JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Three Dimensional Geometry

Let Q be the foot of the perpendicular from the point

P (7, - 2, 13)

on the plane containing the lines

\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}

and

\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}

. Then

(P Q)^{2}

is equal to ____.

Answer: 96

Solution

First, find the equation of the plane containing both lines.

Line 1 passes through

(- 1, 1, 3)

with direction

d_{1} = (6, 7, 8)

.
Line 2 passes through

(1, 2, 3)

with direction

d_{2} = (3, 5, 7)

.

The normal to the plane is

n = d_{1} \times d_{2} = (9, - 18, 9) = 9 (1, - 2, 1)

.

Using point

(- 1, 1, 3)

, the plane equation is:

1 (x + 1) - 2 (y - 1) + 1 (z - 3) = 0

x - 2 y + z = 0

Distance from

P (7, - 2, 13)

to this plane:

P Q = \frac{∣7 - 2 ( - 2 ) + 13∣}{1 + 4 + 1} = \frac{∣7 + 4 + 13∣}{6} = \frac{24}{6}

(P Q)^{2} = \frac{576}{6} = 96

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer