In this article, first, the definition of the projection of a point on a straight line (onto an axis) is given and an explanatory figure is given. Next, a method for finding the coordinates of the projection of a point on a straight line in the introduced rectangular coordinate system on the plane and in three-dimensional space, examples are shown with detailed explanations.
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Point to line projection - definition.
Since everyone geometric figures consist of points, and the projection of a figure is a set of projections of all points of this figure, then to project a figure onto a straight line, you must be able to project the points of this figure onto this straight line.
So what is called the projection of a point onto a line?
Definition.
Point to line projection Is either the point itself, if it lies on a given straight line, or the base of a perpendicular dropped from this point onto a given straight line.
In the figure below, point H 1 is the projection of point M 1 onto line a, and point M 2 is the projection of point M 2 itself onto line a, since M 2 lies on line a.
This definition of the projection of a point on a straight line is valid both for the case on a plane and for the case in three-dimensional space.
On the plane, in order to build a projection of the point M 1 onto the straight line a, you need to draw a straight line b, which passes through the point M 1 and is perpendicular to the straight line a. Then the point of intersection of lines a and b is the projection of point М 1 onto line a.
In three-dimensional space, the projection of point М 1 onto line a is the point of intersection of line a and a plane passing through point М 1 perpendicular to line a.
Finding the coordinates of the projection of a point on a line - theory and examples.
Let's start by finding the coordinates of the projection of a point on a line, when the projected point and line are given in the rectangular coordinate system Oxy on the plane. After that, we will show how the coordinates of the projection of a point on a straight line in a rectangular coordinate system Oxyz in three-dimensional space are found.
The coordinates of the projection of a point onto a straight line on the plane.
Let Oxy be fixed on the plane, a point is given, a straight line a and it is required to determine the coordinates of the projection of the point М 1 onto the straight line a.
Let's solve this problem.
Let us draw a straight line b through the point M 1, perpendicular to the straight line a, and denote the point of intersection of lines a and b as H 1. Then H 1 is the projection of the point М 1 onto the line a.
The construction carried out logically implies algorithm that allows you to find the coordinates of the projection of a point on the line a:
Let's figure out how to find the coordinates of the projection of a point on a straight line when solving an example.
Example.
On the plane relative to the rectangular coordinate system Oxy, a point and a straight line a are given, which corresponds to the general equation of a straight line of the form 
Decision.
We know the equation of the straight line a from the condition, so we can proceed to the second step of the algorithm.
We obtain the equation of the straight line b, which passes through the point М 1 and is perpendicular to the straight line a. To do this, we need the coordinates of the direction vector of line b. Since line b is perpendicular to line a, the normal vector of line a is the direction vector of line b. Obviously, the normal vector of the line is a vector with coordinates, therefore, the direction vector of the line b is a vector. Now we can write the canonical equation of the straight line b, since we know the coordinates of the point through which it passes, and the coordinates of its direction vector:.
It remains to find the coordinates of the point of intersection of straight lines a and b, which will give the desired coordinates of the projection of the point M 1 on the straight line a. To do this, we first pass from the canonical equations of the straight line b to its general equation:. Now we will compose a system of equations from the general equations of the lines a and b, after which we will find its solution (if necessary, refer to the article): 
Thus, the projection of a point onto a line has coordinates.
Answer:
Example.
Three points are specified on the plane in the rectangular Oxy coordinate system. Find the coordinates of the projection of point M 1 on line AB.
Decision.
To find the coordinates of the projection of the point M 1 on the straight line AB, we will act according to the obtained algorithm.
Let's write the equation of a straight line passing through two given points and:
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Now we can go from the obtained canonical equation of the straight line AB to the general equation of the straight line AB and continue the solution by analogy with the previous example. But let's consider another way of finding the equation of the straight line b passing through the point M 1 perpendicular to the straight line AB.
From the canonical equation of the straight line AB we obtain the equation of the straight line with the slope: 
... The slope of the line AB is equal, and the slope of the line b, which is perpendicular to the line AB, is (see the condition for the perpendicularity of the lines). Then the equation of the straight line b passing through the point and having the slope has the form.
To determine the coordinates of the projection of a point on the line AB, it remains to solve the system of equations 
:
Answer:
Let's dwell separately on finding the coordinates of the projection of a point on the coordinate lines Ox and Oy, as well as on lines parallel to them.
