Question

Consider a pyramid $OPQRS$ located in the first octant $(x\ge 0,y\ge 0,z\ge 0)$ with $O$ as origin, and $OP$ and $OR$ along the $x$-axis and the $y$-axis, respectively. The base $OPQR$ of the 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 $\sqrt{\frac{15}{2}}$

$s=(\frac{3}{2},\frac{3}{2},3)$

$\Rightarrow \overline{OQ}=3\hat{i}+3\hat{j}$

$\Rightarrow \overline{OS}=\frac{3}{2}\hat{i}+\frac{3}{2}\hat{j}+3\hat{k}$
$\cos \theta =\frac{\frac{1}{2}+\frac{1}{2}}{\sqrt{2}\sqrt{\frac{1}{2}+\frac{1}{4}+1}}=\frac{1}{\sqrt{2}\sqrt{\frac{3}{2}}}=\frac{1}{\sqrt{3}}$
$\overline{n}=\overline{OQ}\times \overline{OS}=(\hat{i}+\hat{j})\times (\hat{i}+\hat{j}+2k)=\hat{k}-2\hat{j}-\hat{k}+2\hat{i}\Rightarrow 2\hat{i}-2\hat{j}$
$x-y=\lambda$

$\Rightarrow$ Equation of the plane: $x=y$

Length of perpendicular from $(3,0,0)$ on $x-y=0$ is $\frac{3}{\sqrt{2}}$

$RS\to \frac{x-0}{\frac{3}{2}}=\frac{y-3}{-\frac{3}{2}}=\frac{z-0}{3}=\lambda$
$\Rightarrow x=\frac{3}{2}\lambda ,y=-\frac{3}{2}\lambda +3,z=3\lambda$
$T$ distance $\Rightarrow \sqrt{\frac{3}{2}-3+9}\Rightarrow \sqrt{\frac{15}{2}}$
$D=\frac{9}{4}{\lambda }^2+(3-\frac{3}{2}\lambda {)}^2+9{\lambda }^2$

$=\frac{27}{2}{\lambda }^2-9\lambda +9$

$\Rightarrow \lambda =\frac{9}{27}=\frac{1}{3}$