3D Geometry
Angle in Triangle in 3D
nta_pyq_2024_jan
Grade 12
Question:
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
$\dfrac{\pi}{6}$
$\cos^{-1}\!\left(\dfrac{7}{18}\right)$
$\cos^{-1}\!\left(\dfrac{1}{18}\right)$
$\dfrac{\pi}{3}$
Step-by-Step Solution
Key Concept: $\vec{PQ}=(1,4,-1)$, $\vec{PR}=(4,1,-1)$. $\cos(\angle QPR)=\frac{\vec{PQ}\cdot\vec{PR}}{|\vec{PQ}||\vec{PR}|}=\frac{4+4+1}{\sqrt{18}\cdot\sqrt{18}}=\frac{9}{18}=\frac{1}{2}$.
$\angle QPR=\pi/3$.
Correct Answer: 4