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If $\mathrm{P}(3,2,6), \mathrm{Q}(1,4,5)$ and $\mathrm{R}(3,5,3)$ are the vertices of $\Delta \mathrm{PQR}$, then $\mathrm{m} \angle \mathrm{PQR}$ is
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The correct answer is:
$90^{\circ}$
We have $P \equiv(3,2,6) ; Q \equiv(1,4,5)$ and $R \equiv(3,5,3)$
d.r. of $\mathrm{PQ}$ are $-2,2,-1$ and d.r. of $\mathrm{QR}$ are $2,1,-2$
We find that : $(-2)(2)+2(1)+(-1)(-2)=-4+2+2=0$
$\therefore \mathrm{PQ} \perp \mathrm{QR} \Rightarrow \mathrm{m} \angle \mathrm{PQR}=90^{\circ}$
d.r. of $\mathrm{PQ}$ are $-2,2,-1$ and d.r. of $\mathrm{QR}$ are $2,1,-2$
We find that : $(-2)(2)+2(1)+(-1)(-2)=-4+2+2=0$
$\therefore \mathrm{PQ} \perp \mathrm{QR} \Rightarrow \mathrm{m} \angle \mathrm{PQR}=90^{\circ}$
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