In this article we will find answers to questions about how to create a projection of a point onto a plane and how to determine the coordinates of this projection. In the theoretical part we will rely on the concept of projection. We will define the terms and provide information with illustrations. Let's consolidate the acquired knowledge by solving examples.

Projection, types of projection

For the convenience of viewing spatial figures, drawings depicting these figures are used.

Definition 1

Projection of a figure onto a plane– drawing of a spatial figure.

Obviously, there are a number of rules used to construct a projection.

Definition 2

Projection– the process of constructing a drawing of a spatial figure on a plane using construction rules.

Projection plane- this is the plane in which the image is constructed.

The use of certain rules determines the type of projection: central or parallel.

A special case of parallel projection is perpendicular projection or orthogonal: in geometry it is mainly used. For this reason, the adjective “perpendicular” itself is often omitted in speech: in geometry they simply say “projection of a figure” and by this they mean constructing a projection using the method of perpendicular projection. In special cases, of course, something else may be agreed.

Let us note the fact that the projection of a figure onto a plane is essentially a projection of all points of this figure. Therefore, in order to be able to study a spatial figure in a drawing, it is necessary to acquire the basic skill of projecting a point onto a plane. What we will talk about below.

Let us recall that most often in geometry, when speaking about projection onto a plane, they mean the use of a perpendicular projection.

Let us make constructions that will give us the opportunity to obtain a definition of the projection of a point onto a plane.

Let's say a three-dimensional space is given, and in it there is a plane α and a point M 1 that does not belong to the plane α. Draw a straight line through the given point M A perpendicular to a given plane α. We denote the point of intersection of straight line a and plane α as H 1; by construction, it will serve as the base of a perpendicular dropped from point M 1 to plane α.

If a point M 2 is given, belonging to a given plane α, then M 2 will serve as a projection of itself onto the plane α.

Definition 3

- this is either the point itself (if it belongs to a given plane), or the base of a perpendicular dropped from a given point to a given plane.

Finding the coordinates of the projection of a point onto a plane, examples

Let the following be given in three-dimensional space: a rectangular coordinate system O x y z, a plane α, a point M 1 (x 1, y 1, z 1). It is necessary to find the coordinates of the projection of point M 1 onto a given plane.

The solution follows obviously from the definition given above of the projection of a point onto a plane.

Let us denote the projection of the point M 1 onto the plane α as H 1 . According to the definition, H 1 is the intersection point of a given plane α and a straight line a drawn through the point M 1 (perpendicular to the plane). Those. The coordinates of the projection of point M1 that we need are the coordinates of the point of intersection of straight line a and plane α.

Thus, to find the coordinates of the projection of a point onto a plane it is necessary:

Obtain the equation of the plane α (if it is not specified). An article about the types of plane equations will help you here;

Determine the equation of a line a passing through point M 1 and perpendicular to the plane α (study the topic about the equation of a line passing through a given point perpendicular to a given plane);

Find the coordinates of the point of intersection of the straight line a and the plane α (article - finding the coordinates of the point of intersection of the plane and the line). The data obtained will be the coordinates we need for the projection of point M 1 onto the plane α.

Let's look at the theory with practical examples.

Example 1

Determine the coordinates of the projection of point M 1 (- 2, 4, 4) onto the plane 2 x – 3 y + z - 2 = 0.

Solution

As we see, the equation of the plane is given to us, i.e. there is no need to compile it.

Let us write down the canonical equations of a straight line a passing through the point M 1 and perpendicular to the given plane. For these purposes, we determine the coordinates of the directing vector of the straight line a. Since line a is perpendicular to a given plane, the direction vector of line a is the normal vector of the plane 2 x - 3 y + z - 2 = 0. Thus, a → = (2, - 3, 1) – direction vector of straight line a.

