If \(|\vec{a}| = 6\) and the vectors \(\vec{a}+3\vec{b}\) and \(\vec{a}-3\vec{b}\) are perpendicular, then \(|\vec{b}|\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
Since the two vectors are perpendicular, their scalar product is zero:
\((\vec{a}+3\vec{b})\cdot(\vec{a}-3\vec{b}) = 0\)
Expanding, the middle terms \(-3\,\vec{a}\cdot\vec{b}\) and \(+3\,\vec{b}\cdot\vec{a}\) cancel because the scalar product is commutative:
\(|\vec{a}|^2-9|\vec{b}|^2 = 0\)
\(36 = 9|\vec{b}|^2\)
\(|\vec{b}|^2 = 4\)
\(|\vec{b}| = 2\)