Question

Let $\overrightarrow{OA}=4\hat{i}+5\hat{j}-3\hat{k}$ and $\overrightarrow{OB}=5\hat{i}+4\hat{j}-3\hat{k}.$, then the vector $\overrightarrow{OC}$ bisecting the angle $AOB$ and $C$ being the point on the line $AB$ is

• A. $4(\hat{i}+\hat{j}-\hat{k})$
• B. $(\hat{i}-\hat{j})$
• C. $(\hat{i}+\hat{j}-\hat{k})$
• D. None of these

Answer 1

Answer: (B)

$(\hat{i}-\hat{j})$

Let the vector $\overrightarrow{OC}=a\hat{i}+b\hat{j}+c\hat{k}$

Therefore

$\frac{\overrightarrow{OC}.\overrightarrow{OA}}{OC.OA}=\frac{\overrightarrow{OC}.\overrightarrow{OB}}{OC.OB}$

$\frac{4a+5b}{\sqrt{a^2+b^2}.5\sqrt{2}}=-\frac{5a+4b}{\sqrt{a^2+b^2}.5\sqrt{2}}$

Therefore

$4a+5b=-(5a+4b)$
$9a=-9b$)

$a=-b$...$\left(i\right)$

Now

$\overrightarrow{OA}.\overrightarrow{OB}=OA.OB.\cos 2\theta$
$\cos 2\theta =\frac{49}{50}$

Therefore

$\cos \theta =\sqrt{\frac{1+cos2\theta }{2}}$

$=\frac{3\sqrt{11}}{10}$

Hence

$\overrightarrow{OC}.\overrightarrow{OA}=OC.OA.cos\theta$

$4a+5b=\sqrt{a^2+b^2}.5\sqrt{2}.\frac{3\sqrt{11}}{10}$

Squaring both sides, we get

$16a^2+25b^2+40ab=(a^2+b^2)\frac{99}{2}$

$32a^2+50b^2+40ab=99a^2+99b^2$
$67a^2+49b^2-40ab=0$...$\left(ii\right)$

Hence from $i$ and $ii$,

$\overrightarrow{OC}$ can be equal to $i-j$.