vector of the line:
Let's compose the equation of the desired straight line using the point and the direction vector 
Note: here the first equation of the system is multiplied by 5, then the 2nd is subtracted term by term from the 1st equation.
Answer:
In order to use the coordinate method, you need to know the formulas well. There are three of them:
At first glance, it looks threatening, but with just a little practice, everything will work great.
Task. Find the cosine of the angle between the vectors a = (4; 3; 0) and b = (0; 12; 5).
Solution. Since the coordinates of the vectors are given to us, we substitute them into the first formula:
Task. Write an equation for a plane passing through the points M = (2; 0; 1), N = (0; 1; 1) and K = (2; 1; 0), if it is known that it does not pass through the origin.
Solution. The general equation of the plane: Ax + By + Cz + D = 0, but since the desired plane does not pass through the origin of coordinates - the point (0; 0; 0) - then we put D = 1. Since this plane passes through the points M, N and K, then the coordinates of these points should turn the equation into a correct numerical equality.
Let's substitute the coordinates of the point M = (2; 0; 1) instead of x, y and z. We have:
A 2 + B 0 + C 1 + 1 = 0 ⇒ 2A + C + 1 = 0;
Similarly, for points N = (0; 1; 1) and K = (2; 1; 0) we obtain the following equations:
A 0 + B 1 + C 1 + 1 = 0 ⇒ B + C + 1 = 0;
A 2 + B 1 + C 0 + 1 = 0 ⇒ 2A + B + 1 = 0;
So we have three equations and three unknowns. Let's create and solve a system of equations:
We found that the equation of the plane has the form: − 0.25x − 0.5y − 0.5z + 1 = 0.
Task. The plane is given by the equation 7x − 2y + 4z + 1 = 0. Find the coordinates of the vector perpendicular to this plane.
Solution. Using the third formula, we get n = (7; − 2; 4) - that's all!
Calculation of vector coordinates
But what if there are no vectors in the problem - there are only points lying on straight lines, and you need to calculate the angle between these straight lines? It's simple: knowing the coordinates of the points - the beginning and end of the vector - you can calculate the coordinates of the vector itself.
To find the coordinates of a vector, you need to subtract the coordinates of the beginning from the coordinates of its end.
This theorem works equally well both on a plane and in space. The expression “subtract coordinates” means that the x coordinate of another point is subtracted from the x coordinate of one point, then the same must be done with the y and z coordinates. Here are some examples:
Task. There are three points in space, defined by their coordinates: A = (1; 6; 3), B = (3; − 1; 7) and C = (− 4; 3; − 2). Find the coordinates of the vectors AB, AC and BC.
Consider the vector AB: its beginning is at point A, and its end is at point B. Therefore, to find its coordinates, we need to subtract the coordinates of point A from the coordinates of point B:
AB = (3 − 1; − 1 − 6; 7 − 3) = (2; − 7; 4).
Similarly, the beginning of the vector AC is the same point A, but the end is point C. Therefore, we have:
AC = (− 4 − 1; 3 − 6; − 2 − 3) = (− 5; − 3; − 5).
Finally, to find the coordinates of vector BC, you need to subtract the coordinates of point B from the coordinates of point C:
BC = (− 4 − 3; 3 − (− 1); − 2 − 7) = (− 7; 4; − 9).
Answer: AB = (2; − 7; 4); AC = (− 5; − 3; − 5); BC = (− 7; 4; − 9)
Pay attention to the calculation of the coordinates of the last BC vector: many people make mistakes when working with negative numbers. This concerns the variable y: point B has coordinate y = − 1, and point C has coordinate y = 3. We get exactly 3 − (− 1) = 4, and not 3 − 1, as many people think. Don't make such stupid mistakes!
Calculation of direction vectors for straight lines
If you carefully read problem C2, you will be surprised to find that there are no vectors there. There are only straight lines and planes.
First, let's look at the straight lines. Everything is simple here: on any line there are at least two distinct points and, conversely, any two distinct points define a unique line...
