Three Dimensional Geometry Question 14
Question 14 - 2024 (29 Jan Shift 2)
Let $\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2)$ and $\mathrm{R}(7,3,2)$ be the vertices of $\triangle \mathrm{PQR}$. Then, the angle $\angle \mathrm{QPR}$ is
(1) $\frac{\pi}{6}$
(2) $\cos ^{-1}\left(\frac{7}{18}\right)$
(3) $\cos ^{-1}\left(\frac{1}{18}\right)$
(4) $\frac{\pi}{3}$
Show Answer
Answer (4)
Solution
Direction ratio of $\mathrm{PR}=(4,1,-1)$
Direction ratio of $\mathrm{PQ}=(1,4,-1)$
Now, $\cos \theta=\left|\frac{4+4+1}{\sqrt{18} \cdot \sqrt{18}}\right|$
$\theta=\frac{\pi}{3}$
