Question

If `vec"a", vec"b", vec"c"` are position vectors of the vertices A, B, C of a triangle ABC, show that the area of the triangle ABC is `1/2 |vec"a" xx vec"b" + vec"b" xx vec"c" + vec"c" xx vec"a"|`. Also deduce the condition for collinearity of the points A, B, and C

Answer 1

Given `vec"a", vec"b", vec"c"` are the position vectors of the verticles of the triangle ABC

∴ `vec"OA" =vec"a"`

`vec"OB" = vec"b"`

`vec"OC" = vec"c"`

`vec"AB" = vec"OB" - vec"OA"`

= `vec"b" - vec"a"`

`vec"AC" = vec"OC" - vec"OA"`

= `vec"c" - vec"a"`

`vec"AB" xx vec"AC" = (vec"b" - vec"a") xx  (vec"c" - vec"a")`

= `vec"b" xx vec"c" - vec"b" xx vec"a" - vec"a" xx vec"c" + vec"a" xx vec"a"`

= `vec"b" xx vec"c" + vec"a" xx vec"b" + vec"c" xx vec"a" + 0`

`|vec"AB" xx vec"AC"| = |vec"a" xx vec"b" + vec"b" xx vec"c" + vec"c" xx vec"a"|`

Area of ΔABC = `1/2 |vec"AB" xx vec"AC"|`

= `1/2 |vec"a" xx vec"b" + vec"b" xx vec"c" + vec"c" xx vec"a"|`

The points A, B, C are collinear if the area of the triangle formed by these points is zero.

∴ `1/2 |vec"a" xx vec"b" + vec"b" xx vec"c" + vec"c" xx vec"a"|` = 0

`|vec"a" xx vec"b" + vec"b" xx vec"c" + vec"c" xx vec"a"|` = 0