Question

If $\overrightarrow{a}$ is a unit vector such that $(\overrightarrow{x}-\overrightarrow{a})(\overrightarrow{x}+\overrightarrow{a})=8$, find $|\overrightarrow{x}|$.

Answer 1

Given $(\overrightarrow{x}-\overrightarrow{a})(\overrightarrow{x}+\overrightarrow{a})=8$ and $\overrightarrow{a}$ is unit vector , which implies $|\overrightarrow{a}|=1$

We have $(\overrightarrow{x}-\overrightarrow{a})(\overrightarrow{x}+\overrightarrow{a})=\overrightarrow{x}.\overrightarrow{x}+\overrightarrow{x}.\overrightarrow{a}-\overrightarrow{x}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{a}=|\overrightarrow{x}{|}^2-|\overrightarrow{a}{|}^2$

$⟹|\overrightarrow{x}{|}^2-|\overrightarrow{a}{|}^2=8$

$⟹|\overrightarrow{x}{|}^2-1=8$

$⟹|\overrightarrow{x}{|}^2=9$

Since $|\overrightarrow{x}|$ is positive, we get $|\overrightarrow{x}|=3$

Therefore the value of $|\overrightarrow{x}|$ is $3$.