In this article we will answer the question: “How to find the coordinates of the point of intersection of a line and a plane if the equations defining the line and the plane are given”? Let's start with the concept of the point of intersection of a line and a plane. Next, we will show two ways to find the coordinates of the point of intersection of a line and a plane. To consolidate the material, consider detailed solutions to the examples.
Page navigation.
The point of intersection of a line and a plane - definition.
There are three possible options for the relative position of the straight line and the plane in space:
- a straight line lies in a plane;
- a straight line is parallel to a plane;
- a straight line intersects a plane.
We are interested in the third case. Let us recall what the phrase “a straight line and a plane intersect” means. A line and a plane are said to intersect if they have only one common point. This common point of intersecting line and plane is called the point of intersection of a line and a plane.
Let's give a graphic illustration.
Finding the coordinates of the intersection point of a line and a plane.
Let us introduce Oxyz in three-dimensional space. Now, each line corresponds to a straight line equation of some type (the article is devoted to them: types of equations of a line in space), each plane corresponds to an equation of a plane (you can read the article: types of equations of a plane), and each point corresponds to an ordered triple of numbers - the coordinates of the point. Further presentation implies knowledge of all types of equations of a line in space and all types of equations of a plane, as well as the ability to move from one type of equations to another. But don’t be alarmed, throughout the text we will provide links to the necessary theory.
Let's first analyze in detail the problem, the solution of which we can obtain based on determining the point of intersection of a straight line and a plane. This task will prepare us for finding the coordinates of the point of intersection of a line and a plane.
Example.
Is the point M 0 with coordinates the intersection point of the line 
and planes 
.
Solution.
We know that if a point belongs to a certain line, then the coordinates of the point satisfy the equations of the line. Similarly, if a point lies in a certain plane, then the coordinates of the point satisfy the equation of this plane. By definition, the intersection point of a line and a plane is a common point of the line and the plane, then the coordinates of the intersection point satisfy both the equations of the line and the equation of the plane.
Thus, to solve the problem, we should substitute the coordinates of the point M 0 into the given equations of the straight line and into the equation of the plane. If in this case all the equations turn into correct equalities, then the point M 0 is the point of intersection of the given line and plane, otherwise the point M 0 is not the point of intersection of the line and the plane.
Substitute the coordinates of the point 
:
All equations turned into correct equalities, therefore, the point M 0 simultaneously belongs to the straight line and planes , that is, M 0 is the intersection point of the indicated straight line and plane.
Answer:
Yes, period 
is the point of intersection of the line and planes .
So, the coordinates of the point of intersection of a line and a plane satisfy both the equations of the line and the equation of the plane. We will use this fact when finding the coordinates of the point of intersection of a line and a plane.
The first method is to find the coordinates of the intersection point of a line and a plane.
Let a straight line a and a plane be given in the rectangular coordinate system Oxyz, and it is known that straight a and the plane intersect at point M 0 .
The required coordinates of the point of intersection of the straight line a and the plane, as we have already said, satisfy both the equations of the straight line a and the equation of the plane, therefore, they can be found as a solution to a system of linear equations of the form 
. This is true, since solving a system of linear equations turns each equation of the system into an identity.
Note that with this formulation of the problem, we actually find the coordinates of the intersection point of three planes specified by the equations , and .
Let's solve an example to consolidate the material.
Example.
A straight line given by the equations of two intersecting planes as 
, intersects the plane 
. Find the coordinates of the point of intersection of the line and the plane.
Solution.
We obtain the required coordinates of the point of intersection of the line and the plane by solving a system of equations of the form 
. In this case, we will rely on the information in the article.
First, let's rewrite the system of equations in the form 
and calculate the determinant of the main matrix of the system (if necessary, refer to the article):
The determinant of the main matrix of the system is nonzero, so the system of equations has a unique solution. To find it, you can use any method. We use : 
This is how we got the coordinates of the point of intersection of the line and the plane (-2, 1, 1).
Answer:
(-2, 1, 1) .
It should be noted that the system of equations has a unique solution if the line a defined by the equations 
, and the plane defined by the equation intersect. If straight line a lies in the plane, then the system has an infinite number of solutions. If straight line a is parallel to the plane, then the system of equations has no solutions.
Example.
