Question

A unit vectors coplanar with $i+j+2k$ and $i+2j+k$ and perpendicular to $i+j+k$ is

• A. $\frac{j-k}{\sqrt{2}}$
• B. $\frac{i+j+k}{\sqrt{3}}$
• C. $\frac{i+j+2k}{\sqrt{6}}$
• D. $\frac{-j+2k}{\sqrt{5}}$

Answer 1

The correct option is A $\frac{j-k}{\sqrt{2}}$

Let the vector be $\overrightarrow{a}=xi+yj+zk$ and $\overrightarrow{b}=i+j+2k,\overrightarrow{c}=i+2j+k$ and $\overrightarrow{d}=i+j+k$

Given $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are coplanar

$\Rightarrow \overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})=0\Rightarrow \overrightarrow{a}.\left(\right(i+j+2k)\times i+2j+k)=0\Rightarrow \overrightarrow{a}.(-3i+j+k)=0\Rightarrow (xi+yj+zk).(-3i+j+k)=0\Rightarrow -3x+y+z=0$ .......(i)

Also $\overrightarrow{a}$ is perpendicular to $\overrightarrow{d}$

$\Rightarrow \overrightarrow{a}.\overrightarrow{d}=0\Rightarrow (xi+yj+zk).(i+j+k)=0\Rightarrow x+y+z=0$ ......(ii)

On solving (i) and (ii), we get

$\Rightarrow x=0$ and $y=-z$
Also $\left|\overrightarrow{a}\right|=1$
$\Rightarrow \sqrt{x^2+y^2+z^2}=1\Rightarrow \sqrt{0+y^2+(-y{)}^2}=1\Rightarrow y=\pm \frac{1}{\sqrt{2}}\Rightarrow z=\mp \frac{1}{\sqrt{2}}$

Hence. the desired vector is

$\overrightarrow{a}=\pm \frac{1}{\sqrt{2}}j\mp \frac{1}{\sqrt{2}}k\Rightarrow \overrightarrow{a}=\frac{j-k}{\sqrt{2}}.\frac{-j+k}{\sqrt{2}}$

So, option A is correct.