Now let’s compose the canonical equations of a line in space passing through the point M 1 (- 2, 4, 4) and having a direction vector a → = (2 , - 3 , 1) :

x + 2 2 = y - 4 - 3 = z - 4 1

To find the required coordinates, the next step is to determine the coordinates of the intersection point of the straight line x + 2 2 = y - 4 - 3 = z - 4 1 and the plane 2 x - 3 y + z - 2 = 0 . For these purposes, we move from the canonical equations to the equations of two intersecting planes:

x + 2 2 = y - 4 - 3 = z - 4 1 ⇔ - 3 · (x + 2) = 2 · (y - 4) 1 · (x + 2) = 2 · (z - 4) 1 · ( y - 4) = - 3 (z + 4) ⇔ 3 x + 2 y - 2 = 0 x - 2 z + 10 = 0

Let's create a system of equations:

3 x + 2 y - 2 = 0 x - 2 z + 10 = 0 2 x - 3 y + z - 2 = 0 ⇔ 3 x + 2 y = 2 x - 2 z = - 10 2 x - 3 y + z = 2

And let's solve it using Cramer's method:

∆ = 3 2 0 1 0 - 2 2 - 3 1 = - 28 ∆ x = 2 2 0 - 10 0 - 2 2 - 3 1 = 0 ⇒ x = ∆ x ∆ = 0 - 28 = 0 ∆ y = 3 2 0 1 - 10 - 2 2 2 1 = - 28 ⇒ y = ∆ y ∆ = - 28 - 28 = 1 ∆ z = 3 2 2 1 0 - 10 2 - 3 2 = - 140 ⇒ z = ∆ z ∆ = - 140 - 28 = 5

Thus, the required coordinates of a given point M 1 on a given plane α will be: (0, 1, 5).

Answer: (0 , 1 , 5) .

Example 2

In a rectangular coordinate system O x y z of three-dimensional space, points A (0, 0, 2) are given; B (2, - 1, 0); C (4, 1, 1) and M 1 (-1, -2, 5). It is necessary to find the coordinates of the projection M 1 onto the plane A B C

Solution

First of all, we write down the equation of a plane passing through three given points:

x - 0 y - 0 z - 0 2 - 0 - 1 - 0 0 - 2 4 - 0 1 - 0 1 - 2 = 0 ⇔ x y z - 2 2 - 1 - 2 4 1 - 1 = 0 ⇔ ⇔ 3 x - 6 y + 6 z - 12 = 0 ⇔ x - 2 y + 2 z - 4 = 0

Let us write down the parametric equations of the line a, which will pass through the point M 1 perpendicular to the plane A B C. The plane x – 2 y + 2 z – 4 = 0 has a normal vector with coordinates (1, - 2, 2), i.e. vector a → = (1, - 2, 2) – direction vector of straight line a.

Now, having the coordinates of the point of the line M 1 and the coordinates of the direction vector of this line, we write the parametric equations of the line in space:

Then we determine the coordinates of the intersection point of the plane x – 2 y + 2 z – 4 = 0 and the straight line

x = - 1 + λ y = - 2 - 2 λ z = 5 + 2 λ

To do this, we substitute into the equation of the plane:

x = - 1 + λ, y = - 2 - 2 λ, z = 5 + 2 λ

Now, using the parametric equations x = - 1 + λ y = - 2 - 2 · λ z = 5 + 2 · λ, we find the values ​​of the variables x, y and z for λ = - 1: x = - 1 + (- 1) y = - 2 - 2 · (- 1) z = 5 + 2 · (- 1) ⇔ x = - 2 y = 0 z = 3

Thus, the projection of point M 1 onto the plane A B C will have coordinates (- 2, 0, 3).

Answer: (- 2 , 0 , 3) .

Let us separately dwell on the issue of finding the coordinates of the projection of a point onto coordinate planes and planes that are parallel to coordinate planes.

Let points M 1 (x 1, y 1, z 1) and coordinate planes O x y, O x z and O y z be given. The coordinates of the projection of this point onto these planes will be, respectively: (x 1, y 1, 0), (x 1, 0, z 1) and (0, y 1, z 1). Let us also consider planes parallel to the given coordinate planes:

C z + D = 0 ⇔ z = - D C , B y + D = 0 ⇔ y = - D B

And the projections of a given point M 1 onto these planes will be points with coordinates x 1, y 1, - D C, x 1, - D B, z 1 and - D A, y 1, z 1.

Let us demonstrate how this result was obtained.

As an example, let's define the projection of point M 1 (x 1, y 1, z 1) onto the plane A x + D = 0. The remaining cases are similar.

