Question

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

Answer 1

The position vectors of the points are given.
$\text{So,}\overrightarrow{AB}=\text{P.V. of}B-\text{P.V. of}A$
$=2\hat{i}+5\hat{j}-(\hat{i}+\hat{j}+\hat{k})$
$=\hat{i}+4\hat{j}-\hat{k}$

$\text{Similarly}\overrightarrow{CD}=\text{P.V. of}D-\text{P.V. of}C$
$=\hat{i}-6\hat{j}-\hat{k}-(3\hat{i}+2\hat{j}-3\hat{k})$
$=-2\hat{i}-8\hat{j}+2\hat{k}$

Now, we have to find the angle between the two vectors. Angle between the vectors is given by
$\cos \theta =\frac{\overrightarrow{AB}\cdot \overrightarrow{CD}}{|\overrightarrow{AB}||\overrightarrow{CD}|}$
$=\frac{(\hat{i}+4\hat{j}-\hat{k})\cdot (-2\hat{i}-8\hat{j}+2\hat{k})}{\sqrt{1^2+4^2+(-1{)}^2}\sqrt{(-2{)}^2+(-8{)}^2+2^2}}$
$\Rightarrow \cos \theta =\frac{-36}{3\sqrt{2}\times 6\sqrt{2}}$
$\Rightarrow \cos \theta =-1$
$\Rightarrow \theta =\pi$
So, the angle between the straight lines $AB$ and $CD$ is $\pi .$
Since the given vectors are anti-parallel to each other $(\theta =\pi ),$ they are collinear.