Class 12 RD Sharma Solutions - Chapter 29 The Plane - Exercise 29.12

Question 1(i): Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses line through yz-plane.

Solution:

We know, the equation of line through the points (x₁,y₁,z₁) and (x₂,y₂,z₂) is 

Therefore, equation of line joining (5, 1, 6) and (3, 4, 1) is 

⇒

Let, , where  is a constant.

⇒ 

Coordinates of any point on the line is in the form of 

Since, the line crosses the yz-plane, the point  must satisfy the equation of plane x=0,

⇒ ⇒ 

Therefore, coordinates of points is given by, putting  we get,

⇒ 

(ii) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses line through zx-plane.

Solution:

We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is 

Therefore, equation of line joining (5, 1, 6) and (3, 4, 1) is

⇒

Let,  ,where  is a constant.

⇒ 

Coordinates of any point on the line is in the form of 

Since, the line crosses the zx-plane, the point  must satisfy the equation of plane y=0,

⇒  ⇒ 

Therefore, the coordinates of point is given by, putting  we get,

⇒ 

Question 2: Find the coordinates of point where the line through (3, -4, -5) and (2, -3, 1) crosses the plane 2x+y+z=7.

Solution:

We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is 

Therefore, line joining the points (3, -4, -5) and (2, -3, 1) is 

⇒ 

Let  where  is constant.

⇒ 

The coordinates of any point on the line is given by 

The line crosses the plane, therefore, point must satisfy the plane equation.

⇒ 

Therefore, The coordinates of point are given by, putting ,

⇒ 

⇒ (1, -2, 7)

Question 3: Find the distance of the point (-1, -5, -10) from the point of intersection of line

 and the plane 

Solution:

Given equation of line is 

⇒ 

Coordinates of any point of line should be in the form of 

We know, the intersection point of line and plane lies on the plane, using this,

⇒ 

⇒ 

⇒ 

Therefore, coordinates of point is given by, putting  ,

⇒ (2, -1, 2)

Therefore, now distance between (-1, -5, -10) and (2, -1, 2) is,

⇒ 

⇒  ⇒ 13 units

Question 4: Find the distance of point (2, 12, 5) from the point of intersection of line

  and 

Solution:

Given equation of line is 

⇒ 

Coordinates of any point of line should be in the form of 

We know, the intersection point of line and plane lies on the plane, using this,

⇒ .

⇒ 

⇒ 

Therefore, coordinates of point is given by, putting ,

⇒ (14, 12, 10).

Therefore, now distance between the points (2, 12, 5) and (14, 12, 10) is,

⇒ 

⇒  ⇒ 13 units

Question 5: Find the distance of point (-1, -5, -10) from the point of intersection of the joining A(2, -1, 2) and B(5, 3, 4) with the plane x-y+z=5.

Solution:

Equation of line joining the points A(2, -1, 2) and B(5, 3, 4) is 

⇒ 

Let, 

⇒ 

Coordinates of any point on the line is given by 

We know, The intersection of line and plane lies on the plane, so,

⇒ 

⇒ 

Therefore, the coordinates of points is, putting 

⇒ (2, -1, 2)

Now, the distance between the points (-1, -5, -10) and (2, -1, 2) is,

⇒ 

⇒  ⇒ 13 units

Question 6: Find the distance of point (3, 4, 4) from the point, where the line joining the points A(3, -4, -5) and B(2, -3, 1) intersects the 2x+y+z=7.

Solution:

Equation of line passing through A(3, -4, -5) and B(2, -3, 1) is given by 

⇒ 

Let 

⇒ 

Coordinates of any point on the line is given by 

We know, The intersection of line and plane lies on the plane, so,

⇒ 

⇒ 

⇒ 

Therefore, the coordinates of points is, putting 

⇒ (1, -2, 7)

Now, the distance between (3, 4, 4) and (1, -2, 7) is,

⇒ 

⇒  = 7 units

Question 7: Find the distance of point (1, -5, 9) from the plane x- y+ z=5 measured along the line x=y=z.

Solution:

Given, The equation of line is x=y=z, it can also be written as,

, where (1, 1, 1) are direction ratios of the line.

Here we have to measure the distance along the line, the equation of line parallel to x=y=z have same direction ratios (1, 1, 1),

So, the equation of line passing through (1, -5, 9) and having direction ratios (1, 1, 1) is,

⇒ 

Let 

Coordinates of any point on the line is given by 

We know, The intersection of line and plane lies on the plane, so,

⇒ 

⇒ 

Therefore, the coordinates of point is given by, putting  = (-9, -15, -1)

Now, distance between the points (1, -5, 9) and (-9, -15, -1) is,

⇒ 

⇒  units.

Summary

Exercise 29.12 in RD Sharma's Class 12 Chapter 29 on The Plane likely focuses on problems related to the angle between two planes. This exercise probably covers calculating the acute angle between intersecting planes using their normal vectors, understanding conditions for perpendicularity and parallelism of planes, and solving problems involving multiple planes and their angular relationships. Students are expected to apply trigonometric concepts and vector operations to determine these angles and relationships in various scenarios.