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If the position vector of a point $P$ with respect to origin $O$ is
$\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and that of a point $Q$ is $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, then
what is the position vector of a point on the bisector of the angle $P O Q$ ?
Options:
$\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and that of a point $Q$ is $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, then
what is the position vector of a point on the bisector of the angle $P O Q$ ?
Solution:
2368 Upvotes
Verified Answer
The correct answer is:
$\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
Let $\overrightarrow{\mathrm{OP}}=\hat{i}+3 \hat{j}-2 \hat{k}$ and $\overrightarrow{\mathrm{OQ}}=3 \hat{i}+\hat{j}-2 \hat{k}$
Let $\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be required position vector of the bisector of the angle $P O Q$ since, it is the bisector of
$\angle P O Q$ therefore. It will make equal angles with $\overline{O P}$
and $\overrightarrow{\mathrm{OQ}}$.
Let Angle between $\hat{i}+3 \hat{j}-2 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$ is
$\theta=\cos ^{-1}\left(\frac{1+3+2}{\sqrt{1+9+4} \sqrt{1+1+1}}\right)$
$=\cos ^{-1}\left(\frac{6}{\sqrt{14} \sqrt{3}}\right)$
and angle between $3 \hat{i}+\hat{j}-2 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$, is
$\phi=\cos ^{-1}\left(\frac{1+3+2}{\sqrt{9+1+4} \sqrt{1+1+1}}\right)=\cos ^{-1}\left(\frac{6}{\sqrt{14} \sqrt{3}}\right)$
Hence, $\theta=\phi$
Let $\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be required position vector of the bisector of the angle $P O Q$ since, it is the bisector of
$\angle P O Q$ therefore. It will make equal angles with $\overline{O P}$
and $\overrightarrow{\mathrm{OQ}}$.
Let Angle between $\hat{i}+3 \hat{j}-2 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$ is
$\theta=\cos ^{-1}\left(\frac{1+3+2}{\sqrt{1+9+4} \sqrt{1+1+1}}\right)$
$=\cos ^{-1}\left(\frac{6}{\sqrt{14} \sqrt{3}}\right)$
and angle between $3 \hat{i}+\hat{j}-2 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$, is
$\phi=\cos ^{-1}\left(\frac{1+3+2}{\sqrt{9+1+4} \sqrt{1+1+1}}\right)=\cos ^{-1}\left(\frac{6}{\sqrt{14} \sqrt{3}}\right)$
Hence, $\theta=\phi$
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