Question

A unit vector coplanar with $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$ and perpendicular to $\hat{i}+\hat{j}+\hat{k}$ is_________

• A. $\frac{-1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$
  
• B. $\frac{-1}{\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}$
  
• C. $\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$
  
• D. None of these

Answer 1

The correct option is A

$\frac{-1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$

Let the unit vector be $x\hat{i}+y\hat{j}+z\hat{k}$

then $x^2+y^2+z^2=1$ $-\left(1\right)$

since this vector is coplaner with $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$

therefore

$\Rightarrow$$\left|\begin{array}{ccc}x& y& z\\ 1& 1& 2\\ 1& 2& 1\end{array}\right|=0$

$\Rightarrow x\left(1-4\right)\left(1-2\right)+z\left(2-1\right)=0$

$\Rightarrow -3x+y+z=0$

$\Rightarrow 3x-y-z=0-\left(2\right)$

also this unit vector is perpendicular to $\hat{i}+\hat{j}+\hat{k}$

so, $x+y+z=0-\left(3\right)$

Adding 2 and 3 we get

$4x=0$

$\Rightarrow$$x=0$

from 3

$x+y+z=0$

$\Rightarrow y+z=0$

$\Rightarrow y=-z$

putting value of $x$ and $y$ in $\left(1\right)$

$x^2+y^2+z^2=1$

$\Rightarrow 0+z^2+z^2=1$

$\Rightarrow 2z^2=1$

$\Rightarrow z^2=\frac{1}{2}$

$\Rightarrow z=\pm \frac{1}{\sqrt{2}}$

if $z=\frac{1}{\sqrt{2}}$$\Rightarrow y=\frac{-1}{\sqrt{2}}$

and if $z=\frac{-1}{\sqrt{2}}$$\Rightarrow y=\frac{1}{\sqrt{2}}$

thus the required unit vector is

$\frac{-1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$ or $\frac{1}{\sqrt{2}}\hat{j}+\frac{-1}{\sqrt{2}}\hat{k}$

Therefore the required vector is $\frac{-1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$