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 [DSSE 1981]

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

Answer 1

The correct option is A

(3, -1, 1)

Let point be (a,b,c), then 2a + 4b - c = 1 .....(i)
and a = 2k + 1, b = -3k + 2 and c = 4k - 3, (where k is constant)
Substituting these values in (i), we get
2(2k + 1) + 4(-3k + 2) - (4k - 3) = 1 $\Rightarrow$ k = 1
Hence required point is (3,-1,1).
Trick : The point must satisfy the lines and plane. Obviously (3, -1,1) satisfy.