Obviously, the projection of a point onto the coordinate line Ox, which corresponds to an incomplete general equation of the straight line, is a point with coordinates. Similarly, the projection of a point onto the coordinate line Oy has coordinates.
Any straight line parallel to the abscissa axis can be specified by an incomplete general equation of the form 
, and a straight line parallel to the ordinate axis is an equation of the form 
... The projections of a point onto straight lines are points with coordinates and, respectively.
Example.
What are the coordinates of the projection of the point on the coordinate line Oy and on the straight line.
Decision.
The projection of a point onto line Oy is a point with coordinates.
Let's rewrite the equation of the line as. Now you can clearly see that the projection of a point on a straight line has coordinates.
Answer:
And.
The coordinates of the projection of a point on a straight line in three-dimensional space.
Now we turn to finding the coordinates of the projection of a point on a straight line relative to the rectangular coordinate system Oxyz introduced in three-dimensional space.
Let a rectangular coordinate system Oxyz be fixed in space, a point 
, straight line a and it is required to find the coordinates of the projection of point М 1 onto line a.
Let's solve this problem.
Let's construct a plane that passes through the point M 1 perpendicular to the straight line a. The projection of point М 1 onto line a is the point of intersection of line a and plane. Thus, we get algorithm to find the coordinates of the projected point on line a:
Let's consider the solution of an example.
Example.
In the rectangular coordinate system Oxyz, a point and a straight line a are given, and the straight line a is determined by the canonical equations of a straight line in space of the form 
... Find the coordinates of the projection of point M 1 on line a.
Decision.
To determine the coordinates of the projection of the point М 1 onto the line a, we will use the obtained algorithm.
The equations of the straight line a are immediately known to us from the condition, so we pass to the second step.
We obtain the equation of a plane that is perpendicular to the line a and passes through the point. To do this, we need to know the coordinates of the normal vector of the plane. Let's find them. The canonical equations of the straight line a show the coordinates of the direction vector of this straight line:. The direction vector of the line a is the normal vector of the plane, which is perpendicular to the line a. I.e, 
is the normal vector of the plane. Then the equation of the plane passing through a point and having a normal vector , has the form.
It remains to find the coordinates of the point of intersection of the straight line a and the plane - they are the required coordinates of the projection of the point on the straight line a. We will show two ways to find them.
First way.
From the canonical equations of the straight line a, we obtain the equations of two intersecting planes that define the straight line a:
Line Intersection Coordinates 
and plane 
we obtain by solving a system of linear equations of the form 
... Let's apply (if you like more or some other method for solving systems of linear equations, then apply it): 
Thus, the point with coordinates is the projection of point М 1 onto line a.
Second way.
Knowing the canonical equations of the straight line a, it is easy to write down its parametric equations of the straight line in space: 
... Let us substitute the plane of the form instead of x, y and z their expressions through the parameter:
Now we can calculate the required coordinates of the intersection point of the straight line a and the plane using the parametric equations of the straight line a when:
This article examines the concept of projection of a point onto a straight line (axis). We will define it using an explanatory figure; we will study the method of determining the coordinates of the projection of a point on a straight line (on a plane or in three-dimensional space); let's look at examples.
In the article "Projection of a point onto a plane, coordinates" we mentioned that the projection of a figure is a generalized concept of perpendicular or orthogonal projection.
All geometrical figures consist of points, respectively, the projection of this figure is a set of projections of all its points. Therefore, in order to be able to project a figure onto a line, it is necessary to get the skill of projecting a point onto a line.
Definition 1
Point to line projection Is either the point itself, if it belongs to a given straight line, or the base of a perpendicular dropped from this point onto a given straight line.
Consider the figure below: point H 1 serves as a projection of point M 1 onto line a, and point M 2 belonging to the line is a projection of itself.
This definition is true for the case on a plane and in three-dimensional space.
In order to obtain the projection of the point M 1 onto the straight line a on the plane, a straight line b is drawn passing through a given point M 1 and perpendicular to the straight line a. Thus, the point of intersection of lines a and b will be the projection of point М 1 onto line a.
In three-dimensional space, the projection of a point onto a straight line will be the intersection point of the straight line a and the plane α passing through the point M 1 and the perpendicular line a.
Finding the coordinates of the projection of a point on a line
Consider this issue in cases of projection on a plane and in three-dimensional space.
