Question:
If \(\vec{a}\) and \(\vec{b}\) are vectors such that \(|\vec{a}| = \sqrt{3}\), \(|\vec{b}| = 2\) and \(\vec{a} \cdot \vec{b} = \sqrt{6}\) then the angle between \(\vec{a}\) and \(\vec{b}\) is
यदि \(\vec{a}\) और \(\vec{b}\) ऐसे सदिश हैं कि \(|\vec{a}| = \sqrt{3}\), \(|\vec{b}| = 2\) और \(\vec{a} \cdot \vec{b} = \sqrt{6}\) है, तो \(\vec{a}\) और \(\vec{b}\) के बीच का कोण क्या है?
Options:
- \(\dfrac{\pi}{6}\)
- \(\dfrac{2\pi}{3}\)
- \(\dfrac{\pi}{4}\)
- \(\dfrac{\pi}{3}\)
- \(\dfrac{\pi}{6}\)
- \(\dfrac{2\pi}{3}\)
- \(\dfrac{\pi}{4}\)
- \(\dfrac{\pi}{3}\)
Correct Answer: c)
Explanation:
\(\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \dfrac{\sqrt{6}}{\sqrt{3} \times 2} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}} \implies \theta = \dfrac{\pi}{4}\).
\(\cos\theta = \dfrac{\sqrt{6}}{2\sqrt{3}} = \dfrac{1}{\sqrt{2}} \implies \theta = \dfrac{\pi}{4}\)।