If \(\vec a, \vec b, \vec c\) are vectors such that \(\vec a + \vec b + \vec c = 0\) and \(|\vec a| = 10, |\vec b| = 4, |\vec c| = 6\), then the angle between the vectors \(\vec b\) and \(\vec c\) is?

Concept:

\(\rm \vec a.\vec b=|\vec a||\vec b|\cos\theta \)

Calculation:

Here, \(\vec a + \vec b + \vec c = 0\)

\(\Rightarrow \vec b + \vec c = -\vec a\)

Taking magnitude and squaring both sides,

\(\rm \Rightarrow | \vec b + \vec c|^2 = |-\vec a|^2\)

\(\rm \Rightarrow |\vec b|^2+|\vec c|^2+2\vec b.\vec c =100\)

\(\rm \Rightarrow 2 |\vec b|.|\vec c|cos θ =100-(16 + 36)\)

\(\\ \rm \Rightarrow cosθ =\frac{48}{2\;\times\; 4 \;\times\; 6}\)

\(\Rightarrow θ =cos^{-1}(1)\)

θ = 0°

Hence, option (1) is correct.