Let us be given a rectangular coordinate system O x y, point M 1 (x 1, y 1) and straight line a. It is necessary to find the coordinates of the projection of the point M 1 on the line a.
Let us draw a straight line b perpendicular to a straight line through a given point М 1 (x 1, y 1). The intersection point is marked as H 1. Point H 1 will be the point of projection of point M 1 onto line a.
From the described construction, we can formulate an algorithm that allows us to find the coordinates of the projection of the point M 1 (x 1, y 1) on the straight line a:
We compose the equation of the straight line (if it is not given). To perform this action, you need the skill of drawing up basic equations on a plane;
We write down the equation of the straight line b (passing through the point М 1 and perpendicular to the straight line a). An article about the equation of a straight line passing through a given point perpendicular to a given straight line will help here;
We define the required projection coordinates as the coordinates of the point of intersection of straight lines a and b. To do this, we solve a system of equations, the components of which are the equations of straight lines a and b.
Example 1
On the O x y plane, points M 1 (1, 0) and a straight line a (general equation - 3 x + y + 7 \u003d 0) are given. It is necessary to determine the coordinates of the projection of the point M 1 on the straight line a.
Decision
The equation of the given straight line is known, therefore, according to the algorithm, we pass to the step of writing the equation of the straight line b. Line b is perpendicular to line a, which means that the normal vector of line a serves as the direction vector of line b. Then the directing vector of the straight line b can be written as b → \u003d (3, 1). Let us write down the canonical equation of the straight line b, since we are also given the coordinates of the point М 1 through which the straight line b passes:
The final step is to determine the coordinates of the point of intersection of straight lines a and b. Let us pass from the canonical equations of the line b to its general equation:
x - 1 3 \u003d y 1 ⇔ 1 (x - 1) \u003d 3 y ⇔ x - 3 y - 1 \u003d 0
Let's compose a system of equations from the general equations of lines a and b and solve it:
3 x + y + 7 \u003d 0 x - 3 y - 1 \u003d 0 ⇔ y \u003d - 3 x - 7 x - 3 y - 1 \u003d 0 ⇔ y \u003d - 3 x - 7 x - 3 (- 3 x - 7 ) - 1 \u003d 0 ⇔ ⇔ y \u003d - 3 x - 7 x \u003d - 2 ⇔ y \u003d - 3 (- 2) - 7 x \u003d - 2 ⇔ y \u003d - 1 x \u003d - 2
Ultimately, we got the coordinates of the projection of the point М 1 (1, 0) onto the straight line 3 x + y + 7 \u003d 0: (- 2, - 1).
Answer: (- 2 , - 1) .
Let us consider in more detail the case when it is necessary to determine the coordinates of the projection of a given point on coordinate lines and lines parallel to them.
Let the coordinate lines O x and O y, as well as the point M 1 (x 1, y 1) be given. It is clear that the projection of a given point onto the coordinate line O x of the form y \u003d 0 will be a point with coordinates (x 1, 0). So the projection of a given point on the coordinate line O y will have coordinates 0, y 1.
Any arbitrary straight line parallel to the abscissa axis can be specified by an incomplete general equation B y + C \u003d 0 ⇔ y \u003d - C B, and a straight line parallel to the ordinate axis - A x + C \u003d 0 ⇔ x \u003d - C A.
Then the projections of the point М 1 (x 1, y 1) onto the lines y \u003d - C B and x \u003d - C A will be points with coordinates x 1, - C B and - C A, y 1.
Example 2
Determine the coordinates of the projection of the point M 1 (7, - 5) on the coordinate line O y, as well as on the line parallel to the line O y 2 y - 3 \u003d 0.
Decision
Let's write down the coordinates of the projection of a given point on the line O y: (0, - 5).
Write the equation of the line 2 y - 3 \u003d 0 as y \u003d 3 2. It becomes clear that the projection of a given point on the line y \u003d 3 2 will have coordinates 7, 3 2.
Answer: (0, - 5) and 7, 3 2.
Let a rectangular coordinate system O x y z, point М 1 (x 1, y 1, z 1) and straight line a be given in three-dimensional space. Let's find the coordinates of the projection of point M 1 on the line a.
