JEE Main · 2015 · 04 Apr 2015 · Offline · Three Dimensional Geometry

The distance of the point

(1, 0, 2)

from the point of intersection of the line

\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}

and the plane

x - y + z = 16

, is :

(A)8
(B) $321$
(C)13✓ Correct
(D) $214$

Answer: (C)

Solution

Write the line parametrically as

x = 2 + 3 t

y = - 1 + 4 t

z = 2 + 12 t

and drop it into the plane

x - y + z = 16

(2 + 3 t) - (- 1 + 4 t) + (2 + 12 t) = 5 + 11 t = 16

, so

t = 1

and the meeting point is

(5, 3, 14)

. Its distance from

(1, 0, 2)

4^{2} + 3^{2} + 1 2^{2} = 169 = 13

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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