Question

If $a=\lambda (\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}),b=\mu .(\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}),$ and $c$ are unit vectors perpendicular to the vector $a$ and coplanar with $a$ and $b$, then a unit vector $d$ perpendicular to both $a$ and $c$ is

• A. $\frac{1}{\sqrt{6}}(2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})$
• B. $\frac{1}{\sqrt{2}}(\overrightarrow{j}+\overrightarrow{k})$
• C. $\frac{1}{\sqrt{6}}(\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k})$
• D. $\frac{1}{\sqrt{2}}(\overrightarrow{j}-\overrightarrow{k})$

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}(\overrightarrow{j}+\overrightarrow{k})$

Let $c=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$ Since $a,b,c$ are coplanar, so
$\lambda \mu =\left|\begin{array}{ccc}x& y& z\\ 1& 1& -1\\ 1& -1& 1\end{array}\right|=0\Rightarrow 0\cdot x-2y-2z=0$
$\Rightarrow y+z=0$
Also $c$ is perpendicular to $a$ so $x+y-z=0$
Therefore $\frac{x}{-2}=\frac{y}{1}=\frac{z}{-1}$ and $x^2+y^2+z^2=1$
(c being a unit vector). We have $x=-\frac{2}{\sqrt{6}}.y=\frac{1}{\sqrt{6}},z=-\frac{1}{\sqrt{6}}$
Thus $c=\frac{1}{\sqrt{6}}(-2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})$ and
$d=\frac{1}{|a\times c|}a\times c=\frac{1}{\sqrt{6}}\frac{\lambda }{|a\times c|}$
$\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 1& -1\\ -2& 1& -1\end{array}\right|$
$=\frac{1}{\sqrt{6}}\frac{\lambda }{|a\times c|}(0.\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k})=\frac{1}{\sqrt{2}}(\overrightarrow{j}+\overrightarrow{k})$