Question
$\text{Find}\ \big|\vec{x}\big|,$ if for a unit unit vector $\vec{a},\ (\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=12.$

Answer

$\text{Given:}\ \vec{a}\ \text{is a unit vector}\ \Rightarrow\ \big|\vec{a}\big|=1\ \ .......(\text{i})$
$\big(\vec{x}-\vec{a}\big)\big(\vec{a}+\vec{a}\big)=12\ $ $\Rightarrow\ \vec{x}.\vec{x}+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\vec{a}.\vec{a}=12$
$\Rightarrow\ \ \big|\vec{x}\big|^2+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12\ $ $\Rightarrow\ \big|\vec{x}\big|^2+\vec{a}.\vec{x}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12$
$ \Rightarrow\ \ \big|\vec{x}\big|^2-\big|\vec{a}\big|^2=12$
$\text{Putting}\ \big|\vec{a}\big|=1\ \text{from eq. (i),}\ \ \ \ \big|\vec{x}\big|^2-1=12$
$\Rightarrow\ \big|\vec{x}\big|^2=13\ \ \Rightarrow\ \ \big|\vec{x}\big|=13$

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