Question

If $\overline{a}$,$\overline{b}$,$\overline{c}$ are vectors such that $\overline{a}+\overline{b}+\overline{c}=0$ and $\left|\begin{array}{c}a\end{array}\right|=7,\left|\begin{array}{c}\overline{b}\end{array}\right|=5,\left|\begin{array}{c}\overline{c}\end{array}\right|=3$, then angle between $\overline{b}$ and $\overline{c}$ is

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

Answer: (A)

${60}^{\circ }$

Given $\overline{a}+\overline{b}+\overline{c}=0,$ we need angle between $\overline{b}$ and $\overline{c}$

So consider $\overline{b}+\overline{c}=-\overline{a}$

$\Rightarrow b^2+{\overline{c}}^2+2\left|\overline{b}\right|\left|\overline{c}\right|\cos \theta ={\overline{a}}^2$

$\Rightarrow \cos \theta =\frac{{\overline{a}}^2-{\overline{b}}^2-{\overline{c}}^2}{2\left|\overline{b}\right|\left|\overline{c}\right|}$

$=\frac{49-25-9}{2\times 5\times 3}=\frac{1}{2}$

$\therefore \theta ={60}^{\circ }$