Find the point of intersection of the line 
and planes 
, if possible.
Solution.
The “if possible” clause means that the line and the plane may not intersect.
. If this system of equations has a unique solution, then it will give us the desired coordinates of the point of intersection of the line and the plane. If this system has no solutions or has infinitely many solutions, then finding the coordinates of the intersection point is out of the question, since the straight line is either parallel to the plane or lies in this plane.
The main matrix of the system has the form 
, and the extended matrix is 
. Let's define A and the rank of matrix T: 
. That is, the rank of the main matrix is equal to the rank of the extended matrix of the system and is equal to two. Therefore, based on the Kronecker-Capelli theorem, it can be argued that the system of equations has an infinite number of solutions.
Thus, straight lies in a plane , and we cannot talk about finding the coordinates of the point of intersection of the line and the plane.
Answer:
It is impossible to find the coordinates of the intersection point of a line and a plane.
Example.
If straight 
intersects the plane, then find the coordinates of the point of their intersection.
Solution.
Let's create a system from the given equations 
. To find its solution we use . The Gauss method will allow us not only to determine whether the written system of equations has one solution, an infinite number of solutions, or does not have any solutions, but also to find solutions if they exist. 
The last equation of the system after the direct passage of the Gauss method became an incorrect equality, therefore, the system of equations has no solutions. From this we conclude that the straight line and the plane do not have common points. Thus, we cannot talk about finding the coordinates of their intersection point.
Answer:
The line is parallel to the plane and they do not have an intersection point.
Note that if line a corresponds to parametric equations of a line in space or canonical equations of a line in space, then it is possible to obtain the equations of two intersecting planes that define line a, and then find the coordinates of the point of intersection of line a and the plane in a parsed way. However, it is easier to use another method, which we now describe.
The line of intersection of two planes is a straight line. Let us first consider the special case (Fig. 3.9), when one of the intersecting planes is parallel to the horizontal plane of projections (α π 1, f 0 α X). In this case, the intersection line a, belonging to the plane α, will also be parallel to the plane π 1, (Fig. 3.9. a), i.e., it will coincide with the horizontal of the intersecting planes (a ≡ h).
If one of the planes is parallel to the frontal plane of projections (Fig. 3.9. b), then the intersection line a belonging to this plane will be parallel to the plane π 2 and will coincide with the frontal of the intersecting planes (a ≡ f).
.
.
Rice. 3.9. A special case of intersection of a general plane with the planes: a - horizontal level; b - frontal level
An example of constructing the point of intersection (K) of straight line a (AB) with plane α (DEF) is shown in Fig. 3.10. To do this, straight line a is enclosed in an arbitrary plane β and the intersection line of planes α and β is determined.
In the example under consideration, straight lines AB and MN belong to the same plane β and intersect at point K, and since straight line MN belongs to a given plane α (DEF), point K is also the point of intersection of straight line a (AB) with plane α. (Fig. 3.11).
.
Rice. 3.10. Constructing the point of intersection of a line and a plane
To solve such a problem in a complex drawing, you must be able to find the point of intersection of a straight line in general position with a plane in general position.
Let's consider an example of finding the point of intersection of straight line AB with the plane of triangle DEF shown in Fig. 3.11.
To find the intersection point through the frontal projection of straight line A 2 B 2, a frontally projecting plane β was drawn which intersected the triangle at points M and N. On the frontal projection plane (π 2), these points are represented by projections M 2, N 2. From the condition of belonging to a straight plane on the horizontal plane of projections (π 1), horizontal projections of the resulting points M 1 N 1 are found. At the intersection of the horizontal projections of lines A 1 B 1 and M 1 N 1, a horizontal projection of their intersection point (K 1) is formed. According to the line of communication and the conditions of membership on the frontal plane of projections, there is a frontal projection of the intersection point (K 2).
.
Rice. 3.11. An example of determining the intersection point of a line and a plane
The visibility of segment AB relative to triangle DEF is determined by the competing point method.
On the plane π 2 two points NEF and 1AB are considered. From the horizontal projections of these points, it can be established that point N is located closer to the observer (Y N >Y 1) than point 1 (the direction of the line of sight is parallel to S). Consequently, the straight line AB, i.e. part of the straight line AB (K 1) is covered by the plane DEF on the plane π 2 (its projection K 2 1 2 is shown by the dashed line). Visibility on the π 1 plane is established similarly.
