This article talks about determining the distance from a point to a plane. Let's analyze it using the coordinate method, which will allow us to find the distance from a given point in three-dimensional space. To reinforce this, let’s look at examples of several tasks.
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The distance from a point to a plane is found through the known distance from a point to a point, where one of them is given, and the other is a projection onto a given plane.
When a point M 1 with a plane χ is specified in space, then a straight line perpendicular to the plane can be drawn through the point. H 1 is their common point of intersection. From this we obtain that the segment M 1 H 1 is a perpendicular drawn from point M 1 to the plane χ, where point H 1 is the base of the perpendicular.
Definition 1
The distance from a given point to the base of a perpendicular drawn from a given point to a given plane is called.
The definition can be written in different formulations.
Definition 2
Distance from point to plane is the length of the perpendicular drawn from a given point to a given plane.
The distance from point M 1 to the χ plane is determined as follows: the distance from point M 1 to the χ plane will be the smallest from a given point to any point on the plane. If point H 2 is located in the χ plane and is not equal to point H 2, then we obtain a right triangle of the form M 2 H 1 H 2 , which is rectangular, where there is a leg M 2 H 1, M 2 H 2 – hypotenuse. This means that it follows that M 1 H 1< M 1 H 2 . Тогда отрезок М 2 H 1 is considered inclined, which is drawn from point M 1 to the plane χ. We have that the perpendicular drawn from a given point to the plane is less than the inclined one drawn from the point to the given plane. Let's look at this case in the figure below.
Distance from a point to a plane - theory, examples, solutions
There are a number of geometric problems whose solutions must contain the distance from a point to a plane. There may be different ways to identify this. To resolve, use the Pythagorean theorem or similarity of triangles. When, according to the condition, it is necessary to calculate the distance from a point to a plane, given in a rectangular coordinate system of three-dimensional space, it is solved by the coordinate method. This paragraph discusses this method.
According to the conditions of the problem, we have that a point in three-dimensional space with coordinates M 1 (x 1, y 1, z 1) with a plane χ is given; it is necessary to determine the distance from M 1 to the plane χ. Several solution methods are used to solve this problem.
First way
This method is based on finding the distance from a point to a plane using the coordinates of point H 1, which are the base of the perpendicular from point M 1 to the plane χ. Next, you need to calculate the distance between M 1 and H 1.
To solve the problem in the second way, use the normal equation of a given plane.
Second way
By condition, we have that H 1 is the base of the perpendicular, which was lowered from point M 1 to the plane χ. Then we determine the coordinates (x 2, y 2, z 2) of point H 1. The required distance from M 1 to the χ plane is found by the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2, where M 1 (x 1, y 1, z 1) and H 1 (x 2, y 2, z 2). To solve, you need to know the coordinates of point H 1.
We have that H 1 is the point of intersection of the χ plane with the line a, which passes through the point M 1 located perpendicular to the χ plane. It follows that it is necessary to compile an equation for a straight line passing through a given point perpendicular to a given plane. It is then that we will be able to determine the coordinates of point H 1. It is necessary to calculate the coordinates of the point of intersection of the line and the plane.
Algorithm for finding the distance from a point with coordinates M 1 (x 1, y 1, z 1) to the χ plane:
Definition 3
- draw up an equation of straight line a passing through point M 1 and at the same time
- perpendicular to the χ plane;
- find and calculate the coordinates (x 2 , y 2 , z 2) of point H 1, which are points
- intersection of line a with plane χ;
- calculate the distance from M 1 to χ using the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + z 2 - z 1 2.
Third way
In a given rectangular coordinate system O x y z there is a plane χ, then we obtain a normal equation of the plane of the form cos α · x + cos β · y + cos γ · z - p = 0. From here we obtain that the distance M 1 H 1 with the point M 1 (x 1 , y 1 , z 1) drawn to the plane χ, calculated by the formula M 1 H 1 = cos α x + cos β y + cos γ z - p . This formula is valid, since it was established thanks to the theorem.
Theorem
If a point M 1 (x 1, y 1, z 1) is given in three-dimensional space, having a normal equation of the plane χ of the form cos α x + cos β y + cos γ z - p = 0, then calculating the distance from the point to plane M 1 H 1 is obtained from the formula M 1 H 1 = cos α · x + cos β · y + cos γ · z - p, since x = x 1, y = y 1, z = z 1.
