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Let $\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{OB}}=-2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ be the position vectors of two points $A$ and $B$. If $C$ is a point on the bisector $\angle \mathrm{AOB}$ and $\mathrm{OC}=\sqrt{42}$, then $\overrightarrow{\mathrm{OC}}=$
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1259 Upvotes
Verified Answer
The correct answer is:
$\hat{i}+5 \hat{j}+4 \hat{k}$
Given $\overrightarrow{O A}=\vec{i}+2 \vec{j}-2 \vec{k}$ and $\overrightarrow{O B}=-2 \vec{i}-3 \vec{j}+6 \vec{k}$
Now unit vector bisector $\angle A O B$
$\begin{aligned}
& =\frac{\vec{i}+2 \vec{j}-2 \vec{k}}{3} \pm \frac{(-2 i-3 j+6 k)}{7} \\
& =\frac{\vec{i}+5 \vec{j}+4 \vec{k}}{21} \text { or } \frac{13 \vec{i}+23 \vec{j}-32 \vec{k}}{21}
\end{aligned}$
Since $\frac{(1+25+16)}{441} x^2=42 \Rightarrow x=21$
So required vector $=\vec{i}+5 \vec{j}+4 \vec{k}$
Now unit vector bisector $\angle A O B$
$\begin{aligned}
& =\frac{\vec{i}+2 \vec{j}-2 \vec{k}}{3} \pm \frac{(-2 i-3 j+6 k)}{7} \\
& =\frac{\vec{i}+5 \vec{j}+4 \vec{k}}{21} \text { or } \frac{13 \vec{i}+23 \vec{j}-32 \vec{k}}{21}
\end{aligned}$
Since $\frac{(1+25+16)}{441} x^2=42 \Rightarrow x=21$
So required vector $=\vec{i}+5 \vec{j}+4 \vec{k}$
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