Question

If Lines $OA,OB$ are drawn from $O$(origin) with direction cosines proportional to $(1,-2,-1),(3,-2,3)$. Then the direction cosines of the normal to the plane $AOB$ is

• A. $(-\frac{1}{\sqrt{29}},-\frac{2}{2\sqrt{29}},\frac{4}{\sqrt{29}})$
• B. $(\frac{2}{\sqrt{29}},-\frac{3}{2\sqrt{29}},\frac{4}{\sqrt{29}})$
• C. $(-\frac{3}{\sqrt{29}},-\frac{4}{2\sqrt{29}},\frac{2}{\sqrt{29}})$
• D. $(-\frac{4}{\sqrt{29}},-\frac{3}{\sqrt{29}},\frac{2}{\sqrt{29}})$

Answer 1

The correct option is D $(-\frac{4}{\sqrt{29}},-\frac{3}{\sqrt{29}},\frac{2}{\sqrt{29}})$

$\overrightarrow{OA}=\hat{i}-2\hat{j}-\hat{k}\overrightarrow{OB}=3\hat{i}-2\hat{j}+3\hat{k}$
$D.r^{\prime }s$ of normal to the plane $AOB$ is :
$\overrightarrow{OA}\times \overrightarrow{OB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& -1\\ 3& -2& 3\end{array}\right|=-8\hat{i}-6\hat{j}+4\hat{k}$
So, $D.c^{\prime }s$ of normal is
$(-\frac{8}{2\sqrt{29}},-\frac{6}{2\sqrt{29}},\frac{4}{2\sqrt{29}})=(-\frac{4}{\sqrt{29}},-\frac{3}{\sqrt{29}},\frac{2}{\sqrt{29}})$