Questions for self-control
1) What is the essence of the competing point method?
2) What properties of a straight line do you know?
3) What is the algorithm for determining the intersection point of a line and a plane?
4) What tasks are called positional?
5) Formulate the conditions for belonging to a straight plane.
We bring to your attention magazines published by the publishing house "Academy of Natural Sciences"
Let's consider the cases: 1) when the projecting surface is intersected by the projecting plane; 2) when the projecting surface intersects a general plane. In both cases, to construct a section on the diagram, we use the projecting figure algorithm (algorithm No. 1). In the first case, the drawing already knows...(Descriptive geometry)
Constructing a line of intersection of two planes at the points of intersection of straight lines with the plane
Figure 2.60 shows the construction of the line of intersection of two triangles ABC And DEF indicating the visible and invisible sections of these triangles. Figure 2.60 Straight K,K2 built on the points of intersection of the sides AC And Sun triangle ABC with a triangle plane DEF....(Engineering graphics)
Special cases
At moderate pressures (Re " 1000 atm.) the liquid phase (for example, water) can be assumed to be incompressible (Re= const). In this case, the system of equations for this incompressible medium can be further simplified and reduced to the following form: where, and by hydrostatic forces (the term ue7) For...(Basics of cavitation processing of multicomponent media)
Special cases of equilibrium in continuous systems Barometric equation
The barometric equation establishes the dependence of gas pressure on altitude. There are numerous methods for deriving this equation, dating back to Laplace. In this case, we will take advantage of the fact that a gas located in a gravity field is a continuous system containing one component - a gas with...(Thermodynamics in modern chemistry)
SPECIAL CASES OF MUTUAL PARALLEL AND PERPENDICULARITY OF A STRAIGHT AND PLANE. SPECIAL CASES OF MUTUAL PERPENDICULARITY OF TWO PLANES
If the plane is projecting, then any projecting line of the same name is parallel to this plane, because in a plane one can always find a projecting line of the same name. So, in Fig. 67 shows the planes: T 1Sh, FJL Sh, G1 Pz. The lines will be parallel to these planes: A|| T (a 1 Pg);...(Descriptive geometry)
GENERAL CASES. METHOD OF INTERMEDIARIES
To find the points of intersection of a straight line with the surface Ф by the method of intermediaries, it is advisable to enclose the straight line in an intermediary plane T that intersects the given surface Ф along exact line- straight or circle. An overview and classification of various types of such planes was given earlier (see....(Descriptive geometry)
METHOD OF INTERMEDIARIES
If both general position planes are given arbitrarily, then the problem can be solved by the method of intermediaries in accordance with algorithm No. 2. Two planes T and T1 are chosen as intermediaries - projecting or level (Fig. 254). In the case of the intersection of two planes, we write algorithm No. 2 as follows: 1. Select T and T1....(Descriptive geometry)
Hello, friends! Today we are looking at a topic from descriptive geometry - intersection of a straight line with a plane And determining line visibility.
We take the assignment from Bogolyubov’s collection, 1989, p. 63, var. 1. We need to construct a complex drawing of triangle ABC and straight line MN using given coordinates. Find the point of meeting (intersection) of the straight line with the opaque plane ABC. Determine the visible sections of the straight line.
Intersection of a straight line with a plane
1. Using the coordinates of points A, B and C, we construct a complex drawing of a triangle and straight line NM. We start drawing with a horizontal projection. We find the coordinates of the projection points using auxiliary lines.
2. We get such a complex drawing.
3. To determine coordinates of the point of intersection of a straight line and a plane Let's do the following.
a) Through the straight line NM we draw an auxiliary plane P, i.e. on the frontal projection we draw a trace of the plane Pv, on the horizontal plane we lower the perpendicular Pn - the horizontal trace of the plane P.
b) Find the frontal projection of the line of intersection of the trace of the plane P with the triangle ABC. This is the segment d'e'. We find the horizontal projection along the connection lines until they intersect with the sides ab (point d) and ac (point e) of the triangle. We connect points d and e.
c) Together with the intersections of de and nm there will be a horizontal projection of the desired point intersection of a straight line with a plane k.
d) We draw a connection line from k to the intersection with d’e’, we obtain a frontal projection of point k’.
e) using the communication lines we find the profile projection of point k’’.
