If \( |\vec{a}|=2,|\vec{b}|=7 \) and \( \vec{a}\times\vec{b}=3\hat{\imath}+2\hat{\jmath}+6\hat{k} \), then angle between \( \vec{a} \) and \( \vec{b} \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\pi}{6} \)
Since \( |\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta \), the modulus of the cross product gives \( \sin\theta \).
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{3^2+2^2+6^2} \)
\( \displaystyle =\sqrt{49}=7 \)
\( \displaystyle 2\times 7\times\sin\theta=7 \)
\( \displaystyle \sin\theta=\frac{1}{2} \)
So θ = \( \frac{\pi}{6} \) or \( \frac{5\pi}{6} \). Only \( \frac{\pi}{6} \) is among the options.