**Show that the following points are collinear:**

A = (3, 2, -4), B = (9, 8, -10), C = (-2, -3, 1)

#### Solution

Let `bar"a", bar"b", bar"c"` be the position vectors of the points.

A = (3, 2, - 4), B = (9, 8, - 10) and C = (- 2, - 3, 1) respectively.

Then, `bar"a" = 3hat"i" + 2hat"j" - 4hat"k", bar"b" = 9hat"i" + 8hat"j" - 10hat"k", bar"c" = - 2hat"i" - 3hat"j" + hat"k"`

`bar"AB" = bar"b" - bar"a"`

`= (9hat"i" + 8hat"j" - 10hat"k") - (3hat"i" + 2hat"j" - 4hat"k")`

`= 6hat"i" + 6hat"j" - 6hat"k"` ......(1)

and `bar"BC" = bar"c" - bar"b"`

`= (- 2hat"i" - 3hat"j" + hat"k") - (9hat"i" + 8hat"j" - 10hat"k")`

`= - 11hat"i" - 11hat"j" + 11hat"k"`

`= - 11(hat"i" + hat"j" + 11hat"k")`

`= 11(hat"i" + hat"j" - hat"k")`

`= - 11/6 (6hat"i" + 6hat"j" - 6hat"k")`

`= - 11/6 bar"AB"` ....[By (1)]

∴ `bar"BC"` is a non-zero scalar multiple of `bar"AB"`

∴ they are parallel to each other.

But they have point B in common.

∴ `bar"BC"` and `bar"AB"` are collinear vectors.

Hence, points A, B and C are collinear.