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Question:

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

Solution:

$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$

$\Rightarrow \vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\bar{a} \cdot \vec{a}=12$

$\Rightarrow|\vec{x}|^{2}-|\vec{a}|^{2}=12$

$\Rightarrow|\vec{x}|^{2}-1=12 \quad[|\vec{a}|=1$ as $\vec{a}$ is a unit vector $]$

$\Rightarrow|\vec{x}|^{2}=13$

$\therefore|\vec{x}|=\sqrt{13}$

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