If \(\vec{a}\) and \(\vec{b}\) are non-zero vectors such that \(|\vec{b}| = 2|\vec{a}|\) and \(\vec{a}+\vec{b}\) is perpendicular to \(\vec{a}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2\pi}{3}\)
Since \(\vec{a}+\vec{b}\) is perpendicular to \(\vec{a}\), their scalar product is zero:
\((\vec{a}+\vec{b})\cdot\vec{a} = 0\)
\(|\vec{a}|^2+\vec{a}\cdot\vec{b} = 0\)
\(\vec{a}\cdot\vec{b} = -|\vec{a}|^2\)
Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). With \(|\vec{b}| = 2|\vec{a}|\):
\( \displaystyle \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{-|\vec{a}|^2}{2|\vec{a}|^2} = -\frac{1}{2} \)
\( \displaystyle \theta = \frac{2\pi}{3} \)