Question

The Cartesian equation of the plane passing through the line of intersection of the plane r .(2$\widehat{i}$ – 3$\widehat{j}$ + 4$\widehat{k}$) = 1 and

r . ($\widehat{i}$ –$\widehat{j}$) + 4 = 0and perpendicular to the plane

r . (2$\widehat{i}$ –$\widehat{j}$+ $\widehat{k}$) + 8 = 0 is –

• A. 3x – 4y + 4z = 5
• B. x – 2y + 4z = 3
• C. 5x – 2y – 12z + 47 = 0
• D. 2x + 3y + 4 = 0

Answer 1

Answer: (C)

5x – 2y – 12z + 47 = 0

Answer 2

Equation of any plane passing through the intersection of the planes

r . (2$\widehat{i}$ – 3$\widehat{j}$ + 4$\widehat{k}$) = 1 and r . ($\widehat{i}$ –$\widehat{j}$) + 4 = 0 is

2x – 3y + 4z – 1 + l (x – y + 4) = 0

or (2 + l)x – (3 + l)y + 4z + 4l – 1 = 0

This plane is perpendicular to the plane

r . (2$\widehat{i}$ –$\widehat{j}$ + $\widehat{k}$) + 8 = 0 if

Ž 2(2 + l) + (3 + l) + 4 = 0

Ž 11 + 3l = 0 Ž l = –11/3.

and the required equation of the plane is

3(2x – 3y + 4z – 1) – 11 (x – y + 4) = 0

Ž 5x – 2y – 12z + 47 = 0.