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

Answer 1

Answer: (A)

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

Let the unit vector be

$\overrightarrow{d}=x\hat{i}+y\hat{j}+z\hat{k}$

then.,

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

And,

Let the given vectors are

$\begin{array}{c}\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}\\ \\ \overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}\\ \\ \overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}\end{array}$

Since, $\overrightarrow{d}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$

So,

$\begin{array}{c}\left|\begin{array}{ccc}x& y& z\\ 1& 1& 2\\ 1& 2& 1\end{array}\right|=0\\ \\ x\left(1-4\right)-y\left(1-2\right)+z\left(2-1\right)=0\\ \\ 3x-y-z=0---\left(2\right)\end{array}$

Since, $\overrightarrow{d}$ is perpendicular to $\overrightarrow{c}$

$x\times 1+y\times 1+z\times 1=0$

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

Adding $\left(2\right)$ and $\left(3\right)$

$4x=0$

$x=0$

Putting $x=0$ in $\left(2\right)$

$z=-y$

putting $x=0,z=-y$ in $\left(1\right)$

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

$0+y^2+y^2=1$

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

Therefore, required unit vector is

$\overrightarrow{d}=0\hat{i}+\frac{1}{\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}$

$=\frac{\hat{j}-\hat{k}}{\sqrt{2}}$