The given plane is parallel to the coordinate plane O y z and i → = (1, 0, 0) is its normal vector. The same vector serves as the direction vector of the line perpendicular to the O y z plane. Then the parametric equations of a straight line drawn through the point M 1 and perpendicular to a given plane will have the form:

x = x 1 + λ y = y 1 z = z 1

Let's find the coordinates of the intersection point of this line and the given plane. Let us first substitute the equalities into the equation A x + D = 0: x = x 1 + λ , y = y 1 , z = z 1 and get: A · (x 1 + λ) + D = 0 ⇒ λ = - D A - x 1

Then we calculate the required coordinates using the parametric equations of the straight line with λ = - D A - x 1:

x = x 1 + - D A - x 1 y = y 1 z = z 1 ⇔ x = - D A y = y 1 z = z 1

That is, the projection of point M 1 (x 1, y 1, z 1) onto the plane will be a point with coordinates - D A, y 1, z 1.

Example 2

It is necessary to determine the coordinates of the projection of point M 1 (- 6, 0, 1 2) onto the coordinate plane O x y and onto the plane 2 y - 3 = 0.

Solution

The coordinate plane O x y will correspond to the incomplete general equation of the plane z = 0. The projection of point M 1 onto the plane z = 0 will have coordinates (- 6, 0, 0).

The plane equation 2 y - 3 = 0 can be written as y = 3 2 2. Now just write down the coordinates of the projection of point M 1 (- 6, 0, 1 2) onto the plane y = 3 2 2:

6 , 3 2 2 , 1 2

Answer:(- 6 , 0 , 0) and - 6 , 3 2 2 , 1 2

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Find the acute angle between the diagonals of a parallelogram constructed using vectors

5) Determine the coordinates of vector c, directed along the bisector of the angle between vectors a and b, if vector c = 3 roots of 42. a=(2;-3;6), b=(-1;2;-2)

Let's find the unit vector e_a codirectional with a:

similarly e_b = b/|b|,

then the desired vector will be directed in the same way as the vector sum e_a+e_b, because (e_a+e_b) is the diagonal of a rhombus, which is bisector of its angle.

Let us denote (e_a+e_b)=d,

Let's find a unit vector that is directed along the bisector: e_c = d/|d|

If |c| = 3*sqrt(42), then c = |c|*e_c. That's all.

Find the linear relationship between these four non-coplanar vectors: p=a+b; q=b-c; r=a-b+c; s=b+(1/2)*c

From the first three equalities, try expressing `a,b,c` in terms of `p,q,r` (start by adding the second and third equations). Then replace `b` and `c` in the last equation with the expressions you found in terms of `p,q,r`.

13) Find the equation of the plane passing through points A(2, -1, 4) and B(3, 2, -1) perpendicular to the plane x + y + 2z – 3 = 0. The required equation of the plane has the form: Ax + By + Cz + D = 0, the normal vector to this plane (A, B, C). The vector (1, 3, -5) belongs to the plane. The plane given to us, perpendicular to the desired one, has a normal vector (1, 1, 2). Because points A and B belong to both planes, and the planes are mutually perpendicular, then the normal vector is (11, -7, -2). Because point A belongs to the desired plane, then its coordinates must satisfy the equation of this plane, i.e. 11×2 + 7×1 - 2×4 + D = 0; D = -21. In total, we get the equation of the plane: 11x - 7y – 2z – 21 = 0.

14) The equation of a plane passing through a line parallel to a vector.

Let the desired plane pass through the straight line (x-x1)/a1 = (y-y1)/b1 = (z-z1)/c1 parallel to the straight line (x-x2)/a2 = (y-y2)/b2 = (z -z2)/c2 .

Then the normal vector of the plane is the vector product of the direction vectors of these lines:

Let the coordinates of the vector product be (A;B;C). The desired plane passes through the point (x1;y1;z1). The normal vector and the point through which the plane passes uniquely determine the equation of the desired plane:

A·(x-x1) + B·(y-y1) + C·(z-z1) = 0

17) Find the equation of the line passing through the point A(5, -1) perpendicular to the line 3x - 7y + 14 = 0.

18) Write an equation for a straight line passing through point M perpendicular to the given plane M(4,3,1) x+3y+5z-42=0

(x - x0) / n = (y - y0) / m = (z - z0) / p

M(x0,y0,z0) - your point M(4,3,1)

(n, m, p) - the directing vector of the line, also known as the normal vector for a given surface (1, 3, 5) (coefficients for the variables x, y, z in the plane equation)

Find the projection of a point onto a plane

Point M(1,-3,2), plane 2x+5y-3z-19=0

Studying the properties of figures in space and on a plane is impossible without knowing the distances between a point and such geometric objects as a straight line and a plane. In this article we will show how to find these distances by considering the projection of a point onto a plane and onto a straight line.