Proof
The proof of the theorem comes down to finding the distance from a point to a line. From this we obtain that the distance from M 1 to the χ plane is the modulus of the difference between the numerical projection of the radius vector M 1 with the distance from the origin to the χ plane. Then we get the expression M 1 H 1 = n p n → O M → - p. The normal vector of the plane χ has the form n → = cos α, cos β, cos γ, and its length is equal to one, n p n → O M → is the numerical projection of the vector O M → = (x 1, y 1, z 1) in the direction determined by the vector n → .
Let's apply the formula for calculating scalar vectors. Then we obtain an expression for finding a vector of the form n → , O M → = n → · n p n → O M → = 1 · n p n → O M → = n p n → O M → , since n → = cos α , cos β , cos γ · z and O M → = (x 1 , y 1 , z 1) . The coordinate form of writing will take the form n → , O M → = cos α · x 1 + cos β · y 1 + cos γ · z 1 , then M 1 H 1 = n p n → O M → - p = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p . The theorem has been proven.
From here we get that the distance from the point M 1 (x 1, y 1, z 1) to the plane χ is calculated by substituting cos α · x + cos β · y + cos γ · z - p = 0 into the left side of the normal equation of the plane instead of x, y, z coordinates x 1, y 1 and z 1, relating to point M 1, taking the absolute value of the obtained value.
Let's look at examples of finding the distance from a point with coordinates to a given plane.
Example 1
Calculate the distance from the point with coordinates M 1 (5, - 3, 10) to the plane 2 x - y + 5 z - 3 = 0.
Solution
Let's solve the problem in two ways.
The first method starts with calculating the direction vector of the line a. By condition, we have that the given equation 2 x - y + 5 z - 3 = 0 is a general plane equation, and n → = (2, - 1, 5) is the normal vector of the given plane. It is used as a direction vector of a straight line a, which is perpendicular to a given plane. It is necessary to write down the canonical equation of a line in space passing through M 1 (5, - 3, 10) with a direction vector with coordinates 2, - 1, 5.
The equation will become x - 5 2 = y - (- 3) - 1 = z - 10 5 ⇔ x - 5 2 = y + 3 - 1 = z - 10 5.
Intersection points must be determined. To do this, gently combine the equations into a system to move from the canonical to the equations of two intersecting lines. Let's take this point as H 1. We get that
x - 5 2 = y + 3 - 1 = z - 10 5 ⇔ - 1 · (x - 5) = 2 · (y + 3) 5 · (x - 5) = 2 · (z - 10) 5 · ( y + 3) = - 1 · (z - 10) ⇔ ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 5 y + z + 5 = 0 ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0
After which you need to enable the system
x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 2 x - y + 5 z - 3 = 0 ⇔ x + 2 y = 1 5 x - 2 z = 5 2 x - y + 5 z = 3
Let us turn to the Gaussian system solution rule:
1 2 0 - 1 5 0 - 2 5 2 - 1 5 3 ~ 1 2 0 - 1 0 - 10 - 2 10 0 - 5 5 5 ~ 1 2 0 - 1 0 - 10 - 2 10 0 0 6 0 ⇒ ⇒ z = 0 6 = 0 , y = - 1 10 10 + 2 z = - 1 , x = - 1 - 2 y = 1
We get that H 1 (1, - 1, 0).
We calculate the distance from a given point to the plane. We take points M 1 (5, - 3, 10) and H 1 (1, - 1, 0) and get
M 1 H 1 = (1 - 5) 2 + (- 1 - (- 3)) 2 + (0 - 10) 2 = 2 30
The second solution is to first bring the given equation 2 x - y + 5 z - 3 = 0 to normal form. We determine the normalizing factor and get 1 2 2 + (- 1) 2 + 5 2 = 1 30. From here we derive the equation of the plane 2 30 · x - 1 30 · y + 5 30 · z - 3 30 = 0. The left side of the equation is calculated by substituting x = 5, y = - 3, z = 10, and you need to take the distance from M 1 (5, - 3, 10) to 2 x - y + 5 z - 3 = 0 modulo. We get the expression:
M 1 H 1 = 2 30 5 - 1 30 - 3 + 5 30 10 - 3 30 = 60 30 = 2 30
Answer: 2 30.
When the χ plane is specified by one of the methods in the section on methods for specifying a plane, then you first need to obtain the equation of the χ plane and calculate the required distance using any method.
Example 2
In three-dimensional space, points with coordinates M 1 (5, - 3, 10), A (0, 2, 1), B (2, 6, 1), C (4, 0, - 1) are specified. Calculate the distance from M 1 to plane A B C.
