Question

Consider a pyramid $OPQRS$ located in the first octant $(x\ge 0,y\ge 0,z\ge 0)$ with origin, and $OP$ and $OR$ along the $x-$axis and the $y-$axis, respectively. The base $OPQR$ of pyramid is a square with $OP=3.$ The point $S$ is directly above the mid-point $T$ of diagonal $OQ$ such that $TS=3.$Then

• A. the acute angle between $OQ$ and $OS$ is $\frac{\pi }{3}$
• B. the equation of the plane containing the triangle $OQS$ is $x-y=0$
• C. the length of the perpendicular from $P$ to the plane containing the triangle $OQS$ is $\frac{3}{\sqrt{2}}$
• D. the perpendicular distance from $O$ to the straight line containing $RS$ is $\sqrt{\frac{15}{2}}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B the equation of the plane containing the triangle $OQS$ is $x-y=0$
C the length of the perpendicular from $P$ to the plane containing the triangle $OQS$ is $\frac{3}{\sqrt{2}}$
D the perpendicular distance from $O$ to the straight line containing $RS$ is $\sqrt{\frac{15}{2}}$

$\cos \theta =\frac{\overrightarrow{OQ}.\overrightarrow{OS}}{|\overrightarrow{OQ}||\overrightarrow{OS}|}=\frac{1}{\sqrt{3}}\theta ={\cos }^{-1}\frac{1}{\sqrt{3}}$
Also, equation of the plane containing the triangle $OQS$
$\left|\begin{array}{ccc}x& y& z\\ 3& 3& 0\\ \frac{3}{2}& \frac{3}{2}& 3\end{array}\right|=0\Rightarrow x-y=0$

and the length of the perpendicular from $P$ to the plane containing the triangle $OQS$ $=|\frac{3-0}{\sqrt{1^2+(-1{)}^2}}|=\frac{3}{\sqrt{2}}$

Then perpendicular distance from $O$ to the straight line containing $RS$
$\overrightarrow{RM}=\overrightarrow{OR}.\frac{\overrightarrow{RS}}{|\overrightarrow{RS}|}$
$\overrightarrow{RM}=\sqrt{\frac{3}{2}}....\left(i\right)$
$\overrightarrow{OR}=\mathrm{3...}\left(ii\right)$
$\overrightarrow{OM}=\sqrt{{\overrightarrow{OR}}^2-{\overrightarrow{RM}}^2}$
$\overrightarrow{OM}=\sqrt{\frac{15}{2}}$