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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
Solution
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
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