• # question_answer The points $O,\,A,\,B,\,C,\,D$ are such that $\overrightarrow{OA}=\mathbf{a},$ $\overrightarrow{OB}=\mathbf{b},\,$ $\overrightarrow{OC}=2\mathbf{a}+3\mathbf{b}$ and $\overrightarrow{OD}=\mathbf{a}-2\mathbf{b}.$ If $|\mathbf{a}|\,=3\,|\mathbf{b}|,$ then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is A) $\frac{\pi }{3}$ B) $\frac{\pi }{4}$ C) $\frac{\pi }{6}$ D) None of these

• We have $\overrightarrow{BD}=\overrightarrow{OD}-\overrightarrow{OB}=\mathbf{a}-2\mathbf{b}-\mathbf{b}=\mathbf{a}-3\mathbf{b}$  and $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=2\mathbf{a}+3\mathbf{b}-\mathbf{a}=\mathbf{a}+3\mathbf{b}.$
• Let $\theta$ be the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}.$
• Then $\cos \theta =\frac{\overrightarrow{BD}\,.\,\overrightarrow{AC}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}=\frac{|\mathbf{a}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}$
• $=\frac{9|\mathbf{b}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|\,|\overrightarrow{AC}|}$,
• $(\because \,\,\,\left| \,\mathbf{a}\, \right|=3|\mathbf{b}|)$
• $\Rightarrow \cos \theta =0{}^\circ \Rightarrow \theta =\frac{\pi }{2}.$