If â is a unit vector such that $(\bar{x}-\hat{a}) \cdot(\bar{x}+\hat{a})=8$, then $|\bar{x}|=$
If â is a unit vector such that $(\bar{x}-\hat{a}) \cdot(\bar{x}+\hat{a})=8$, then $|\bar{x}|=$
$\pm 3$
$2 \sqrt{2}$
3
$\pm \sqrt{7}$
Solution
We have $(\overline{\mathrm{x}}-\hat{\mathrm{a}}) \cdot(\overline{\mathrm{x}}+\hat{\mathrm{a}})=8$
$\therefore|\overline{\mathrm{x}}|^2-|\hat{\mathrm{a}}|^2=8 \Rightarrow|\overline{\mathrm{x}}|^2=8+1=9 \Rightarrow|\overline{\mathrm{x}}|=3$