Did anyone understand what was written in the previous paragraph? I didn’t understand it myself, so I’ll explain it more simply: in problem C2, straight lines are always defined by a pair of points. If we introduce a coordinate system and consider a vector with the beginning and end at these points, we obtain the so-called direction vector for the line:
Why is this vector needed? The fact is that the angle between two straight lines is the angle between their direction vectors. Thus, we move from incomprehensible straight lines to specific vectors whose coordinates are easy to calculate. How easy is it? Take a look at the examples:
Task. In the cube ABCDA 1 B 1 C 1 D 1 the lines AC and BD 1 are drawn. Find the coordinates of the direction vectors of these lines.
Since the length of the edges of the cube is not specified in the condition, we set AB = 1. We introduce a coordinate system with the origin at point A and the x, y, z axes directed along the straight lines AB, AD and AA 1, respectively. The unit segment is equal to AB = 1.
Now let's find the coordinates of the direction vector for straight line AC. We need two points: A = (0; 0; 0) and C = (1; 1; 0). From here we get the coordinates of the vector AC = (1 − 0; 1 − 0; 0 − 0) = (1; 1; 0) - this is the direction vector.
Now let's look at the straight line BD 1. It also has two points: B = (1; 0; 0) and D 1 = (0; 1; 1). We obtain the direction vector BD 1 = (0 − 1; 1 − 0; 1 − 0) = (− 1; 1; 1).
Answer: AC = (1; 1; 0); BD 1 = (− 1; 1; 1)
Task. In a regular triangular prism ABCA 1 B 1 C 1, all edges of which are equal to 1, straight lines AB 1 and AC 1 are drawn. Find the coordinates of the direction vectors of these lines.
Let us introduce a coordinate system: the origin is at point A, the x axis coincides with AB, the z axis coincides with AA 1, the y axis forms the OXY plane with the x axis, which coincides with the ABC plane.
First, let's look at the straight line AB 1. Everything is simple here: we have points A = (0; 0; 0) and B 1 = (1; 0; 1). We obtain the direction vector AB 1 = (1 − 0; 0 − 0; 1 − 0) = (1; 0; 1).
Now let's find the direction vector for AC 1. Everything is the same - the only difference is that point C 1 has irrational coordinates. So A = (0; 0; 0), so we have:
Answer: AB 1 = (1; 0; 1);
A small but very important note about the last example. If the beginning of the vector coincides with the origin of coordinates, the calculations are greatly simplified: the coordinates of the vector are simply equal to the coordinates of the end. Unfortunately, this is only true for vectors. For example, when working with planes, the presence of the origin of coordinates on them only complicates the calculations.
Calculation of normal vectors for planes
Normal vectors are not those vectors that are fine or feel good. By definition, a normal vector (normal) to a plane is a vector perpendicular to a given plane.
In other words, a normal is a vector perpendicular to any vector in a given plane. You've probably come across this definition - however, instead of vectors we were talking about straight lines. However, it was shown just above that in problem C2 you can operate with any convenient object - be it a straight line or a vector.
Let me remind you once again that every plane is defined in space by the equation Ax + By + Cz + D = 0, where A, B, C and D are some coefficients. Without losing the generality of the solution, we can assume D = 1 if the plane does not pass through the origin, or D = 0 if it does. In any case, the coordinates of the normal vector to this plane are n = (A; B; C).
So, the plane can also be successfully replaced by a vector - the same normal. Every plane is defined in space by three points. We already discussed how to find the equation of the plane (and therefore the normal) at the very beginning of the article. However, this process causes problems for many, so I’ll give a couple more examples:
Task. In the cube ABCDA 1 B 1 C 1 D 1 a section A 1 BC 1 is drawn. Find the normal vector for the plane of this section if the origin of coordinates is at point A, and the x, y and z axes coincide with the edges AB, AD and AA 1, respectively.
Since the plane does not pass through the origin, its equation looks like this: Ax + By + Cz + 1 = 0, i.e. coefficient D = 1. Since this plane passes through points A 1, B and C 1, the coordinates of these points turn the equation of the plane into the correct numerical equality.