Let us construct a plane α passing through point М 1 and perpendicular to the straight line a. The projection of a given point onto line a will be the point of intersection of line a and plane α. Based on this, we present an algorithm for finding the coordinates of the projection of the point M 1 (x 1, y 1, z 1) on the straight line a:
Let us write the equation of the straight line a (if it is not given). To solve this problem, you need to read the article on the equations of a straight line in space;
Let's compose the equation of the plane α passing through the point М 1 and perpendicular to the straight line a (see the article "Equation of the plane passing through the given point perpendicular to the given straight line");
Let's find the required coordinates of the projection of the point М 1 (x 1, y 1, z 1) on the straight line a - these will be the coordinates of the intersection point of the straight line α and the plane α (for help - the article "Coordinates of the point of intersection of a straight line and a plane").
Example 3
A rectangular coordinate system O x y z is given, and it contains point М 1 (0, 1, - 1) and straight line a. Line a corresponds to canonical equations of the form: x + 2 3 \u003d y - 6 - 4 \u003d z + 1 1. Determine the coordinates of the projection of point M 1 on line a.
Decision
We use the above algorithm. The equations of the straight line a are known, so we skip the first step of the algorithm. Let us write down the equation of the plane α. To do this, we define the coordinates of the normal vector of the plane α. From the given canonical equations of the straight line a, we select the coordinates of the direction vector of this straight line: (3, - 4, 1), which will be the normal vector of the plane α, perpendicular to the straight line a. Then n → \u003d (3, - 4, 1) is the normal vector of the plane α. Thus, the equation of the plane α will have the form:
3 (x - 0) - 4 (y - 1) + 1 (z - (- 1)) \u003d 0 ⇔ 3 x - 4 y + z + 5 \u003d 0
Now we will find the coordinates of the intersection point of the straight line a and the plane α, for this we use two methods:
- The specified canonical equations make it possible to obtain the equations of two intersecting planes defining the straight line a:
x + 2 3 \u003d y - 6 - 4 \u003d z + 1 1 ⇔ - 4 (x + 2) \u003d 3 (y - 6) 1 (x + 2) \u003d 3 (z + 1) 1 ( y - 6) \u003d - 4 (z + 1) ⇔ 4 x + 3 y - 10 \u003d 0 x - 3 z - 1 \u003d 0
To find the intersection points of the straight line 4 x + 3 y - 10 \u003d 0 x - 3 z - 1 \u003d 0 and the plane 3 x - 4 y + z + 5 \u003d 0, solve the system of equations:
4 x + 3 y - 10 \u003d 0 x - 3 z - 1 \u003d 0 3 x - 4 y + z + 5 \u003d 0 ⇔ 4 x + 3 y \u003d 10 x - 3 z \u003d 1 3 x - 4 y + z \u003d - five
In this case, we use Cramer's method, but it is possible to apply any convenient one:
∆ \u003d 4 3 0 1 0 - 3 3 - 4 1 \u003d - 78 ∆ x \u003d 10 3 0 1 0 - 3 - 5 - 4 1 \u003d - 78 ⇒ x \u003d ∆ x ∆ \u003d - 78 - 78 \u003d 1 ∆ y \u003d 4 10 0 1 1 - 3 3 - 5 1 \u003d - 156 ⇒ y \u003d ∆ y ∆ \u003d - 156 - 78 \u003d 2 ∆ z \u003d 4 3 10 1 0 1 3 - 4 - 5 \u003d 0 ⇒ z \u003d ∆ z ∆ \u003d 0 - 78 \u003d 0
Thus, the projection of a given point onto line a is a point with coordinates (1, 2, 0)
- Based on the specified canonical equations, it is easy to write the parametric equations of a straight line in space:
x + 2 3 \u003d y - 6 - 4 \u003d z + 1 1 ⇔ x \u003d - 2 + 3 λ y \u003d 6 - 4 λ z \u003d - 1 + λ
Substitute in the equation of the plane, which has the form 3 x - 4 y + z + 5 \u003d 0, instead of x, y and z their expressions through the parameter:
3 (- 2 + 3 λ) - 4 (6 - 4 λ) + (- 1 + λ) + 5 \u003d 0 ⇔ 26 λ \u003d 0 ⇔ λ \u003d 1
Let us calculate the required coordinates of the point of intersection of the straight line a and the plane α using the parametric equations of the straight line a at λ \u003d 1:
x \u003d - 2 + 3 1 y \u003d 6 - 4 1 z \u003d - 1 + 1 ⇔ x \u003d 1 y \u003d 2 z \u003d 0
Thus, the projection of a given point onto line a has coordinates (1, 2, 0)
Answer: (1 , 2 , 0)
Finally, we note that the projections of the point М 1 (x 1, y 1, z 1) onto the coordinate lines O x, O y and O z will be points with coordinates (x 1, 0, 0), (0, y 1, 0 ) and (0, 0, z 1), respectively.
