Question
If $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ 2\hat{\text{i}}+5\hat{\text{J}},\ 3\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\hat{\text{i}}-6\hat{\text{j}}-\hat{\text{k}}$ 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 $\vec{\text{AB}}$ and $\vec{\text{CD}}$ are collinear or not.

Answer

Given:
The position vector of A is $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
The position vector of B is $2\hat{\text{i}}+5\hat{\text{j}}.$
Therefore, $\vec{\text{AB}}=(2-1)\hat{\text{i}}+(5-1)\hat{\text{j}}+(0-1)\hat{\text{k}}=\hat{\text{i}}+\hat4{\text{j}}-\hat{\text{k}}$
The position vector of C is $3\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and
The postion vector of D is $\hat{\text{i}}-6\hat{\text{j}}-\hat{\text{k}}.$
Therefore, $\vec{\text{CD}}=(1-3)\hat{\text{i}}+(-6-2)\hat{\text{j}}\\+(-1+3)\hat{\text{k}}=-2\hat{\text{i}}-\hat8{\text{j}}+2\hat{\text{k}}$
$\cos\theta=\frac{\vec{\text{AB}}.\vec{\text{CD}}}{|\vec{\text{AB}}||\vec{\text{CD}}|}$
$\Rightarrow\cos\theta=\frac{-2-32-2}{\sqrt{18}\sqrt{72}}=-1$
$\Rightarrow\theta=180^\circ$
Since, angle between Line AB and CD is 180°, therefore $\vec{\text{AB}}$ and $\vec{\text{CD}}$ are collinear.

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