A 0 + B 0 + C 1 + 1 = 0 ⇒ C + 1 = 0 ⇒ C = − 1;
Similarly, for points B = (1; 0; 0) and C 1 = (1; 1; 1) we obtain the following equations:
A 1 + B 0 + C 0 + 1 = 0 ⇒ A + 1 = 0 ⇒ A = − 1;
A 1 + B 1 + C 1 + 1 = 0 ⇒ A + B + C + 1 = 0;
But we already know the coefficients A = − 1 and C = − 1, so it remains to find the coefficient B:
B = − 1 − A − C = − 1 + 1 + 1 = 1.
We obtain the equation of the plane: − A + B − C + 1 = 0. Therefore, the coordinates of the normal vector are equal to n = (− 1; 1; − 1).
Task. In the cube ABCDA 1 B 1 C 1 D 1 there is a section AA 1 C 1 C. Find the normal vector for the plane of this section if the origin of coordinates is at point A and the x, y and z axes coincide with the edges AB, AD and AA 1 respectively.
In this case, the plane passes through the origin, so the coefficient D = 0, and the equation of the plane looks like this: Ax + By + Cz = 0. Since the plane passes through points A 1 and C, the coordinates of these points turn the equation of the plane into the correct numerical equality.
Let us substitute the coordinates of point A 1 = (0; 0; 1) instead of x, y and z. We have:
A 0 + B 0 + C 1 = 0 ⇒ C = 0;
Similarly, for the point C = (1; 1; 0) we obtain the equation:
A 1 + B 1 + C 0 = 0 ⇒ A + B = 0 ⇒ A = − B;
Let us set B = 1. Then A = − B = − 1, and the equation of the entire plane has the form: − A + B = 0. Therefore, the coordinates of the normal vector are equal to n = (− 1; 1; 0).
Generally speaking, in the above problems you need to create a system of equations and solve it. You will get three equations and three variables, but in the second case one of them will be free, i.e. take arbitrary values. That is why we have the right to set B = 1 - without prejudice to the generality of the solution and the correctness of the answer.
Very often in Problem C2 you need to work with points that bisect a segment. The coordinates of such points are easily calculated if the coordinates of the ends of the segment are known.
So, let the segment be defined by its ends - points A = (x a; y a; z a) and B = (x b; y b; z b). Then the coordinates of the middle of the segment - let’s denote it by point H - can be found using the formula:
In other words, the coordinates of the middle of a segment are the arithmetic mean of the coordinates of its ends.
Task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in a coordinate system so that the x, y and z axes are directed along edges AB, AD and AA 1, respectively, and the origin coincides with point A. Point K is the middle of edge A 1 B 1 . Find the coordinates of this point.
Since point K is the middle of the segment A 1 B 1, its coordinates are equal to the arithmetic mean of the coordinates of the ends. Let's write down the coordinates of the ends: A 1 = (0; 0; 1) and B 1 = (1; 0; 1). Now let's find the coordinates of point K:
Task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in a coordinate system so that the x, y and z axes are directed along the edges AB, AD and AA 1, respectively, and the origin coincides with point A. Find the coordinates of the point L at which they intersect diagonals of the square A 1 B 1 C 1 D 1 .
From the planimetry course we know that the point of intersection of the diagonals of a square is equidistant from all its vertices. In particular, A 1 L = C 1 L, i.e. point L is the middle of the segment A 1 C 1. But A 1 = (0; 0; 1), C 1 = (1; 1; 1), so we have:
Answer: L = (0.5; 0.5; 1)
Option 2.13
1.(C03.RP) Create an equation of a line passing through a point perpendicular to the line 
.
Vector 
- normal line vector 
,
Let's write the equation AB:
Answer: 
.
2.(8T3.RP) Create a general equation for a line passing through a point 
and the point of intersection of the lines 
And 
.
Find the coordinates of the point IN– point of intersection of lines 
And 
:
multiplied the second equation by -2, and now add them
We got the coordinates. IN(
).
