Question

A unit vector coplanar with $\overrightarrow{1}+\overrightarrow{j}+3\overline{k}and\overline{i}+3\overline{j}+\overrightarrow{k}$ and perpendicular to $\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$ is

• A. $\frac{1}{\sqrt{2}}(j+k)$
• B. $\frac{1}{\sqrt{3}}(\overline{i}-\overrightarrow{j}+\overrightarrow{k})$
• C. $\frac{1}{\sqrt{2}}(j-k)$
• D. $\frac{1}{\sqrt{3}}(\overline{i}+\overrightarrow{j}-\overrightarrow{k})$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{2}}(j-k)$

Let the vector be $a\hat{i}+b\hat{j}+c\hat{k}=\overrightarrow{v}$

then $(a\hat{i}+b\hat{j}+c\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=0$

$a+b+c=0$ ...(1)

Now, it is co planer with the two vectors

So, $\left|\begin{array}{ccc}a& b& c\\ 1& 1& 3\\ 1& 3& 1\end{array}\right|=0$

$-8a+2b+2c=0$ ...(2)

On solving $e{q}^n$ 1 and 2

$-10a=0$

$\Rightarrow a=0$

Put $a=0$ in eqn (1)

$b+c=0$

$b=-c$

$\therefore \overrightarrow{v}=b\hat{j}-b\hat{k}$

$\therefore$ unit vector $=\frac{1}{\sqrt{2}b}(b\hat{j}-b\hat{k})$

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