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.

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