Coordinates of the point of intersection of a line and a plane K found. This point is also called the meeting point of the line and the plane.
Determining line visibility
For determining line visibility let's use the method competing points.
In relation to our drawing, the competing points will be:
— points: d’ belonging to a’b’ and e’ belonging to n’m’ (frontally competing),
— points: g belonging to bc and h belonging to nm (horizontally competing),
— points: l’’ belonging to b’’c’’ and p’’ belonging to n’’m’’ (profile-competing).
Of the two competing points, the one whose height is greater will be visible. The visibility limit is limited by point K.
For a pair of points d’ and e’, we determine visibility as follows: lower the perpendicular to the intersection with ab and nm on the horizontal projection, find points d and f. We see that the y coordinate for point f is greater than that of d → point f is visible → straight line nm is visible in section f’k’, and invisible in section k’m’.
We reason similarly for a pair of points g and h: on the frontal projection, the z coordinate of point h’ is greater than that of g’ → point h’ is visible, g’ is not → straight line nm is visible on the segment hk, but invisible on the segment kn.
And for a pair of points l''p'': on the frontal projection the x-coordinate is greater at point p', which means it covers point l'' on the profile projection → p'' is visible, l'' is not → straight line segment n' 'k'' is visible, k''m'' is invisible.
This chapter talks about how to find the coordinates of the point of intersection of a line with a plane given the equations that define this plane. The concept of the point of intersection of a line with a plane and two ways of finding the coordinates of the point of intersection of a line with a plane will be considered.
For an in-depth study of the theory, it is necessary to begin consideration with the concept of a point, a straight line, a plane. The concept of a point and a straight line is considered both on the plane and in space. For a detailed consideration, it is necessary to turn to the topic of straight lines and planes in space.
There are several variations in the location of the line relative to the plane and space:
- a straight line lies in a plane;
- a straight line is parallel to a plane;
- a straight line intersects a plane.
If we consider the third case, we can clearly see that when a straight line and a plane intersect, they form a common point, which is called the point of intersection of the straight line and the plane. Let's look at this case using an example.
Finding the coordinates of the intersection point of a line and a plane
A rectangular coordinate system O x y z of three-dimensional space was introduced. Each straight line has its own equation, and each plane corresponds to its own given equation, each point has a certain number of real numbers - coordinates.
To understand in detail the topic of intersection coordinates, you need to know all types of straight line equations in space and plane equations. in this case, knowledge about the transition from one type of equation to another will be useful.
Consider a problem that is based on a given intersection of a line and a plane. it comes down to finding the coordinates of the intersections.
Example 1
Calculate whether point M 0 with coordinates - 2, 3, - 5 can be the intersection point of the line x + 3 - 1 = y - 3 = z + 2 3 with the plane x - 2 y - z + 3 = 0.
Solution
When a point belongs to a certain line, the coordinates of the intersection point are the solution to both equations. From the definition we have that at intersection a common point is formed. To solve the problem, you need to substitute the coordinates of the point M 0 into both equations and calculate. If it is the intersection point, then both equations will correspond.
Let's imagine the coordinates of the point - 2, 3, - 5 and get:
2 + 3 - 1 = 3 - 3 = - 5 + 2 3 ⇔ - 1 = - 1 = - 1 - 2 - 2 3 - (- 5) + 3 = 0 ⇔ 0 = 0
Since we obtain the correct equalities, we conclude that point M 0 is the point of intersection of the given line with the plane.
Answer: the given point with coordinates is the intersection point.
If the coordinates of the intersection point are solutions to both equations, then they intersect.
The first method is to find the coordinates of the intersection of a line and a plane.
When a straight line a is specified with a plane α of a rectangular coordinate system, it is known that they intersect at the point M 0. First, let's look for the coordinates of a given intersection point for a given plane equation, which has the form A x + B y + C z + D = 0 with a straight line a, which is the intersection of the planes A 1 x + B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0. This method of defining a line in space is discussed in the article equations of a line and equations of two intersecting planes.
