JEE Main 202429 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2024 · 29 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let P ( 3, 2, 3 ) , Q ( 4, 6, 2 ) and R ( 7, 3, 2 ) be the vertices...

Let

P ( 3, 2, 3 ) , Q ( 4, 6, 2 )

and

R ( 7, 3, 2 )

be the vertices of

∆ PQR

. Then, the angle

∠ QPR

Options

A $\dfrac{\pi}{6}$
B $\cos^{- 1}\left(\dfrac{7}{18}\right)$
C $\cos^{- 1}\left(\dfrac{1}{18}\right)$
D $\dfrac{\pi}{3}$

Correct answer

D. $\dfrac{\pi}{3}$

Step-by-step solution

Given,

Let P ( 3, 2, 3 ) , Q ( 4, 6, 2 ) and R ( 7, 3, 2 ) be the vertices of ∆ PQR . Then, the angle ∠ QPR is

$⇒ \overset{\to}{QP} = - \overset{^}{i} - 4\overset{^}{j} + \overset{^}{k}$ and $\overset{\to}{RP} = - 4\overset{^}{i} - \overset{^}{j} + \overset{^}{k}$

$cos ∠ QPR = \dfrac{\overset{\to}{QP} \cdot \overset{\to}{RP}}{\left|\overset{\to}{QP}\right|\left|\overset{\to}{RP}\right|}$

$⇒ cos ∠ QPR = \dfrac{4 + 4 + 1}{\sqrt{1 + 16 + 1}\sqrt{16 + 1 + 1}}$

$⇒ cos ∠ QPR = \dfrac{9}{18} = \dfrac{1}{2}$

$⇒ ∠ QPR = \dfrac{\pi}{3}$

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