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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
The points $O,A,B,C,D$ are such that $\overrightarrow {OA}  = \vec a,\,\overrightarrow {OB}  = \vec b,\,$ $\overrightarrow {OC}  = \,2\vec a + 3\vec b\,$ and$ \,\overrightarrow {OD}  = \,\vec a - 2\vec b.\,\,$ If $ \,\left| {\vec a} \right|\, = 3\left| {\vec b,} \right|$ then the angle between $\overrightarrow {BD} $ and $\overrightarrow {AC} $ is
  • A
    $\frac{\pi }{3}$
  • B
    $\frac{\pi }{4}$
  • C
    $\frac{\pi }{6}$
  • ✓
    $\frac{\pi }{2}$
✓

Answer

Correct option: D.
$\frac{\pi }{2}$
d
We have $\overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OB}}=a-2 b-b=a-3 b$ and

$\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=2 a+3 b-a=a+3 b$

Let $\theta$ be the angle between $\overrightarrow{\mathrm{BD}}$ and $\overrightarrow{\mathrm{AC}}$

Then $\cos \theta  = \frac{{\overrightarrow {{\rm{BD}}}  \cdot \overrightarrow {\overline {{\rm{AC}}} } }}{{|\overrightarrow {{\rm{BD}}} ||\overrightarrow {{\rm{AC}}} |}} = \frac{{|{\rm{a}}{|^2} - 9|{\rm{b}}{|^2}}}{{|\overrightarrow {{\rm{BD}}} \overrightarrow {{\rm{AC}}} |}}$

$=\frac{9|\mathrm{b}|^{2}-9|\mathrm{b}|^{2}}{|\overrightarrow{\mathrm{BD}}||\overrightarrow{\mathrm{AC}}|},(\therefore|\mathrm{a}|=3|\mathrm{b}|)$

$\Rightarrow \cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}$

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