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The projection of a point onto a straight line is quite simple, and when performing some operations, the zero approximation is calculated as the projection of a point onto a tangent line. Consider this particular case of a general problem.
Let a straight line be given
and point. We will assume that the line vector w has an arbitrary length. The straight line passes through the point at which the parameter t is equal to zero and has the direction of the vector w. It is required to find the projection of a point onto a straight line. This problem has only one solution. Let's construct a vector from a point of a straight line to a point and calculate the scalar product of this vector and a vector of a straight line w. In fig. 4.5.1 shows the direction vector of the straight line w, its starting point Co and the projection; set point. If we divide this dot product by the length of the vector w, then we get the length of the projection of the vector onto a straight line.
Figure: 4.5.1. Projection of a point onto a straight line
If we divide this scalar product by the square of the length of the vector w, then we obtain the length of the projection of the vector onto a straight line in units of the length of the vector w, i.e., we obtain the parameter t for the projection of a point onto a straight line.
Thus, the parameter of the projection of a point on a straight line and the radius vector of the projection; calculated by the formulas
(4.5.3)
If the length of the vector w is equal to one, then in (4.5.2) it is not required to perform division by The distance from the point to its projection onto the curve is generally calculated as the length of the vector. The distance from a point to its projection onto a straight line can be determined without calculating the projection of the point, but using the formula
Special cases.
The projection of a point onto analytical curves can also be found without using numerical methods. For example, to find the projection of a point onto a conical section, you need to translate the projected point into the local coordinate system conical section, project this point onto the plane of the conical section and find the parameter of the two-dimensional projection of the given point.
General case.
Suppose it is required to find all the projections of a point onto a curved line.Each desired point of the curve satisfies the equation
(4.5.5)
This equation contains one unknown quantity - the parameter t. As already mentioned, we will divide the solution of this problem into two stages. At the first stage, we will determine the zero approximations of the parameters of the projections of the point onto the curve, and at the second stage, we will find the exact values \u200b\u200bof the parameters of the curve that determine the projections of the given point onto the curved line
1-12. Projection of a point onto a plane or line
FORMULATION OF THE PROBLEM.Find the coordinates of the projection P "of the point P (^ PiURCHzp) onto the plane Ax + By - \\ - Cz - \\ - D \u003d O,
SOLUTION PLAN. The projection P "of point P onto a plane is the base of the perpendicular dropped from point P onto this plane.
1. We compose the equations of a straight line passing through point P perpendicular to this plane. For this, we take the normal vector of the plane as the directing vector of the straight line: a \u003d n \u003d (A, B, C). Then the canonical equations of the line have the form
X \u003d At \u200b\u200b- \\ - xp, y \u003d Bt - \\ - yp, Z \u003d z Ct - \\ - Zp.
3. Substituting x ^ y ^ z into the equation of the plane and solving it with respect to t, we find the value of the parameter t \u003d to, at which the intersection of the straight line and the plane occurs.
4. We substitute the found value ^ o into the parametric equations of the straight line and obtain the sought coordinates of the pointR".
COMMENT. The problem of finding the coordinates of the projection of a point onto a line is solved in a similar way.
EXAMPLE. Find the coordinates of the projection P "of the point P (1,2, -1) on the plane Zzh - 2 / 4-22: - 4 \u003d 0.
1. We compose the equations of a straight line passing through point P perpendicular to this plane. For this, we take the normal vector of the plane as the directing vector of the straight line: a \u003d n \u003d
Ch. 1. Ansiitic geometry |
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\u003d (3, -1.2). Then the canonical equations of the line have the form |
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U-2 _ z-hl |
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2. Find the coordinates of the POINT intersection P "of this line with a given |
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noah plane. We put |
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x- ~ 1 __ y-2 __ Z + 1 _ |
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Then the parametric equations of the straight line have the form |
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3. Substituting these expressions for x ^ y and z into the equation of the plane, we find the value of the parameter ^ at which the intersection of the straight line and the plane occurs:
3 (3t + 1) - l (-t + 2) + 2 (2t - 1) - 27 \u003d O \u003d\u003e to \u003d 2.