Equation of a straight line for two-dimensional and three-dimensional spaces

Calculation of the distances of a point to a straight line and a plane is carried out using its projection onto these objects. To be able to find these projections, you should know in what form the equations for lines and planes are given. Let's start with the first ones.

A straight line is a collection of points, each of which can be obtained from the previous one by transferring it to vectors parallel to each other. For example, there is a point M and N. The vector MN¯ connecting them takes M to N. There is also a third point P. If the vector MP¯ or NP¯ is parallel to MN¯, then all three points lie on the same line and form it.

Depending on the dimension of space, the equation defining the line can change its form. Thus, the well-known linear dependence of the coordinate y on x in space describes a plane that is parallel to the third axis z. In this regard, in this article we will consider only the vector equation for the line. It has the same appearance for a plane and three-dimensional space.

In space, a straight line can be defined by the following expression:

(x; y; z) = (x 0 ; y 0 ; z 0) + α*(a; b; c)

Here, the coordinate values ​​with zero indices correspond to a certain point belonging to the line, u¯(a; b; c) are the coordinates of the direction vector that lies on this line, α is an arbitrary real number, by changing which you can get all the points of the line. This equation is called a vector equation.

The above equation is often written in expanded form:

In a similar way, you can write the equation for a line located in a plane, that is, in two-dimensional space:

(x; y) = (x 0 ; y 0) + α*(a; b);

Plane equation

To be able to find the distance from a point to projection planes, you need to know how a plane is defined. Just like a straight line, it can be represented in several ways. Here we will consider only one: the general equation.

Suppose that the point M(x 0 ; y 0 ; z 0) belongs to the plane, and the vector n¯(A; B; C) is perpendicular to it, then for all points (x; y; z) of the plane the equality will be valid:

A*x + B*y + C*z + D = 0, where D = -1*(A*x 0 + B*y 0 + C*z 0)

It should be remembered that in this general plane equation, the coefficients A, B and C are the coordinates of the vector normal to the plane.

Calculation of distances by coordinates

Before moving on to considering projections onto the plane of a point and onto a straight line, it is worth recalling how to calculate the distance between two known points.

Let there be two spatial points:

A 1 (x 1 ; y 1 ; z 1) and A 2 (x 2 ; y 2 ​​; z 2)

Then the distance between them is calculated by the formula:

A 1 A 2 = √((x 2 -x 1) 2 +(y 2 -y 1) 2 +(z 2 -z 1) 2)

Using this expression, the length of the vector A 1 A 2 ¯ is also determined.

For the case on the plane, when two points are defined by just a pair of coordinates, we can write a similar equality without the presence of a term with z in it:

A 1 A 2 = √((x 2 -x 1) 2 +(y 2 -y 1) 2)

Now let us consider various cases of projection on a plane of a point onto a straight line and onto a plane in space.

Point, line and distance between them

Suppose there is a point and a line:

P 2 (x 1 ; y 1);

(x; y) = (x 0 ; y 0) + α*(a; b)

The distance between these geometric objects will correspond to the length of the vector, the beginning of which lies at the point P 2, and the end is at a point P on the specified line for which the vector P 2 P ¯ is perpendicular to this line. Point P is called the projection of point P 2 onto the line under consideration.

Below is a figure that shows point P 2, its distance d to the line, as well as the direction vector v 1 ¯. Also, an arbitrary point P 1 is selected on the line and a vector is drawn from it to P 2. Point P here coincides with the place where the perpendicular intersects the line.

It can be seen that the orange and red arrows form a parallelogram, the sides of which are the vectors P 1 P 2 ¯ and v 1 ¯, and the height is d. It is known from geometry that to find the height of a parallelogram, its area should be divided by the length of the base on which the perpendicular is lowered. Since the area of ​​a parallelogram is calculated as the vector product of its sides, we obtain a formula for calculating d:

d = ||/|v 1 ¯|

All vectors and coordinates of points in this expression are known, so you can use it without performing any transformations.

This problem could have been solved differently. To do this, write two equations:

• the scalar product of P 2 P ¯ by v 1 ¯ must be equal to zero, since these vectors are mutually perpendicular;
• the coordinates of point P must satisfy the equation of the line.

These equations are enough to find the coordinates P, and then the length d using the formula given in the previous paragraph.