Solution
First you need to write down the equation of the plane passing through the given three points with coordinates M 1 (5, - 3, 10), A (0, 2, 1), B (2, 6, 1), C (4, 0, - 1) .
x - 0 y - 2 z - 1 2 - 0 6 - 2 1 - 1 4 - 0 0 - 2 - 1 - 1 = 0 ⇔ x y - 2 z - 1 2 4 0 4 - 2 - 2 = 0 ⇔ ⇔ - 8 x + 4 y - 20 z + 12 = 0 ⇔ 2 x - y + 5 z - 3 = 0
It follows that the problem has a solution similar to the previous one. This means that the distance from point M 1 to plane A B C has a value of 2 30.
Answer: 2 30.
Finding the distance from a given point on a plane or to a plane to which they are parallel is more convenient by applying the formula M 1 H 1 = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p. From this we obtain that the normal equations of planes are obtained in several steps.
Example 3
Find the distance from a given point with coordinates M 1 (- 3, 2, - 7) to the coordinate plane O x y z and the plane given by the equation 2 y - 5 = 0.
Solution
The coordinate plane O y z corresponds to an equation of the form x = 0. For the O y z plane it is normal. Therefore, it is necessary to substitute the values x = - 3 into the left side of the expression and take the absolute value of the distance from the point with coordinates M 1 (- 3, 2, - 7) to the plane. We get a value equal to - 3 = 3.
After the transformation, the normal equation of the plane 2 y - 5 = 0 will take the form y - 5 2 = 0. Then you can find the required distance from the point with coordinates M 1 (- 3, 2, - 7) to the plane 2 y - 5 = 0. Substituting and calculating, we get 2 - 5 2 = 5 2 - 2.
Answer: The required distance from M 1 (- 3, 2, - 7) to O y z has a value of 3, and to 2 y - 5 = 0 has a value of 5 2 - 2.
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Job type: 14
Condition
In a regular triangular pyramid DABC with base ABC, the side of the base is 6\sqrt(3), and the height of the pyramid is 8. On the edges AB, AC and AD, points M, N and K are marked, respectively, such that AM=AN=\frac(3\sqrt(3))(2) And AK=\frac(5)(2).
A) Prove that the planes MNK and DBC are parallel.
b) Find the distance from point K to the DBC plane.
Show solutionSolution
A) Planes MNK and DBC are parallel if two intersecting lines of one plane are respectively parallel to two intersecting lines of another plane. Let's prove it. Consider the lines MN and KM of the MNK plane and the lines BC and DB of the DBC plane.
In triangle AOD: \angle AOD = 90^\circ and by the Pythagorean theorem AD=\sqrt(DO^2 +AO^2).
Let's find AO using the fact that \bigtriangleup ABC is correct.
AO=\frac(2)(3)AO_1, where AO_1 is the height of \bigtriangleup ABC, AO_1 = \frac(a\sqrt(3))(2), where a is the side of \bigtriangleup ABC.
AO_1 = \frac(6\sqrt(3) \cdot \sqrt(3))(2)=9, then AO=6, AD=\sqrt(8^2 + 6^2)=10.
1. Since \frac(AK)(AD)=\frac(5)(2) : 10=\frac(1)(4), \frac(AM)(AB)=\frac(3\sqrt(3))(2) : 6\sqrt(3)=\frac(1)(4) and \angle DAB is general, then \bigtriangleup AKM \sim ADB.
From the similarity it follows that \angle AKM = \angle ADB. These are the corresponding angles for straight lines KM and BD and secant AD. So KM \parallel BD.
2. Since \frac(AN)(AC)=\frac(3 \sqrt(3))(2 \cdot 6 \sqrt(3))=\frac(1)(4), \frac(AM)(AB)=\frac(1)(4) and \angle CAB is common, then \bigtriangleup ANM \sim \bigtriangleup ACB.
From the similarity it follows that \angle ANM = \angle ACB. These angles are corresponding to the lines MN and BC and the secant AC. This means MN \parallel BC.
Conclusion: since two intersecting lines KM and MN of the plane MNK are respectively parallel to two intersecting lines BD and BC of the plane DBC, then these planes are parallel - MNK \parallel DBC.
b) Let's find the distance from point K to the plane BDC.
Since the plane MNK is parallel to the plane DBC, the distance from point K to the plane DBC is equal to the distance from point O_2 to the plane DBC and it is equal to the length of the segment O_2 H. Let us prove this.
