If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three vectors such that \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\), where \(\vec{a}\) and \(\vec{b}\) are unit vectors and \(|\vec{c}|=2\), then the angle between the vectors \(\vec{b}\) and \(\vec{c}\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(180^{\circ}\)
Since \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\):
\(\vec{a} = -\left(\vec{b} + \vec{c}\right)\)
Squaring the magnitudes (taking the dot product of each side with itself):
\(|\vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\,\vec{b}\cdot\vec{c}\)
With \(|\vec{a}| = |\vec{b}| = 1\), \(|\vec{c}| = 2\) and \(\vec{b}\cdot\vec{c} = |\vec{b}|\,|\vec{c}|\cos\theta\):
\(1 = 1 + 4 + 2(1)(2)\cos\theta\)
\(4\cos\theta = -4\)
\(\cos\theta = -1\)
\(\theta = 180^{\circ}\)
Indeed \(\vec{c} = -2\vec{b}\) and then \(\vec{a} = \vec{b}\), a unit vector, so the data are consistent.