Question:
If $\vec{a}$ is a unit vector such that $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=15$ find $|\vec{x}|$.
Solution:
$(\vec{x}-\vec{a})(\vec{x}+\vec{a})=15$
$\left.\Rightarrow\left|\vec{x}_{\left.\right|^{2}}-\right| \vec{a}\right|^{2}=15$
$\Rightarrow|\vec{x}|^{2}=|\vec{a}|^{2}+15$
Now, a is a unit vector,
$\Rightarrow|\vec{a}|=1$
$\Rightarrow|\vec{x}|^{2}=1^{2}+15$
$\Rightarrow|\vec{x}|^{2}=16$
$\Rightarrow|\vec{x}|=4$
Ans: $\left.\right|^{2} \mid=4$
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