The coordinates of the straight line a and the plane α we need must satisfy both equations. Thus, a system of linear equations is specified, having the form
A x + B y + C z + D = 0 A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0
Solving the system implies turning each identity into a true equality. It should be noted that with this solution we determine the coordinates of the intersection of 3 planes of the form A x + B y + C z + D = 0, A 1 x + B 1 y + C 1 z + D 1 = 0, A 2 x + B 2 y + C 2 z + D 2 = 0 . To consolidate the material, we will consider solving these problems.
Example 2
The straight line is defined by the equation of two intersecting planes x - y + 3 = 0 5 x + 2 z + 8 = 0, and intersects another one 3 x - z + 7 = 0. It is necessary to find the coordinates of the intersection point.
Solution
We obtain the necessary coordinates by compiling and solving a system that has the form x - y + 3 = 0 5 x + 2 z + 8 = 0 3 x - z + 7 = 0.
You should pay attention to the topic of solving systems of linear equations.
Let's take a system of equations of the form x - y = - 3 5 x + 2 z = - 8 3 x - z = - 7 and carry out calculations using the determinant of the main matrix of the system. We get that
∆ = 1 - 1 0 5 0 2 3 0 - 1 = 1 0 (- 1) + (- 1) 2 3 + 0 5 0 - 0 0 3 - 1 2 0 - (- 1) · 5 · (- 1) = - 11
Since the determinant of the matrix is not equal to zero, the system has only one solution. To do this, we will use Cramer's method. It is considered very convenient and suitable for this occasion.
∆ x = - 3 - 1 0 - 8 0 2 - 7 0 - 1 = (- 3) 0 (- 1) + (- 1) 2 (- 7) + 0 (- 8) 0 - - 0 0 (- 7) - (- 3) 2 0 - (- 1) (- 8) (- 1) = 22 ⇒ x = ∆ x ∆ = 22 - 11 = - 2 ∆ y = 1 - 3 0 5 - 8 2 3 - 7 - 1 = 1 · (- 8) · (- 1) + (- 3) · 2 · 3 + 0 · 5 · (- 7) - - 0 · ( - 8) 3 - 1 2 (- 7) - (- 3) 5 (- 1) = - 11 ⇒ y = ∆ y ∆ = - 11 - 11 = 1 ∆ z = 1 - 1 - 3 5 0 - 8 3 0 - 7 = 1 0 (- 7) + (- 1) (- 8) 3 + (- 3) 5 0 - - (- 3) 0 3 - 1 · (- 8) · 0 - (- 1) · 5 · (- 7) = - 11 ⇒ z = ∆ z ∆ = - 11 - 11 = 1
It follows that the coordinates of the point of intersection of a given line and plane have the value (- 2, 1, 1).
Answer: (- 2 , 1 , 1) .
A system of equations of the form A x + B y + C z + D = 0 A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0 has only one solution. When line a is defined by equations such as A 1 x + B 1 y + C 1 z + D 1 = 0 A 2 x + B 2 y + C 2 z + D 2 = 0, and the plane α is given by A x + B y + C z + D = 0, then they intersect. When a straight line lies in a plane, the system produces an infinite number of solutions. If they are parallel, the equation has no solutions, since there are no common points of intersection.
Example 3
Find the intersection point of the straight line z - 1 = 0 2 x - y - 2 = 0 and the plane 2 x - y - 3 z + 1 = 0.
Solution
The given equations must be converted into the system z - 1 = 0 2 x - y - 2 = 0 2 x - y - 3 z + 1 = 0. When it has a unique solution, we will obtain the required intersection coordinates at the point. Provided that if there are no solutions, then they are parallel, or the straight line lies in the same plane.
We obtain that the main matrix of the system is A = 0 0 1 2 - 1 0 2 - 1 - 3, the extended matrix is T = 0 0 1 1 2 - 1 0 2 2 - 1 - 3 - 1. We need to determine the rank of the matrix A and T using the Gaussian method:
1 = 1 ≠ 0 , 0 1 - 1 0 = 1 ≠ 0 , 0 0 1 2 - 1 0 2 - 1 - 3 = 0 , 0 1 1 - 1 0 2 - 1 - 3 - 1 = 0
Then we find that the rank of the main matrix is equal to the rank of the extended one. Let us apply the Kronecker-Capelli theorem, which shows that the system has an infinite number of solutions. We obtain that the line z - 1 = 0 2 x - y - 2 = 0 belongs to the plane 2 x - y - 3 z + 1 = 0, which indicates the impossibility of their intersection and the presence of a common point.
