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If $\hat{i}+\hat{j}+\hat{k}$ , $2\hat{i}+5\hat{j}$ , $3\hat{i}+2\hat{j}-3\hat{k}$ and $\hat{i}-6\hat{j}-\hat{k}$ respectively are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether $\vec{AB}$ and $\vec{CD}$ are collinear or not.

Answers (1)

Given position vectors :

$\vec{A}=\hat{i}+\hat{j}+\hat{k}$ , $\vec{B}=2\hat{i}+5\hat{j}$ , $\vec{C}=3\hat{i}+2\hat{j}-3\hat{k}$ , $\vec{D}=\hat{i}-6\hat{j}-\hat{k}$

$\vec{AB}=position\: vector\: o\!f\: \vec{B}-position\: vector\: o\!f\: \vec{A}$

$=2\hat{i}+5\hat{j}-\hat{i}-\hat{j}-\hat{k}$

$=2\hat{i}+4\hat{j}-\hat{k}$

$\vec{CD}=position\: vector\: o\!f\: \vec{D}-position\: vector\: o\!f\: \vec{C}$

$=\hat{i}-6\hat{j}-\hat{k}-3\hat{i}-2\hat{j}+3\hat{k}$

$=-2\hat{i}-8\hat{j}+2\hat{k}$

To find the angle between the lines $\vec{AB}$ and $\vec{CD}$

$\cos \theta =\frac{\overrightarrow{AB}\cdot \overrightarrow{CD}}{|\overrightarrow{AB}||\overrightarrow{CD}|}$ [ Dot product formula]

$\therefore |\overrightarrow{AB}|=\sqrt{1+16+1}=\sqrt{18}$ $\therefore |\overrightarrow{C\!D}|=\sqrt{4+64+4}=\sqrt{72}=2\sqrt{18}$

And $\overrightarrow{AB}\cdot \overrightarrow{CD}=-2-32-2=-36$

$\cos \theta =\frac{-36}{\sqrt{18}\times 2\times \sqrt{18}}=\frac{-36}{2\times 18}=1$

$\cos \theta =1$

$\therefore \theta =\pi =180^{\circ}$

$\Rightarrow$ AB and CD are collinear.

Posted by

Ravindra Pindel

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