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The position vector of $\mathrm{A}$ and $\mathrm{B}$ are $2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
The length of the internal bisector of $\angle \mathrm{BOA}$ of triangle $\mathrm{AOB}$ is
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The length of the internal bisector of $\angle \mathrm{BOA}$ of triangle $\mathrm{AOB}$ is
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Verified Answer
The correct answer is:
$\sqrt{\frac{136}{9}}$
We have, $\mathrm{OA}=|2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}|=3 ; \mathrm{OB}=|2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|=6$ Since the internal bisector divides opposite side in the
ratio of adjacent sides
$$
\therefore \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{3}{6}=\frac{1}{2}
$$
where $\mathrm{OC}$ is the bisector of $\angle \mathrm{BOA}$.
$\therefore$ Position vector of $C$ is
$$
\begin{array}{l}
\frac{2(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})}{2+1}=2 \hat{\mathrm{i}}+\frac{8}{3} \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\
\therefore \mathrm{OC}=\left|2 \hat{\mathrm{i}}+\frac{8}{3} \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\right|=\sqrt{\frac{136}{9}}
\end{array}
$$
ratio of adjacent sides
$$
\therefore \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{3}{6}=\frac{1}{2}
$$
where $\mathrm{OC}$ is the bisector of $\angle \mathrm{BOA}$.
$\therefore$ Position vector of $C$ is
$$
\begin{array}{l}
\frac{2(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})}{2+1}=2 \hat{\mathrm{i}}+\frac{8}{3} \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\
\therefore \mathrm{OC}=\left|2 \hat{\mathrm{i}}+\frac{8}{3} \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\right|=\sqrt{\frac{136}{9}}
\end{array}
$$
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