Answer: there are no coordinates of the intersection point.
Example 4
Given the intersection of the straight line x + z + 1 = 0 2 x + y - 4 = 0 and the plane x + 4 y - 7 z + 2 = 0, find the coordinates of the intersection point.
Solution
It is necessary to assemble the given equations into a system of the form x + z + 1 = 0 2 x + y - 4 = 0 x + 4 y - 7 z + 2 = 0. To solve, we use the Gaussian method. With its help we will determine all available solutions in a short way. To do this, let's write
x + z + 1 = 0 2 x + y - 4 x + 4 y - 7 z + 2 = 0 ⇔ x + z = - 1 2 x + y = 4 x + 4 y - 7 z = - 2 ⇔ ⇔ x + z = - 1 y - 2 z = 6 4 y - 8 z = - 1 ⇔ x + z = - 1 y - 2 z = 6 0 = - 25
Having applied the Gauss method, it became clear that the equality is incorrect, since the system of equations has no solutions.
We conclude that the straight line x + z + 1 = 0 2 x + y - 4 = 0 with the plane x + 4 y - 7 z + 2 = 0 have no intersections. It follows that it is impossible to find the coordinates of the point, since they do not intersect.
Answer: there are no points of intersection, since the line is parallel to the plane.
When a straight line is given by a parametric or canonical equation, then from here you can find the equation of the intersecting planes that define the straight line a, and then look for the necessary coordinates of the intersection point. There is another method that is used to find the coordinates of the intersection point of a line and a plane.
The second method of finding a point begins with specifying a straight line a intersecting the plane α at the point M 0. It is necessary to find the coordinates of a given intersection point for a given plane equation A x + B y + C z + D = 0. We define straight line a by parametric equations of the form x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ, λ ∈ R.
When substitution is made into the equation A x + B y + C z + D = 0 x = x 1 + a x · λ , y = y 1 + a y · λ , z = z 1 + a z · λ , the expression takes the form of an equation with an unknown λ. It is necessary to resolve it with respect to λ, then we obtain λ = λ 0, which corresponds to the coordinates of the point at which they intersect. The coordinates of a point are calculated from x = x 1 + a x · λ 0 y = y 1 + a y · λ 0 z = z 1 + a z · λ 0 .
This method will be discussed in more detail using the examples given below.
Example 5
Find the coordinates of the point of intersection of the line x = - 1 + 4 · λ y = 7 - 7 · λ z = 2 - 3 · λ, λ ∈ R with the plane x + 4 y + z - 2 = 0.
Solution
To solve the system, it is necessary to make a substitution. Then we get that
1 + 4 λ + 4 7 - 7 λ + 2 - 3 λ - 2 = 0 ⇔ - 27 λ + 27 = 0 ⇔ λ = 1
Let's find the coordinates of the point of intersection of the plane with the straight line using parametric equations with the value λ = 1.
x = - 1 + 4 1 y = 7 - 7 1 z = 2 - 3 1 ⇔ x = 3 y = 0 z = - 1
Answer: (3 , 0 , - 1) .
When a line of the form x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ, λ ∈ R belongs to the plane A x + B y + C z + D = 0, then it is necessary to substitute there equation of the plane of expression x = x 1 + a x · λ, y = y 1 + a y · λ, z = z 1 + a z · λ, then we obtain an identity of this form 0 ≡ 0. If the plane and the line are parallel, we obtain an incorrect equality, since there are no points of intersection.
If a straight line is given by a canonical equation having the form x - x 1 a x = y - y 1 a y = z - z 1 a z , then it is necessary to move from canonical to parametric when searching for the coordinates of the point of intersection of the straight line with the plane A x + B y + C z + D = 0, that is, we get x - x 1 a x = y - y 1 a y = z - z 1 a z ⇔ x = x 1 + a x · λ y = y 1 + a y · λ z = z 1 + a z · λ and We will apply the necessary method to find the coordinates of the point of intersection of a given line and a plane in space.
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