BC \perp AO_1 and BC \perp DO_1 (as the heights of triangles ABC and DBC), which means BC is perpendicular to the plane ADO_1, and then BC is perpendicular to any line of this plane, for example, O_2 H. By construction, O_2H\perp DO_1, which means O_2H is perpendicular two intersecting straight lines of the BCD plane, and then the segment O_2 H is perpendicular to the BCD plane and equal to the distance from O_2 to the BCD plane.
In a triangle O_2HO_1:O_2H=O_(2)O_(1)\sin\angle HO_(1)O_(2).
O_(2)O_(1)=AO_(1)-AO_(2).\, \frac(AO_2)(AO_1)=\frac(1)(4), AO_(2)=\frac(AO_1)(4)=\frac(9)(4).
O_(2)O_(1)=9-\frac(9)(4)=\frac(27)(4).
\sin \angle DO_(1)A= \frac(DO)(DO_(1))= \frac(8)(\sqrt(64+3^2))= \frac(8)(\sqrt(73)).
O_2H=\frac(27)(4) \cdot \frac(8)(\sqrt(73))=\frac(54)(\sqrt(73)).
Answer
\frac(54)(\sqrt(73))
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 14
Topic: Distance from a point to a plane
Condition
ABCDA_1B_1C_1D_1 is a regular quadrangular prism.
a) Prove that the plane BB_1D_1 \perp AD_1C .
b) Knowing AB = 5 and AA_1 = 6, find the distance from point B_1 to plane AD_1C.
Show solutionSolution
a) Since this prism is regular, then BB_1 \perp ABCD, hence BB_1 \perp AC. Since ABCD is a square, then AC \perp BD . Thus AC \perp BD and AC \perp BB_1 . Since lines BD and BB_1 intersect, then, according to the sign of perpendicularity of a line and a plane, AC \perp BB_1D_1D. Now based on the perpendicularity of the planes AD_1C \perp BB_1D_1.
b) Let us denote by O the point of intersection of the diagonals AC and BD of the square ABCD. Planes AD_1C and BB_1D_1 intersect along straight line OD_1. Let B_1H be a perpendicular drawn in the plane BB_1D_1 to the straight line OD_1. Then B_1H \perp AD_1C . Let E=OD_1 \cap BB_1 . For similar triangles D_1B_1E and OBE (the equality of the corresponding angles follows from the condition BO \parallel B_1D_1) we have \frac (B_1E)(BE)=\frac(B_1D_1)(BO)=\frac(2)1.
This means B_1E=2BE=2 \cdot 6=12. Since B_1D_1=5\sqrt(2) , then the hypotenuse D_1E= \sqrt(B_1E^(2)+B_1D_1^(2))= \sqrt(12^(2)+(5\sqrt(2))^(2))= \sqrt(194). Next, we use the area method in triangle D_1B_1E to calculate the height B_1H lowered onto the hypotenuse D_1E:
S_(D_1B_1E)=\frac1(2)B_1E \cdot B_1D_1=\frac1(2)D_1E \cdot B_1H; 12 \cdot 5\sqrt(2)=\sqrt(194) \cdot B_1H;
B_1H=\frac(60\sqrt(2))(\sqrt(194))=\frac(60)(\sqrt(97))=\frac(60\sqrt(97))(97).
Answer
\frac(60\sqrt(97))(97)
Source: “Mathematics. Preparation for the Unified State Exam 2016. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 14
Topic: Distance from a point to a plane
Condition
ABCDA_1B_1C_1D_1 is a rectangular parallelepiped. Edges AB=24, BC=7, BB_(1)=4 .
a) Prove that the distances from points B and D to the plane ACD_(1) are the same.
b) Find this distance.
Show solutionSolution
A) Consider the triangular pyramid D_1ACD.
In this pyramid, the distance from point D to the base plane ACD_1-DH is equal to the height of the pyramid drawn from point D to the base ACD_1.
V_(D_1ABC)=\frac1(3)S_(ACD_1) \cdot DH, from this equality we obtain
DH=\frac(3V_(D_1ACD))(S_(ACD_1)).
Consider the pyramid D_1ABC. The distance from point B to the plane ACD_1 is equal to the height lowered from the top of B to the base of ACD_1. Let's denote this distance BK. Then V_(D_1ABC)=\frac1(3)S_(ACD_1) \cdot BK, from this we get BK=\frac(3V_(D_1ABC))(S_(ACD_1)).\: But V_(D_1ACD) = V_(D_1ABC) , since if we consider ADC and ABC as bases in pyramids, then the height D_1D is total and S_(ADC)=S_(ABC) ( \bigtriangleup ADC=\bigtriangleup ABC on two legs). So BK=DH.
b) Find the volume of the pyramid D_1ACD.
