If \(\vec{a}\times\vec{b}=3\hat{\imath}+2\hat{\jmath}+6\hat{k}\) and \(\vec{a}\cdot\vec{b}=7\sqrt{3}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\pi}{6}\)
If \(\theta\) is the angle between the vectors, the cross product and dot product give
\(|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta\)
\(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\)
Dividing the first by the second:
\( \displaystyle \tan\theta = \frac{|\vec{a}\times\vec{b}|}{\vec{a}\cdot\vec{b}} \)
From the given cross product:
\(|\vec{a}\times\vec{b}| = \sqrt{3^2+2^2+6^2}\)
\( = \sqrt{49} = 7\)
So
\( \displaystyle \tan\theta = \frac{7}{7\sqrt{3}} = \frac{1}{\sqrt{3}} \)
Since \(\vec{a}\cdot\vec{b}\) is positive, \(\theta\) is acute, and the only such angle is \(\theta = \frac{\pi}{6}\).