Question

The point where the line $\frac{x - 1}{2} = \frac{y - 2}{- 3} = \frac{z + 3}{4}$ meets the plane $2x + 4y - z = 1$, is

• A. (3, –1, 1)
• B. (3, 1, 1)
• C. (1, 1, 3)
• D. (1, 3, 1)

Answer 1

Answer: (A)

(3, –1, 1)

Answer 2

Let point be (a, b, c), then .....(i)

and $l + 2 m + 3 n = 0$ and ,

(where k is constant)

Substituting these values in (i), we get

$l = - ( 2 m + 3 n )$

Hence required point is (3, –1, 1).

Trick : The point must satisfy the lines and plane. Obviously (3, – 1, 1) satisfies.