Question

Consider two fixed points $A(1,0,0)$ and $P(1,0,1)$. If $OA$ (where ' $O$ ' is origin) is rotated through ' $O$ ' in xy-plane by a variable angle $\theta$, to reach a new position $OB$, then which of the following statement(s) is(are) correct?

• A. Maximum volume of tetrahedron OAPB is $\frac{1}{6}$.
• B. Maximum volume of tetrahedron $OAPB$ is unity.
• C. The distance of plane OPB from point A is $\frac{|\sin \theta |}{\sqrt{\left(1+{\sin }^2\theta \right)}}$.
• D. Equation of the plane OPB is $(z-x)\sin \theta +y\cos \theta =0$.

Answer 1

Answer: (A), (C), (D)

Volume of tetrahedron $OAPB=\frac{1}{6}|[\overrightarrow{OA}\overrightarrow{OB}\overrightarrow{OP}]|$
$=\frac{1}{6}\left|\begin{array}{ccc}1& 0& 0\\ \cos \theta & \sin \theta & 0\\ 1& 0& 1\end{array}\right||=|\frac{\sin \theta }{6}\mid$

Also, $h=\overrightarrow{OP}\times \overrightarrow{OB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 1\\ \cos \theta & \sin \theta & 0\end{array}\right|=-\sin \theta \hat{i}-\hat{j}(-\cos \theta )+\hat{k}(\sin \theta )$
As, $r.h=0\Rightarrow -x\sin \theta +y\cos \theta +z\sin \theta =0$
$d=\frac{|-\sin \theta |}{\sqrt{\left({\sin }^2\theta +{\cos }^2\theta +{\sin }^2\theta \right)}}=\frac{|\sin \theta |}{\sqrt{\left(1+{\sin }^2\theta \right)}}$