JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Three Dimensional Geometry

The distance of the point

(- 1, 2, - 2)

from the line of intersection of the planes

2 x + 3 y + 2 z = 0

and

x - 2 y + z = 0

is :

(A) $\frac{5}{2}$
(B) $\frac{42}{2}$
(C) $\frac{1}{2}$
(D) $\frac{34}{2}$ ✓ Correct

Answer: (D)

Solution

The line of intersection has direction

(2, 3, 2) \times (1, - 2, 1) = (7, 0, - 7)

, parallel to

(1, 0, - 1)

, and passes through the origin since both planes do. For the foot

Q = (t, 0, - t)

of the perpendicular from

P (- 1, 2, - 2)

, the condition

P Q \cdot (1, 0, - 1) = 0

gives

(t + 1) + (t - 2) = 0

, so

t = \frac{1}{2}

and

Q = (\frac{1}{2}, 0, - \frac{1}{2})

. The distance is

∣ P Q ∣ = (\frac{3}{2})^{2} + 2^{2} + (\frac{3}{2})^{2} = \frac{34}{4} = \frac{34}{2}

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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