Question

The points O,A,B,C,D are such that $\overline{OA}=a,\overline{OB}=b,\overline{OC}=2a+3b$ and $\overline{OD}=a-2b$. If $|a|=3|b|$, then the angle between $\overline{BD}$ ,$\overline{AC}$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

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