Let Q be the foot of perpendicular drawn from the point $\mathrm{P}(1,2,3)$ to the plane $\mathrm{x}+2 \mathrm{y}+\mathrm{z}=14$. If R is a point on the plane such that $\angle \mathrm{PRQ}=60^{\circ}$, then the area of $\triangle \mathrm{PQR}$ is equal to:

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Three Dimensional Geometry

Let Q be the foot of perpendicular drawn from the point

P (1, 2, 3)

to the plane

x + 2 y + z = 14

. If R is a point on the plane such that

∠ PRQ = 6 0^{\circ}

, then the area of

△ PQR

is equal to:

(A) $\frac{3}{2}$
(B) $3$ ✓ Correct
(C) $23$
(D) $3$

Answer: (B)

Solution

The perpendicular distance from

P (1, 2, 3)

to the plane

x + 2 y + z = 14

PQ = \frac{1 + 4 + 3 - 14}{1 ^{2} + 2 ^{2} + 1 ^{2}} = \frac{6}{6} = 6

. In the right triangle

PQR

(Q is the foot of perpendicular, so

∠ PQR = 9 0^{\circ}

), we have

∠ PRQ = 6 0^{\circ}

. Thus

QR = PQ cot 6 0^{\circ} = 6 \cdot \frac{1}{3} = 2

. Area

= \frac{1}{2} PQ \cdot QR = \frac{1}{2} 6 \cdot 2 = 3

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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