If \(|\vec{a}| = \sqrt{3}\), \(|\vec{b}| = 2\) and \(\vec{a}\cdot\vec{b} = \sqrt{6}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{4}\)
By the definition of the scalar product, \(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\). So
\( \displaystyle \cos\theta = \frac{\sqrt{6}}{\sqrt{3}\times 2} \)
\( \displaystyle = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)
\( \displaystyle \theta = \frac{\pi}{4} \)