If a is a unit vector and $(\vec x-\vec a).(\vec x+\vec a) = 15$, then the value of $|\vec x|$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
Given: $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 15$
Use identity: $(\vec{p} - \vec{q}) \cdot (\vec{p} + \vec{q}) = \vec{p} \cdot \vec{p} - \vec{q} \cdot \vec{q}$
So, $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = \vec{x} \cdot \vec{x} - \vec{a} \cdot \vec{a}$
$|\vec{x}|^2 - |\vec{a}|^2 = 15$
Since $\vec{a}$ is a unit vector, $|\vec{a}|^2 = 1$
$|\vec{x}|^2 - 1 = 15$
$|\vec{x}|^2 = 16$
$|\vec{x}| = 4$