Subject - Mathematics:

General

MCQ - 198-8930

Question:

The line $\frac{x + 3}{3}$ = $\frac{y - 2}{2}$ = $\frac{z + 1}{1}$ and the plane 4 x + 5 y + 3 z – 5 = 0 intersect at a point

(3, 1, -2)
(3, -2, 1)
(2, -1, 3)
(3, -2, 1)

Correct Answer: B

Explanation:

Any pt. on the line is (3 r – 3, -2 r + 2 r – 1).
This lies on the plane 4 x + 5 y + 3 z – 5 = 0
if4 ( 3r – 3) + 5 (-2r + 2) + 3(r – 1) –5 = 0
i.e. 5 r – 10 = 0 ∴ r = 2
∴ reqd. pt. is (6 – 3, -4 + 2, 2 – 1)
i.e. (3, -2, 1).

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