JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Three Dimensional Geometry

If the line,

\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}

meets the plane,

x + 2 y + 3 z = 15

at a point

P

, then the distance of

P

from the origin is:

(A) $\frac{5}{2}$
(B) $25$
(C) $\frac{9}{2}$ ✓ Correct
(D) $\frac{7}{2}$

Answer: (C)

Solution

Write a point on the line as

(1 + 2 λ, - 1 + 3 λ, 2 + 4 λ)

and feed it into

x + 2 y + 3 z = 15

(1 + 2 λ) + 2 (- 1 + 3 λ) + 3 (2 + 4 λ) = 20 λ + 5 = 15

, so

λ = \frac{1}{2}

. That places

P

(2, \frac{1}{2}, 4)

, whose distance from the origin is

4 + \frac{1}{4} + 16 = \frac{81}{4} = \frac{9}{2}

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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