Let's write the equation AB:
Answer: 
.
3.(T43.RP) Write the general equation of the plane passing through the points 
,
perpendicular to the plane 
.
The general equation of the plane is A(x-x 1 )+B(y-y 1 )+C(z-z 1 ) =0
M 1 (4,-3,3), then we can write:
A(x-4)+B(y+3)+C(z-3)=0
Because the plane passes through the point M 2 (1,1,-2), then we can write:
A(x-1)+B(y-1)+C(z+2)=0
The desired plane is perpendicular to the plane given by the equation: According to the condition of perpendicularity of planes:
A 1 A 2 +B 1 B 2 +C 1 C 2 =0
1 × A+(-3)× B+5× C=0
A=3B-5C
Let's substitute into the lower equation
4.(303) Find the distance from the point 
to a straight line 
.
Find the intersection point of the perpendicular passing through the point A. Let's call her N(x, y, z) .
AN:3(x-2)+4(y+1)+2z=0 3x+4y+2z-2=0
The parametric equations of the line have the form:
T. N(4,-3,1)
5.(5B3.RP) Find those parameter values 
And 
, for which the straight lines 
And 
parallel.
To calculate the direction vector we use the formula:
Let's calculate the direction vector of the line
Because A||B
We get a system of equations:
Answer: A=0, B=-1.
6.(733) Direct 
parallel to the plane, intersects the line 
and passes through the point 
. Find the ordinate of the point of intersection of a line with a plane 
.
We'll find k:
Let's write down the parametric equations of the straight line:
Let's substitute x,y,z into the equation L and get the t value.
T. IN(8;-8;5) belongs to L
Let us write the parametric equations L:
Let's substitute these values into the equation:
Find the ordinate of the intersection point 
Answer: -2.5.
7.(983). Find the radius of a circle centered at a point 
, if it touches the line 
.
In order to find the radius of a circle, you can find the distance from point A to a given line and this distance will be equal to the radius.
Let's use the formula:
8. Given a curve.
8.1. Prove that this curve is an ellipse.
8.2.(TT3.RP) Find the coordinates of the center of its symmetry.
8.3.(4B3.RP) Find its major and minor semi-axes of the curve.
8.4.(2P3) Write down the equation of the focal axis.
8.5. Construct this curve.
The canonical equation of the ellipse has the form
Let us bring the equation of the curve to canonical form:
Because does not contain what you are looking for xy, then we remain in the old coordinate system.
Taking the point for a new beginning 
, apply the coordinate transformation formulas 
This corresponds to the general form of the equation of an ellipse, in which the semi-major axis is 4 and the semi-minor axis is 2.
The focal radius vectors of a given ellipse correspond to the equation
9. Given a curve 
.
9.1. Prove that this curve is a parabola.
9.2.(L33). Find the value of its parameter 
.
9.3.(2T3.RP). Find the coordinates of its vertex.
9.4.(7B3). Write the equation of its axis of symmetry.
9.5. Construct this curve.
The canonical equation of a parabola is: y 2 =2px
In our example
Those. this curve is a parabola, symmetrical about the ordinate axis.
In this case 2р=-12
p=-6, therefore the branches of the parabola face down.
The vertex of the parabola is at point (-3;-2)
Equation of the axis of symmetry of this parabola: x=-3
10. Given a curve.
10.1. Prove that this curve is a hyperbola.
10.2.(793.RP). Find the coordinates of its center of symmetry.
10.3.(8D3.RP). Find the real and imaginary semi-axes.
10.4.(PS3.RP). Write the equation of the focal axis.
10.5. Construct this curve.
The canonical equation of a hyperbola has the form 
Let's transform the equation using the formulas for rotating the coordinate axis:
We get: 
Let's find l from the condition:
those. let us equate the coefficient at x`y` to zero
solutions normal
Basic educational program of basic general education table of contents
Main educational program... Vectors. Length (module) vector. Equality vectors. Collinear vectors. Coordinates vector. Multiplication vector by number, amount vectors, decomposition vector ... solution child development tasks that are missing in the content of education Fine ...
