Question
If ${\vec a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8$, then find $\left| {\vec x} \right|$.

Answer

$\left| {\vec a} \right| = 1$
$\left( {\vec x - \vec a} \right).\left( {\vec x + \vec a} \right) = 8$
${\left| {\vec x} \right|^2} - |\vec a|^2 = 8$
$|\vec x|^2-1=8$
${\left| {\vec x} \right|^2} = 9$
$\left| {\vec x} \right| = 3$

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