4. Substituting the found value to \u003d 2 into the parametric equations of the straight line, we obtain x0 \u003d 7, yo \u003d 0, ^ o \u003d 1.
Thus, the point of intersection of the straight line and the plane and, therefore, the projection of the point P onto the plane has coordinates (7,0,1).
Answer. The projection P "has coordinates (7,0,1).
STATEMENT OF TASKS. Find coordinates |
the projection of the point I ^ on the plane |
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4x + bu -f 4z - |
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2x + 6y "-2g - \\ - 11 |
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4 x - 5 2 / - g - 7 |
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f-f-42 / + Z2: 4-5 \u003d 0. |
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2x -h Yuu + lOz - |
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2х -МО2 / -f- lOz - |
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Answers. 1. (2.3 / 2.2). 2. (-3 / 2, -3 / 2, -1 / 2). 3. (2, -1 / 2, -3 / 2). 4. (-1 / 2,1,1). 5. (1, -1 / 2, -1 / 2). 6. (3/2, -1 / 2.0). 7. (1/2, -1, -1 / 2). 8. (1/2, -1 / 2.1 / 2). 9. (1/2, -1 / 2.1 / 2). 10. (1.1 / 2.0).
1.13. Symmetry about a line or plane
FORMULATION OF THE PROBLEM.Find the coordinates of the point Q, symmetric
SOLUTION PLAN. The sought point Q lies on a straight line perpendicular to the given one and intersecting it at point P ". Since point P" divides the segment PQ in half, the coordinates w, yd and ZQ of POINT Q are determined from the conditions
2 "^, ur" \u003d |
2 ~ ^. ^ R "\u003d |
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where xp, yp, zp |
The coordinates of the point P and xp ^^ ypf ^ zp / are the coordinates |
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its projection P "on this line.
1. Find the projection of the pointP on this line, i.e. point Р "(see problem 1.12). For this:
a) compose the equation of the plane passing through point P perpendicular to this straight line. As the normal vector n of this plane, we can take the direction vector of this line, i.e. n \u003d a \u003d (l ^ m ^ n). We get
1 (x - Xp) + t (y - UR) -f n (z - zp) \u003d 0;
b) find the coordinates of the point of intersection P "of this plane with a given straight line. For this, we write the equations of the straight line in parametric form
X \u003d H - \\ - xo, y \u003d mt - \\ - yo, Z \u003d nt - \\ - ZQ.
Substituting x ^ y ^ z into the equation of the plane and solving it with respect to t, we find the value of the parameter t \u003d to, at which the intersection of the straight line and the plane occurs;
c) the found value to is substituted into the parametric equations of the straight line and we obtain the sought coordinates of the point P ".
2. Coordinates of point Q, symmetrical point Р with respect to a given straight line, we determine from conditions (1). We get
XQ \u003d 2хр / - Хр, yq \u003d 2yr "- ur, ZQ \u003d 22; р / - zp.
COMMENT. The problem of finding the coordinates of a point symmetric to a given one with respect to the plane is solved in a similar way.
EXAMPLE. Find the coordinates of the point Q, symmetric to the point P (2, -1,2) relative to the straight line
X - 1 _ y __ Z - \\ - 1
DECISION.
1. Find the projection of the pointP on this line, i.e. point P ". For this:
a) compose the equation of the plane passing through point P perpendicular to this straight line. As the normal vector n of this plane, we can take the direction vector of this straight line: n \u003d a \u003d (1,0, -2). Then
Substituting these expressions for x, y, and z into the equation of the plane, we find the value of the parameter t at which the intersection of the straight line and the plane occurs: to \u003d -1;
c) substituting the found value to \u003d -1 into the parametric equations of the straight line, we obtain
zp / \u003d 0, r / p / \u003d 0, zpr \u003d 1.
Thus, the point of intersection of the straight line and the plane and, consequently, the projection of the point P onto the straight line is P "(0,0,1).
2. The coordinates of the point Q, symmetric to the point P relative to this straight line, are determined from conditions (1):
XQ \u003d 2хр "- Хр \u003d -2,
VQ \u003d 2ur / - 2 / p \u003d 1,
ZQ \u003d 2zpf - zp \u003d 0.
Answer. Point Q has coordinates (-2,1,0).
Conditions of the TASKS. Find the coordinates of a point symmetric to the point P from the support of a given straight line.
X - 1 |
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