Educational program of basic general education (FSOS LLC)
Educational program... vectors direct solutions... ensuring rational organization of the motor regime, normal physical development and motor readiness...
Sample basic educational program
Program... vectors, establish perpendicularity direct. The graduate will have the opportunity to: master the vector method for solutions... ensuring rational organization of the motor regime, normal physical development and motor readiness...
Equation of a plane. How to write an equation of a plane?
Mutual arrangement of planes. Tasks
Spatial geometry is not much more complicated than “flat” geometry, and our flights in space begin with this article. To master the topic, you need to have a good understanding of vectors, in addition, it is advisable to be familiar with the geometry of the plane - there will be many similarities, many analogies, so the information will be digested much better. In a series of my lessons, the 2D world opens with an article Equation of a straight line on a plane. But now Batman has left the flat TV screen and is launching from the Baikonur Cosmodrome.
Let's start with drawings and symbols. Schematically, the plane can be drawn in the form of a parallelogram, which creates the impression of space: 
The plane is infinite, but we have the opportunity to depict only a piece of it. In practice, in addition to the parallelogram, an oval or even a cloud is also drawn. For technical reasons, it is more convenient for me to depict the plane in exactly this way and in exactly this position. Real planes, which we will consider in practical examples, can be located in any way - mentally take the drawing in your hands and rotate it in space, giving the plane any slope, any angle.
Designations: planes are usually denoted in small Greek letters, apparently so as not to confuse them with straight line on a plane or with straight line in space. I'm used to using the letter . In the drawing it is the letter “sigma”, and not a hole at all. Although, the holey plane is certainly quite funny.
In some cases, it is convenient to use the same Greek letters with lower subscripts to designate planes, for example, .
It is obvious that the plane is uniquely defined by three different points that do not lie on the same line. Therefore, three-letter designations of planes are quite popular - by the points belonging to them, for example, etc. Often the letters are enclosed in parentheses: so as not to confuse the plane with another geometric figure.
For experienced readers I will give quick access menu:
- How to create an equation of a plane using a point and two vectors?
- How to create an equation of a plane using a point and a normal vector?
and we will not languish in long waits:
General plane equation
The general equation of the plane has the form , where the coefficients are not equal to zero at the same time.
A number of theoretical calculations and practical problems are valid both for the usual orthonormal basis and for affine basis space (if the oil is oil, return to the lesson Linear (non) dependence of vectors. Basis of vectors). For simplicity, we will assume that all events occur in an orthonormal basis and a Cartesian rectangular coordinate system.
Now let’s practice our spatial imagination a little. It’s okay if yours is bad, now we’ll develop it a little. Even playing on nerves requires training.
In the most general case, when the numbers are not equal to zero, the plane intersects all three coordinate axes. For example, like this:
I repeat once again that the plane continues indefinitely in all directions, and we have the opportunity to depict only part of it.
Let's consider the simplest equations of planes:
How to understand this equation? Think about it: “Z” is ALWAYS equal to zero, for any values of “X” and “Y”. This is the equation of the "native" coordinate plane. Indeed, formally the equation can be rewritten as follows: 
, from where you can clearly see that we don’t care what values “x” and “y” take, it is important that “z” is equal to zero.
Likewise:
– equation of the coordinate plane;
– equation of the coordinate plane.
Let's complicate the problem a little, consider a plane (here and further in the paragraph we assume that the numerical coefficients are not equal to zero). Let's rewrite the equation in the form: . How to understand it? “X” is ALWAYS, for any values of “Y” and “Z”, equal to a certain number. This plane is parallel to the coordinate plane. For example, a plane is parallel to a plane and passes through a point.
Likewise:
– equation of a plane that is parallel to the coordinate plane;
– equation of a plane that is parallel to the coordinate plane.
