Question

If $\overrightarrow{a}=\overline{i}+\overline{j}+\overline{k},\overline{c}=\overline{j}-\overline{k},$, then find vector $\overline{b}$ such that $\overline{a}.\overline{b}=\overline{0},\overline{c}.\overline{b}=0and\left|\overline{b}\right|=\sqrt{6}$

Answer 1

Let vector $\overline{b}=p\hat{i}+q\hat{j}+r\hat{k}$

$\overline{a}.\overline{b}=0\Rightarrow (\hat{i}+\hat{j}+\hat{k}).(p\hat{i}+q\hat{j}+r\hat{k})=0$

$\Rightarrow (p+q+r)=0$----- $\left(1\right)$

$\overline{c}.\overline{b}=0\Rightarrow (\hat{j}-\hat{k}).(p\hat{i}+q\hat{j}+r\hat{k})=0$

$\Rightarrow q-r=0$

$\Rightarrow q=r$------ $\left(2\right)$

$|b|=\sqrt{6}\Rightarrow \sqrt{p^2+q^2+r^2}=\sqrt{6}$

$\Rightarrow p^2+q^2+r^2=6$

$\Rightarrow p^{2}2r^2=6$ ------- $\left(3\right)$

$p+q+r=0\Rightarrow (p+2r)=0$

$\Rightarrow P=-2r$ ------- $\left(4\right)$

Put $\left(4\right)$ in $\left(1\right)$

$(-2r{)}^2+2r^2=6$

$\Rightarrow 6r^2=6\Rightarrow r=\pm 1$

if $r=1$

$p=-2$

$q=1$

$\Rightarrow \overline{b}=-2\hat{i}+\hat{j}+\hat{k}$

if $r=-1$

$p=2$

$q=1$

$\Rightarrow \overline{b}=2\hat{i}+\hat{j}+\hat{k}$

$\therefore \overline{b}=\pm (2\hat{i}-\hat{j}-\hat{k})$