Find |x|, if for a unit vector a,(x-a) (x+a)=12 - Vidyadip

Question

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

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Answer

Given, vector $\vec a $ is a unit vector
$\therefore|\vec{\mathrm{a}}|=1$
$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$
$\Rightarrow \vec{\mathrm{x}} \cdot \vec{\mathrm{x}}+\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}-\vec{\mathrm{a}} \cdot \vec{\mathrm{x}}-\vec{\mathrm{a}} \cdot \vec{\mathrm{a}}=12$
$\Rightarrow|\vec{\mathrm{x}}|^{2}+\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}-\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}-|\vec{\mathrm{a}}|^{2}=12$ [Since $\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}=\vec{\mathrm{a}} \cdot \vec{\mathrm{x}}$]
$\Rightarrow|\vec{\mathrm{x}}|^{2}-|\vec{\mathrm{a}}|^{2}=12$
$\Rightarrow|\vec{\mathrm{x}}|^{2}-1^{2}=12$
$\Rightarrow|\vec{\mathrm{x}}|^{2}=12+1=13$
$\Rightarrow|\vec{\mathrm{x}}|=\sqrt{13}$
$\therefore|\vec{\mathrm{x}}|=\sqrt{13}$

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