JEE Main 2022 · 29 Jul · Shift 2 · Mathematics Q14

← JEE Main 2022 · 29 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 14 of 90

Let Q be the foot of perpendicular drawn from the point P (1, 2, 3) to the plane $x+2y+z=14$ . If R is a point on the plane such that ∠PRQ = 60°, then the area of ΔPQR is equal to :

\dfrac{\sqrt3}2

\sqrt3

2\sqrt3

Answer and worked solution

Official answer: B

PQ $=\dfrac{|1+4+3-14|}{\sqrt6}=\sqrt6$ . With the angle 60° at R, $QR=\dfrac{PQ}{\tan60°}=\sqrt2$ .

Area $\frac12\sqrt6\cdot\sqrt2=\sqrt3$ .

(B)

Topics:Angles and distances

← Previous question Next question →

JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 14