The angle between \(\vec{a}\) and \(\vec{b}\), where \(\left|\vec{a}\right|=\sqrt{3}\) and \(\left|\vec{b}\right|=2\) and \(\vec{a}\cdot\vec{b}=\sqrt{6}\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\pi/4\)
From the dot product, \(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\):
\( \displaystyle \cos\theta = \frac{\sqrt{6}}{\sqrt{3}\times 2} \)
\( \displaystyle = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)
Hence \(\theta = \frac{\pi}{4}\).
NCERT (2026-27 edition):
Product of Two Vectors