The task of finding the distance between a line and a point

We will show how to use this theoretical information to solve a specific problem. Suppose the following point and line are known:

(x; y) = (3; 1) - α*(0; 2)

It is necessary to find the points of projection onto a straight line on the plane, as well as the distance from M to the straight line.

Let us denote the projection to be found by the point M 1 (x 1 ; y 1). Let's solve this problem in two ways, described in the previous paragraph.

Method 1. The direction vector v 1 ¯ has coordinates (0; 2). To construct a parallelogram, we select some point belonging to the line. For example, a point with coordinates (3; 1). Then the vector of the second side of the parallelogram will have the coordinates:

(5; -3) - (3; 1) = (2; -4)

Now you need to calculate the product of vectors defining the sides of the parallelogram:

We substitute this value into the formula and get the distance d from M to the straight line:

Method 2. Now let’s find in another way not only the distance, but also the coordinates of the projection M onto the straight line, as required by the condition of the problem. As mentioned above, to solve the problem it is necessary to create a system of equations. It will look like:

(x 1 -5)*0+(y 1 +3)*2 = 0;

(x 1 ; y 1) = (3; 1)-α*(0; 2)

Let's solve this system:

The projection of the original coordinate point has M 1 (3; -3). Then the required distance is:

d = |MM 1 ¯| = √(4+0) = 2

As you can see, both solution methods gave the same result, which indicates the correctness of the mathematical operations performed.

Projection of a point onto a plane

Now let's consider what the projection of a point given in space onto a certain plane is. It is easy to guess that this projection is also a point that, together with the original one, forms a vector perpendicular to the plane.

Let us assume that the projection onto the plane of point M has the following coordinates:

The plane itself is described by the equation:

A*x + B*y + C*z + D = 0

Based on these data, we can create an equation for a line intersecting the plane at a right angle and passing through M and M 1:

(x; y; z) = (x 0 ; y 0 ; z 0) + α*(A; B; C)

Here, the variables with zero indices are the coordinates of point M. The position on the plane of point M 1 can be calculated based on the fact that its coordinates must satisfy both written equations. If these equations are not enough to solve the problem, then you can use the condition of parallelism between MM 1 ¯ and the guide vector for a given plane.

Obviously, the projection of a point belonging to the plane coincides with itself, and the corresponding distance is zero.

Problem with a point and a plane

Let a point M(1; -1; 3) and a plane be given, which is described by the following general equation:

It is necessary to calculate the coordinates of the projection onto the plane of the point and calculate the distance between these geometric objects.

First, let's construct the equation of a straight line passing through M and perpendicular to the indicated plane. It looks like:

(x; y; z) = (1; -1; 3) + α*(-1; 3; -2)

Let us denote the point where this line intersects the plane as M 1 . The equalities for the plane and the line must be satisfied if the coordinates M 1 are substituted into them. Writing the equation of the line explicitly, we obtain the following four equalities:

X 1 + 3*y 1 -2*z 1 + 4 = 0;

y 1 = -1 + 3*α;

From the last equality we get the parameter α, then we substitute it into the penultimate and second expressions, we get:

y 1 = -1 + 3*(3-z 1)/2 = -3/2*z 1 + 3.5;

x 1 = 1 - (3-z 1)/2 = 1/2*z 1 - 1/2

We substitute the expression for y 1 and x 1 into the equation for the plane, we have:

1*(1/2*z 1 - 1/2) + 3*(-3/2*z 1 + 3.5) -2*z 1 + 4 = 0

Where do we get it from:

y 1 = -3/2*15/7 + 3.5 = 2/7;

x 1 = 1/2*15/7 - 1/2 = 4/7

We have determined that the projection of point M onto a given plane corresponds to the coordinates (4/7; 2/7; 15/7).

Now let's calculate the distance |MM 1 ¯|. The coordinates of the corresponding vector are:

MM 1 ¯(-3/7; 9/7; -6/7)

The required distance is:

d = |MM 1 ¯| = √126/7 ≈ 1.6

Three point projection

During the production of drawings, it is often necessary to obtain projections of sections onto mutually perpendicular three planes. Therefore, it is useful to consider what the projections of a certain point M with coordinates (x 0 ; y 0 ; z 0) onto three coordinate planes will be equal to.

It is not difficult to show that the xy plane is described by the equation z = 0, the xz plane corresponds to the expression y = 0, and the remaining yz plane is denoted by x = 0. It is not difficult to guess that the projections of a point on 3 planes will be equal:

for x = 0: (0; y 0; z 0);

for y = 0: (x 0 ; 0 ; z 0);

for z = 0: (x 0 ; y 0 ; 0)

Where is it important to know the projection of a point and its distance to planes?

