If \(\vec{a}+\vec{b}+\vec{c} = \vec{0}\), \(|\vec{a}| = 3\), \(|\vec{b}| = 8\) and \(|\vec{c}| = 7\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2\pi}{3}\)
From \(\vec{a}+\vec{b}+\vec{c} = \vec{0}\), we have \(\vec{a}+\vec{b} = -\vec{c}\). Squaring the magnitudes:
\(|\vec{a}+\vec{b}|^2 = |\vec{c}|^2\)
\(|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b} = |\vec{c}|^2\)
\(9+64+2\,\vec{a}\cdot\vec{b} = 49\)
\(\vec{a}\cdot\vec{b} = -12\)
Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\):
\( \displaystyle \cos\theta = \frac{-12}{3\times 8} = -\frac{1}{2} \)
\( \displaystyle \theta = \frac{2\pi}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{3}\) — Incorrect. This takes \(\cos\theta = +\displaystyle\frac{1}{2}\); the sign is lost. With \(\theta = \displaystyle\frac{\pi}{3}\), \(|\vec{a}+\vec{b}|^2 = 9+64+24 = 97\), not 49.
Option (2) \(\displaystyle\frac{\pi}{6}\) — Incorrect. \(\cos\displaystyle\frac{\pi}{6} = \frac{\sqrt{3}}{2}\), not \(-\displaystyle\frac{1}{2}\).
Option (4) \(\displaystyle\frac{5\pi}{6}\) — Incorrect. \(\cos\displaystyle\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\), not \(-\displaystyle\frac{1}{2}\).