Let \(\vec{a}\) and \(\vec{b}\) be unit vectors such that the vectors \(\vec{a}+4\vec{b}\) and \(3\vec{a}-2\vec{b}\) are perpendicular. Then the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{\pi}{3}\)
Since the two vectors are perpendicular, their scalar product is zero. Expanding by the distributive law:
\((\vec{a}+4\vec{b})\cdot(3\vec{a}-2\vec{b}) = 0\)
\(3|\vec{a}|^2-2\,\vec{a}\cdot\vec{b}+12\,\vec{b}\cdot\vec{a}-8|\vec{b}|^2 = 0\)
With \(|\vec{a}| = |\vec{b}| = 1\):
\(3+10\,\vec{a}\cdot\vec{b}-8 = 0\)
\( \displaystyle \vec{a}\cdot\vec{b} = \frac{1}{2} \)
For unit vectors, \(\vec{a}\cdot\vec{b} = \cos\theta\):
\( \displaystyle \cos\theta = \frac{1}{2}, \quad \theta = \frac{\pi}{3} \)