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

3D Geometry PYQ — JEE Main 2024 · 29 Jan · Shift 2 · Q39 with Solution

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Correct answer: D

To find $\angle QPR$ , we need the angle between the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ .
Compute the vectors:

\overrightarrow{PQ}=Q-P=(4-3,\,6-2,\,2-3)=(1,4,-1)

\overrightarrow{PR}=R-P=(7-3,\,3-2,\,2-3)=(4,1,-1)

Use the dot product formula:

\overrightarrow{PQ}\cdot\overrightarrow{PR}=|PQ|\,|PR|\cos\theta

where $\theta=\angle QPR$ .

Compute the dot product:

(1)(4)+(4)(1)+(-1)(-1)=4+4+1=9

Compute the magnitudes:

|PQ|=\sqrt{1^2+4^2+(-1)^2}=\sqrt{1+16+1}=\sqrt{18}

|PR|=\sqrt{4^2+1^2+(-1)^2}=\sqrt{16+1+1}=\sqrt{18}

Therefore,

\cos\theta=\frac{9}{\sqrt{18}\cdot\sqrt{18}}=\frac{9}{18}=\frac{1}{2}

So,

\theta=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}

Check options:

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

Hence, the correct answer is D.

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