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If $\mathbf{P}, \mathbf{Q}, \mathbf{R}$ and $\mathbf{S}$ are the points with position vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}-3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ respectively, then the angle between $\mathbf{P Q}$ and $\mathbf{R S}$ is
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1860 Upvotes
Verified Answer
The correct answer is:
$0$
Given position vectors
$\begin{aligned} & \mathbf{O P}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{O Q}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \mathbf{O R}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} \text { and } \mathbf{O S}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\end{aligned}$
So,
$\mathbf{P Q}=\mathbf{O Q}-\mathbf{O P}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
and $\quad \mathbf{R S}=\mathbf{O S}-\mathbf{O R}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
As $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\therefore$ Angle between $\mathbf{P Q}$ and $\mathbf{R S}$ is 0 . Hence, option (a) is correct.
$\begin{aligned} & \mathbf{O P}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{O Q}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \mathbf{O R}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} \text { and } \mathbf{O S}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\end{aligned}$
So,
$\mathbf{P Q}=\mathbf{O Q}-\mathbf{O P}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
and $\quad \mathbf{R S}=\mathbf{O S}-\mathbf{O R}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
As $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\therefore$ Angle between $\mathbf{P Q}$ and $\mathbf{R S}$ is 0 . Hence, option (a) is correct.
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