Question

The direction ratios of the normal to the plane passing through the points (1, -2, 3), (-1, 2, -1) and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ is

• A. (2, 3, 4)
• B. (14, 0, 7)
• C. (-2, 0, -1)
• D. (2, 0, -1)

Answer 1

Answer: (D)

Any plane through (1, -2, 3) is
A(x -1) + B(y + 2) + C(z - 3) = 0 ...(1)
The point (-1, 2, -1) lies in this plane if -2A + 4B- 4C = 0
i.e. if A - 2B+ 2C = 0 ...(2)
The plane (1) is parallel to the given line with d.r., 2, 3, 4
if 2A + 3B+ 4C = 0 ...(3)
From (2) and (3), we have
$\frac{A}{-8-6}=\frac{B}{4-4}=\frac{C}{3+4}$
$\Rightarrow \frac{A}{-14}=\frac{B}{0}=\frac{C}{7}\Rightarrow A:B:C=2:0:-1$