Question

The coordinates of the vertices $P,Q$ and $R$ of a triangle $PQR$ are $(1,-1,1),(3,-2,2)$ and $(0,2,6)$ respectively. If $\angle RQP=\theta$, then what is $\angle PRQ$ equal to

• A. ${30}^o+\theta$
• B. ${45}^o-\theta$
• C. ${60}^o-\theta$
• D. ${90}^o-\theta$

Answer 1

Answer: (D)

${90}^o-\theta$

Given $P(1,-1,1),Q(3,-2,2),R(0,2,6)$

$\therefore \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=3\hat{i}-2\hat{j}+2\hat{k}-(\hat{i}-\hat{j}+\hat{k})=2\hat{i}-\hat{j}+\hat{k}$ and

$\overrightarrow{PR}=\overrightarrow{OR}-\overrightarrow{OP}=0\hat{i}+2\hat{j}+6\hat{k}-(\hat{i}-\hat{j}+\hat{k})=-\hat{i}+3\hat{j}+5\hat{k}$

$\Rightarrow \overrightarrow{PQ}\cdot \overrightarrow{PR}=-2-3+5=0$

$\Rightarrow \angle QPR={90}^{\circ }$

We have $\angle RQP=\theta$

$\therefore \angle PRQ={180}^{\circ }-({90}^{\circ }+\theta )={90}^{\circ }-\theta$