Let's add members: . The equation can be rewritten as follows: , that is, “zet” can be anything. What does it mean? “X” and “Y” are connected by the relation, which draws a certain straight line in the plane (you will find out equation of a line in a plane?). Since “z” can be anything, this straight line is “replicated” at any height. Thus, the equation defines a plane parallel to the coordinate axis
Likewise:
– equation of a plane that is parallel to the coordinate axis;
– equation of a plane that is parallel to the coordinate axis.
If the free terms are zero, then the planes will directly pass through the corresponding axes. For example, the classic “direct proportionality”: . Draw a straight line in the plane and mentally multiply it up and down (since “Z” is any). Conclusion: the plane defined by the equation passes through the coordinate axis.
We complete the review: the equation of the plane 
passes through the origin. Well, here it is quite obvious that the point satisfies this equation.
And finally, the case shown in the drawing: – the plane is friendly with all coordinate axes, while it always “cuts off” a triangle, which can be located in any of the eight octants.
Linear inequalities in space
To understand the information you need to study well linear inequalities in the plane, because many things will be similar. The paragraph will be of a brief overview nature with several examples, since the material is quite rare in practice.
If the equation defines a plane, then the inequalities
ask half-spaces. If the inequality is not strict (the last two in the list), then the solution of the inequality, in addition to the half-space, also includes the plane itself.
Example 5
Find the unit normal vector of the plane 
.
Solution: A unit vector is a vector whose length is one. Let us denote this vector by . It is absolutely clear that the vectors are collinear: 
First, we remove the normal vector from the equation of the plane: .
How to find a unit vector? In order to find the unit vector, you need every divide the vector coordinate by the vector length.
Let's rewrite the normal vector in the form and find its length:
According to the above:
Answer: 
Verification: what was required to be verified.
Readers who carefully studied the last paragraph of the lesson probably noticed that the coordinates of the unit vector are exactly the direction cosines of the vector:
Let's take a break from the problem at hand: when you are given an arbitrary non-zero vector, and according to the condition it is required to find its direction cosines (see the last problems of the lesson Dot product of vectors), then you, in fact, find a unit vector collinear to this one. Actually two tasks in one bottle.
The need to find the unit normal vector arises in some problems of mathematical analysis.
We’ve figured out how to fish out a normal vector, now let’s answer the opposite question:
How to create an equation of a plane using a point and a normal vector?
This rigid construction of a normal vector and a point is well known to the dartboard. Please extend your hand forward and mentally choose arbitrary point spaces, for example, a small cat in a sideboard. Obviously, through this point you can draw a single plane perpendicular to your hand.
The equation of a plane passing through a point perpendicular to the vector is expressed by the formula:
Methods for defining a plane.
Mutual arrangement of planes.
Two planes in space can coincide. In this case they have at least three points in common.
Two planes in space can intersect. The intersection of two planes is a straight line, which is established by the axiom: if two planes have a common point, then they have a common straight line on which all the common points of these planes lie.
In this case, the concept of an angle between intersecting planes arises. Of particular interest is the case when the angle between the planes is ninety degrees. Such planes are called perpendicular.
Finally, two planes in space can be parallel, that is, have no common points.
Also interesting are cases when several planes intersect along one straight line and several planes intersect at one point.
Let us list the main ways to define a specific plane in space.
Firstly, a plane can be defined by fixing three points in space that do not lie on the same straight line. This method is based on the axiom: through any three points that do not lie on the same line, there is a single plane.
If in three-dimensional space If a rectangular coordinate system is fixed and a plane is specified by specifying the coordinates of its three different points that do not lie on the same straight line, then we can write the equation of the plane passing through the three given points.
The next two methods of defining a plane are a consequence of the previous one. They are based on corollaries of the axiom about a plane passing through three points:
· a plane, and only one, passes through a straight line and a point not lying on it;
A single plane passes through two intersecting lines.
The fourth method of defining a plane in space is based on defining parallel lines. Recall that two lines in space are called parallel if they lie in the same plane and do not intersect. Thus, by indicating two parallel lines in space, we will determine the only plane in which these lines lie.
