If \(\vec{a}, \vec{b}\) and \(\vec{c}\) be vectors such that \(\vec{a}+\vec{b}+\vec{c}=\vec{0}, |\vec{a}| = 3, |\vec{b}| = 5\) and \(|\vec{c}|= 7\), then the angle between \(\vec{a}\) and \(\vec{b}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{3}\)
Since \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), we have \(\vec{a} + \vec{b} = -\vec{c}\). Squaring the magnitudes, using \(|\vec{u}|^2 = \vec{u}\cdot\vec{u}\):
\(|\vec{a} + \vec{b}|^2 = |\vec{c}|^2\)
\(|\vec{a}|^2 + |\vec{b}|^2 + 2\,\vec{a}\cdot\vec{b} = |\vec{c}|^2\)
\(9 + 25 + 2\,\vec{a}\cdot\vec{b} = 49\)
\(\displaystyle \vec{a}\cdot\vec{b} = \frac{15}{2}\)
If \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\), by the dot product,
\(\displaystyle \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{15/2}{3 \times 5} = \frac{1}{2}\)
\(\displaystyle \theta = \frac{\pi}{3}\)