Question

Find the angle between the vectors if \(|\vec{a}|=|\vec{b}|=3\sqrt{2}\) and \(\vec{a}.\vec{b}=9\sqrt{3}\).

(a) \(\frac{π}{6}\)

(b) \(\frac{π}{5}\)

(c) \(\frac{π}{3}\)

(d) \(\frac{π}{2}\)

This question was addressed to me in an interview.

Asked question is from Product of Two Vectors-2 in section Vector Algebra of Mathematics – Class 12

Answer 1

The correct answer is (a) \(\frac{π}{6}\)

Explanation: We know that, \(\vec{a}.\vec{b}=|\vec{a}|.|\vec{b}| \,cos⁡θ\)

Given that, \(|\vec{a}|=|\vec{b}|=3\sqrt{2} \,and \,\vec{a}.\vec{b}=9\sqrt{3}\)

\(cos⁡θ=\frac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}=\frac{9\sqrt{3}}{(3\sqrt{2})^2}=\frac{\sqrt{3}}{2}\)

\(θ=cos^{-1}\frac{\sqrt{3}}{2}=\frac{π}{6}\).