If \(\vec{a}\) is a unit vector and $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8$ then \(|\vec{x}|\) is:

3

Given $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8$ & $\vec{a}$ is a unit vector

which means $|\vec{a}|=1$

So, $(\vec{x}-\vec{a})(\vec{x}+\vec{a})⇒|\vec{x}|^2-|\vec{a}|^2$

$⇒|\vec{x}|^2-|\vec{a}|^2=8$

$⇒|\vec{x}|^2-1=8⇒|\vec{x}|^2=8+1⇒9$

$∴ |\vec{x}|$ is positive, so $|\vec{x}|=3$

Option 2 is correct.