JEE Main · 2018 · 08 Apr 2018 · Offline · Three Dimensional Geometry

L_{1}

is the line of intersection of the planes

2 x - 2 y + 3 z - 2 = 0, x - y + z + 1 = 0

and

L_{2}

is the line of intersection of the planes

x + 2 y - z - 3 = 0, 3 x - y + 2 z - 1 = 0

, then the distance of the origin from the plane, containing the lines

L_{1}

and

L_{2}

, is :

Answer: (C)

Solution

The direction of

L_{1}

(2, - 2, 3) \times (1, - 1, 1) = (1, 1, 0)

, and the direction of

L_{2}

(1, 2, - 1) \times (3, - 1, 2) = (3, - 5, - 7)

. Finding a point on

L_{1}

from its two planes, elimination gives

z = 4

and then

x - y = - 5

, so

(- 5, 0, 4)

lies on it. The plane containing both lines has normal

(1, 1, 0) \times (3, - 5, - 7) = (- 7, 7, - 8)

, so it is

- 7 (x + 5) + 7 y - 8 (z - 4) = 0

, that is

7 x - 7 y + 8 z + 3 = 0

. The distance of the origin from it is

\frac{∣3∣}{49 + 49 + 64} = \frac{3}{162} = \frac{1}{3 2}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option