Determining the position of the projection of points onto a given plane is important when finding quantities such as surface area and volume for inclined prisms and pyramids. For example, the distance from the top of the pyramid to the base plane is the height. The latter is included in the formula for the volume of this figure.

The considered formulas and methods for determining projections and distances from a point to a straight line and plane are quite simple. It is only important to remember the corresponding forms of the equations of a plane and a straight line, as well as to have good spatial imagination in order to successfully apply them.

When solving geometric problems in space, the problem of determining the distance between a plane and a point often arises. In some cases this is necessary for a comprehensive solution. This value can be calculated by finding the projection onto the plane of the point. Let's look at this issue in more detail in the article.

Equation to describe a plane

Before moving on to consider the question of how to find the projection of a point onto a plane, you should become familiar with the types of equations that define the latter in three-dimensional space. More details below.

A general equation that defines all points that belong to a given plane is the following:

A*x + B*y + C*z + D = 0.

The first three coefficients are the coordinates of the vector, which is called the guide for the plane. It coincides with the normal for it, that is, it is perpendicular. This vector is denoted by n¯(A; B; C). The free coefficient D is uniquely determined from knowledge of the coordinates of any point belonging to the plane.

The concept of point projection and its calculation

Suppose that some point P(x 1 ; y 1 ; z 1) and a plane are given. It is defined by the equation in general form. If we draw a perpendicular line from P to a given plane, then it is obvious that it will intersect the latter at one specific point Q (x 2 ; y 2 ​​; z 2). Q is called the projection of P onto the plane under consideration. The length of the segment PQ is called the distance from point P to the plane. Thus PQ itself is perpendicular to the plane.

How can you find the coordinates of the projection of a point onto a plane? It's not difficult to do this. First, you need to create an equation for a straight line that will be perpendicular to the plane. Point P will belong to it. Since the normal vector n¯(A; B; C) of this line must be parallel, the equation for it in the appropriate form will be written as follows:

(x; y; z) = (x 1; y 1; z 1) + λ*(A; B; C).

Where λ is a real number, which is usually called a parameter of the equation. By changing it, you can get any point on the line.

After the vector equation for a line perpendicular to the plane has been written, it is necessary to find the common intersection point for the geometric objects under consideration. Its coordinates will be the projection P. Since they must satisfy both equalities (for the line and for the plane), the problem is reduced to solving the corresponding system of linear equations.

The concept of projection is often used in the study of drawings. They depict lateral and horizontal projections of the part on the zy, zx, and xy planes.

Calculating the distance from a plane to a point

As noted above, knowing the coordinates of the projection onto the plane of a point allows you to determine the distance between them. Using the notation introduced in the previous paragraph, we find that the required distance is equal to the length of the segment PQ. To calculate it, it is enough to find the coordinates of the vector PQ¯, and then calculate its module using the known formula. The final expression for d distance between the P point and the plane takes the form:

d = |PQ¯| = √((x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2).

The resulting value of d is presented in units in which the current Cartesian xyz coordinate system is specified.

Sample task

Let's say there is a point N(0; -2; 3) and a plane, which is described by the following equation:

You need to find the projection points onto the plane and calculate the distance between them.

First of all, let's create an equation for a straight line that intersects the plane at an angle of 90 o. We have:

(x; y; z) = (0; -2; 3) + λ*(2; -1; 1).

Writing this equality explicitly, we arrive at the following system of equations:

Substituting the coordinate values ​​from the first three equalities into the fourth, we obtain the value λ, which determines the coordinates of the common point of the line and the plane:

2*(2*λ) - (-2 - λ) + λ + 3 + 4 = 0 =>

6*λ + 9 = 0 =>

λ = 9/6 = 3/2 = 1.5.

Let's substitute the found parameter into and find the coordinates of the projection of the starting point onto the plane:

(x; y; z) = (0; -2; 3) + 1.5*(2; -1; 1) = (3; -3.5; 4.5).

To calculate the distance between the geometric objects specified in the problem statement, we apply the formula for d:

d = √((3 - 0) 2 + (-3.5 + 2) 2 + (4.5 - 3) 2) = 3.674.

In this problem we showed how to find the projection of a point onto an arbitrary plane and how to calculate the distance between them.