if `hat"i" + hat"j" + hat"k", 2hat"i" + 5hat"j", 3hat"i" + 2 hat"j" - 3hat"k" and hat"i" - 6hat"j" - hat"k"` respectively are the position vectors 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.

#### Solution

**Given:**

The position vector of A is `hat"i"+hat"j"+hat"k"`.

The position vector of B is `hat2"i"+5hat"j"`.

Therefore, `vec"AB"= (2-1)hat"i"+(5-1)hat"j"+(0-1)hat"k" = hat"i"+ 4hat"j" - hat"k"`

The position vector of C is `3hat"i"+2 hat"j"- 3hat"k"` and

The position vector of D is `hat"i"- 6hat"j"- hat"k"`.

Therefore, `vec"CD" = (1 -3) hat"i"+(-6 -2)hat"j"+ (-1+3)hat"k" = -2hat"i"- 8hat"j"+2hat"k"`

Let `theta` be the angle between AB and CD, then

⇒ `cos theta = (vec"AB".vec"CD")/(|vec"AB"| |vec"CD"|`

⇒ `cos theta = (-2-32-2)/(sqrt18 sqrt72)= -1`

⇒ `theta = 180°`

since angle between Line AB and CD is 180°, therefore `vec"AB"` and `vec"CD"` are collinear.