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 $\Delta OQS$ is x - y = 0.
• C. the length of the perpendicular from P to the plane containing the $\Delta OQS$ 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 $\Delta OQS$ is x - y = 0.

C

the length of the perpendicular from P to the plane containing the $\Delta OQS$ is $\frac{3}{\sqrt{2}}$

D

the perpendicular distance from O to the straight line containing RS is $\sqrt{\frac{15}{2}}$

Given, square base OP=OR=3
$\therefore P(3,0,0),R=(0,3,0)$

Also, mid-point of OQ is $T(\frac{3}{2},\frac{3}{2},0)$
Since, S is derectly above the mid-point T of diagonal OQ and ST = 3.
i.e.~~~$s(\frac{3}{2},\frac{3}{2},3)$
Here, DR's of OQ (3, 3, 0) and DR's of $OS(\frac{3}{2},\frac{3}{2},3)$.
$\therefore cos\theta =\frac{\frac{9}{2}+\frac{9}{2}}{\sqrt{9+9+0}\sqrt{\frac{9}{4}+\frac{9}{4}+9}}=\frac{9}{\sqrt{18}.\sqrt{\frac{27}{2}}}=\frac{1}{\sqrt{3}}$
$\therefore Option\left(a\right)isincorrect.$
Now, equation of the plane containing the $\Delta OQS$ is
$\left|\begin{array}{ccc}xyz& & \\ 330& & \\ \frac{3}{2}\frac{3}{2}3& & \end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}xyz& & \\ 110& & \\ 112& & \end{array}\right|=0\Rightarrow x(2-0)-y(2-0)+z(1-1)=0\Rightarrow 2x-2y=0orx-y=0\therefore Option\left(b\right)iscorrect$
Now, length of the perpendicular from P(3, 0, 0) to the plane containing
$\Delta OQS$ is
$\frac{|3-0|}{\sqrt{1+1}}=\frac{3}{\sqrt{2}}$
$\therefore Option\left(c\right)iscorrect.$
Here, equation of RS is
$\frac{x-0}{\frac{3}{2}}=\frac{y-3}{\frac{-3}{2}}=\frac{z-0}{3}=\lambda \Rightarrow x=\frac{3}{2}\lambda ,y=-\frac{3}{2}\lambda +3,z=3\lambda$
To find the distance from O(0, 0, 0) to RS.
Let M be the foot of perpendicular.

$\because \overline{OM}\angle \overline{RS}\Rightarrow \overline{OM}.\overline{RS}=0\Rightarrow \frac{9\lambda }{4}-\frac{3}{2}(3-\frac{3\lambda }{2})+3\left(3\lambda \right)=0\Rightarrow \lambda =\frac{1}{3}\therefore M(\frac{1}{2},\frac{5}{2},1)\Rightarrow OM=\sqrt{\frac{1}{4}+\frac{25}{4}+1}=\sqrt{\frac{30}{4}}=\sqrt{\frac{15}{2}}\therefore Option\left(d\right)iscorrect.$