Question

If $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k},$ $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{c}$ is a unit vector perpendicular to the vector $\overrightarrow{a}$ and coplanar with $\overrightarrow{a}$ and $\overrightarrow{b},$ then direction cosines of a vector which is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{c}$ are

• A. $(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
• B. $(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$
• C. $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$
• D. $(-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$

Answer 1

The correct option is A $(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

$\overrightarrow{c}$is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}.$
$\therefore \overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}=(x+y)\hat{i}+(x-y)\hat{j}+(y-x)\hat{k}\cdots \left(1\right)$
Since $|\overrightarrow{c}|=1,$
$\Rightarrow 3x^2+3y^2-2xy=1\cdots \left(2\right)$
Since $\overrightarrow{c}\cdot \overrightarrow{a}=0,$
$\therefore x+y+x-y+x-y=0\Rightarrow 3x=y$

From $\left(2\right),$
$x=\pm \frac{1}{2\sqrt{6}};y=\pm \frac{3}{2\sqrt{6}}$
From $\left(1\right),$
$\overrightarrow{c}=\pm (\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$

So, $\overrightarrow{d}=\pm \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{2}{\sqrt{6}}& -\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ 1& 1& -1\end{array}\right|$
$\Rightarrow \overrightarrow{d}=\pm (0,\frac{3}{\sqrt{6}},\frac{3}{\sqrt{6}})$
$\therefore$ Unit vector $\hat{d}=\frac{\overrightarrow{d}}{|\overrightarrow{d}|}=\pm (0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$