Question

If `vec"a", vec"b", vec"c"` determine the vertices of a triangle, show that `1/2[vec"b" xx vec"c" + vec"c" xx vec"a" + vec"a" xx vec"b"]` gives the vector area of the triangle. Hence deduce the condition that the three points `vec"a", vec"b", vec"c"` are collinear. Also find the unit vector normal to the plane of the triangle.

Answer 1

Since, `vec"a", vec"b"` and `vec"c"` are the vertices of ΔABC

∴ `vec"AB" = vec"b" - vec"a"`

`vec"BC" = vec"c" - vec"b"`

And `vec"AC" = vec"c" - vec"a"`

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

= `1/2|(vec"b" - vec"a") xx (vec"c" - vec"a")|`

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

= `1/2 |vec"b" xx vec"c" + vec"a" xx vec"b" + vec"c" xx vec"a"|`   ......`[(because vec"a" xx vec"b" = - vec"b" xx vec"a"),(vec"c" xx vec"a" = - vec"a" xx vec"c"),(vec"a" xx vec"a" = vec0)]`

For three vectors are collinear, area of ΔABC = 0

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

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

Which is the condition of collinearity of `vec"a", vec"b"` and `vec"c"`.

Let `hat"n"` be the unit vector normal to the plane of the ΔABC

 ∴ `hat"n" = (vec"AB" xx vec"AC")/|vec"AB" xx vec"AC"|`

⇒ `(vec"a" xx vec"b" + vec"b" xx vec"c" + vec"c" xx vec"a")/|vec"a" xx vec"b" + vec"b" xx vec"c" + vec"c" xx vec"a"|`