Height D_1D=4 .
S_(ACD)=\frac1(2)AD \cdot DC=\frac1(2) \cdot24 \cdot 7=84.
V=\frac1(3)S_(ACD) \cdot D_1D=\frac1(3) \cdot84 \cdot4=112.
The area of face ACD_1 is \frac1(2)AC \cdot D_1P.
AD_1= \sqrt(AD^(2)+DD_1^(2))= \sqrt(7^(2)+4^(2))= \sqrt(65), \: AC= \sqrt(AB^(2)+BC^(2))= \sqrt(24^(2)+7^(2))= 25
Knowing that the leg of a right triangle is the mean proportional to the hypotenuse and the segment of the hypotenuse enclosed between the leg and the altitude drawn from the vertex of the right angle, in triangle ADC we have AD^(2)=AC\cdot AP, \: AP=\frac(AD^(2))(AC)=\frac(7^(2))(25)=\frac(49)(25).
In a right triangle AD_1P according to the Pythagorean theorem D_1P^(2)= AD_1^(2)-AP^(2)= 65-\left (\frac(49)(25) \right)^(2)= \frac(38\:224)(25^(2)), D_1P=\frac(4\sqrt(2\:389))(25).
S_(ACD_1)=\frac1(2) \cdot25 \cdot\frac(4\sqrt(2\:389))(25)=2\sqrt(2\:389).
DH=\frac(3V)(S_(ACD_1))=\frac(3 \cdot112)(2\sqrt(2\:389))=\frac(168)(\sqrt(2\:389)).
- We construct a plane through point Aβ II α .
- Building the third plane, perpendicular to parallel planes α And β
- On the line of intersection of the planes, select point B and drop a perpendicular from point B.
- Segment BN - the distance between the planes is equal to the distance from point A to the planeα . AH = BN.
2. Given a cube ABCDA 1 B 1 C 1 D 1 . The length of the edge of the cube is 1. Find the distance from point A to plane CB 1 D 1.
Solution [, 250Kb]. The following algorithm will help us with this task:
- Through point A we construct a plane perpendicular to the plane α
- We lower the perpendicular to the line of intersection of the planes AH. AR – the required distance from point A to the plane α .
For example, in the above problem, I found the distance from point A to plane A 1 BT, expressing twice the volume of the pyramid ABTA 1 with the base ABT.
Given a cube ABCDA 1 B 1 C 1 D 1 with edge 1. Find the distance from point A to plane A 1 BT, where T is the midpoint of segment AD.
Solution [, 193Kb].
4. In a regular quadrangular prism ABCDA 1 B 1 C 1 D 1 with base side 12 and height 21, point M is taken on edge AA 1 so that AM = 8. Point K is taken on edge BB 1 so that B 1 K=8. Find the distance from point A 1 to plane D 1 MK.
Solution [, 347Kb].
5. In a regular triangular prism ABCA 1 B 1 C 1, the sides of the base are equal to 2, and the side edges are equal to 3. Point D is the middle of the edge CC 1. Find the distance from vertex C to plane ADB 1.
Solution [, 285Kb].
6. The base of the right prism ABCA 1 B 1 C 1 is the isosceles triangle ABC, AB = AC = 5, BC = 6. The height of the prism is 3. Find the distance from the middle of the edge B 1 C 1 to the plane BCA 1.
Solution [, 103Kb].
7. The base of the right prism ABCA 1 B 1 C 1 is a right triangle ABC with right angle C. BC = 3. The height of the prism is 4. Find the distance from point B to the plane ACB 1.
Solution [, 127Kb].
8. The base of the prism ABCDA 1 B 1 C 1 D 1 is a rhombus ABCD, AB = 10, ВD = 12. The height of the prism is 6. Find the distance from the center of the face A 1 B 1 C 1 D 1 to the plane BDC 1.
Solution [, 148Kb].
9. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are equal to 1. Find the distance from point B to the plane DEA 1.
Solution [, 194Kb].
10. Given a regular tetrahedron ABCD with edge . Find the distance from vertex A to plane BDC.
Solution [, 119Kb].
11. In the DABC pyramid, all edges are equal to a. Let O denote the center of the base ABC, and K the midpoint of the height DO of the pyramid. Find the distance from point K to edge ABD.
Solution [