# If $$\hat i+ \hat j+\hat k,\: 2\hat i+5\hat j,\: 3\hat i+2\hat j=3\hat k\: and \: \hat i-6\hat j-\hat k$$ are the position vectorsof the points A,B,C and D, find the angle between $$\overrightarrow{AB}\: and \: \overrightarrow{CD}$$. Deduce that $$\overrightarrow{AB}\: and \: \overrightarrow{CD}$$ are collinear.

Toolbox:
• $\cos\theta=\large\frac{\overrightarrow{AB}.\overrightarrow{CD}}{\mid \overrightarrow{AB}\mid\mid\overrightarrow {CD}\mid}$
Step 1:
Let $\overrightarrow{OA}=\hat i+\hat j+\hat k$
$\overrightarrow{OB}=2\hat i+5\hat j$
$\overrightarrow{OC}=2\hat i+2\hat j-3\hat k$
$\overrightarrow{OD}=\hat i-6\hat j-\hat k$
$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\quad\;\;=(2\hat i+5\hat j)-(\hat i+\hat j+\hat k)$
$\quad\;\;=(\hat i+4\hat j-\hat k)$
$\overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}$
$\quad\;\;=(\hat i-6\hat j-\hat k)-(3\hat i+2\hat j-3\hat k)$
$\quad\;\;=(-2\hat i-8\hat j+2\hat k)$
Step 2:
Angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$
$\cos\theta=\large\frac{\overrightarrow{AB}.\overrightarrow{CD}}{\mid \overrightarrow{AB}\mid\mid\overrightarrow {CD}\mid}$
$\qquad=\large\frac{(\hat i+4\hat j-\hat k)(-2\hat i-8\hat k+2\hat k)}{\sqrt{1^2+4^2+1^2}\sqrt{(-3)^2+(8)^2+(4)^2}}$
$\cos\theta=\large\frac{-2-32-2}{\sqrt{18}\sqrt{2}}$
$\qquad=\large\frac{-36}{36}$
$\qquad=-1$
$\cos \theta=1$
$\Rightarrow \theta=\pi$
$\therefore \overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.