Question

Let the plane x – y – z = 2 is rotated through 90° about its line, of intersection with the plane x + 2y + z = 2. Find the equation of the plane in the new position.

Answer 1

Given planes are

x – y – z = 2 ...(i)

and x + 2y + z = 2 ...(ii)

The equation of any plane passing through the line of intersection of planes (i) and (ii) can be written as

(x – y – z – 2) + λ(x + 2y + z – 2) = 0 ...(iii)

The direction of the cosines of the normal to the plane (iii) are 

The direction cosines of the normal to the plane (i) are

Since the angle between the planes (i) and (ii) is 90°, so,

Hence, the required equation of the plane is

(x – y – z – 2) + 3/2(x + 2y + z – 2) = 0 

2(x – y – z – 2) + 3(x + 2y + z – 2) = 0 

5x + 4y + z = 10