Question

Find a vector of magnitude $4$ unite perpendicular to each of vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ ,where $\overrightarrow{a}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$ and $\overrightarrow{b}=\overrightarrow{i}+\overrightarrow{2j}+3\overrightarrow{k}$

Answer 1

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}$

Let $\overrightarrow{x}=l\hat{i}+m\hat{j}+n\hat{k}$

Given $\sqrt{l^2+m^2+n^2}=4$ ...... $\left(1\right)$

$\overrightarrow{a}+\overrightarrow{b}=2\hat{i}+3\hat{j}+4\hat{k}$

$\overrightarrow{x}.(\overrightarrow{a}+\overrightarrow{b})=0$

$2l+3m+4n=0$ ..... $\left(2\right)$

$\overrightarrow{a}-\overrightarrow{b}=-\hat{j}-2\hat{k}$

$\overrightarrow{x}(\overrightarrow{a}-\overrightarrow{b})=0$

$-m-2n=0$ ....... $\left(3\right)$

Put $m=-2n$ in $\left(2\right)$

$2l-6n+9n=0$

$l=n=-\frac{m}{2}$

Put in $\left(1\right)$

$\sqrt{l^2+(-2l{)}^2+l^2=4}$

$6l^2=16$

$l=\frac{4}{\sqrt{6}}=\frac{2\sqrt{2}}{\sqrt{3}}$

$m=\frac{-4\sqrt{2}}{\sqrt{3}}n=\frac{2\sqrt{2}}{\sqrt{3}}$

$\overrightarrow{x}=\frac{1}{\sqrt{3}}(2\sqrt{2}\hat{i}-4\sqrt{2}\hat{j}+2\sqrt{2}\hat{k})$