If in three-dimensional space relative to a rectangular coordinate system a plane is specified in the indicated way, then we can create an equation for a plane passing through two parallel lines.
The sign of parallelism of two planes gives us another way to define a plane. Let us recall the formulation of this feature: if two intersecting lines of one plane are respectively parallel to two lines of another plane, then such planes are parallel. Therefore, we can specify a specific plane if we specify the point through which it passes and the plane to which it is parallel.
I know high school In geometry lessons, the following theorem is proven: through a fixed point in space there passes a single plane perpendicular to a given line. Thus, we can define a plane if we specify the point through which it passes and a line perpendicular to it.
If a rectangular coordinate system is fixed in three-dimensional space and a plane is specified in the indicated way, then it is possible to construct an equation for a plane passing through a given point perpendicular to a given straight line.
Instead of a line perpendicular to the plane, you can specify one of the normal vectors of this plane. In this case, it is possible to write a general equation of the plane.
A good understanding of a straight line begins from the moment when, along with its image, images of its guide and normal vectors simultaneously appear. Likewise, when referring to a plane in space, it must be represented along with its normal vector. Why is that? Yes, because in many cases it is more convenient to use the normal vector of the plane than the plane itself.
First, we will give the definition of a normal vector of a plane, give examples of normal vectors and the necessary graphic illustrations. Next, we will place the plane in a rectangular coordinate system in three-dimensional space and learn to determine the coordinates of the normal vector of the plane from its equation.
2.1. Normal plane vector - definition, examples, illustrations.
Definition. Normal plane vector is any non-zero vector lying on a line perpendicular to a given plane.
From the definition it follows that there is an infinite number of normal vectors of a given plane.
Since all normal vectors given plane lie on parallel lines, then all normal vectors of the plane are collinear. In other words, if is a normal vector of the plane, then the vector for some non-zero real value t is also a normal vector of the plane.
It should also be noted that any normal vector of a plane can be considered as a direction vector of a line perpendicular to this plane.
Sets of normal vectors parallel planes coincide, since a straight line perpendicular to one of the parallel planes is also perpendicular to the second plane.
From the definition perpendicular planes and the definition of the normal vector of a plane it follows that the normal vectors of perpendicular planes are perpendicular.
An example of a normal plane vector. Let a rectangular coordinate system Oxyz be fixed in three-dimensional space. Coordinate vectors are normal vectors of the Oyz, Oxz and Oxy planes, respectively. This is true because the vectors are non-zero and lie on the coordinate lines Ox, Oy and Oz, respectively, which are perpendicular coordinate planes Oyz, Oxz and Oxy respectively.
2.2. Coordinates of the normal vector of a plane - finding the coordinates of the normal vector of a plane using the equation of the plane.
Let's find the coordinates of the normal vector of the plane if we know the equation of the plane in the rectangular coordinate system Oxyz.
The general equation of a plane of the form defines in the rectangular coordinate system Oxyz a plane whose normal vector is the vector . Thus, in order to find the coordinates of the normal vector of a plane, it is enough for us to have before our eyes the general equation of this plane.
Example. Find the coordinates of any normal vector of the plane.
Solution. We are given the general equation of a plane, the coefficients of the variables x, y and z represent the corresponding coordinates of the normal vector of this plane. Therefore, is one of the normal vectors of a given plane. The set of all normal vectors of this plane can be specified as , where t is an arbitrary real number other than zero.
Example. The plane is given by the equation. Determine the coordinates of its direction vectors.
Solution. We are given an incomplete equation of the plane. To make the coordinates of its direction vector visible, we rewrite the equation in the form . Thus, the normal vector of this plane has coordinates , and the set of all normal vectors will be written as .
The equation of a plane in segments of the form , like the general equation of a plane, allows you to immediately write down one of the normal vectors of this plane - it has coordinates.
In conclusion, we will say that using the normal vector of a plane, various problems can be solved. The most common are problems to prove the parallelism or perpendicularity of planes, problems to draw up an equation of a plane, as well as problems to find the angle between planes and to find the angle between a straight line and a plane.