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Question
If G is the centroid of a triangle ABC, prove that `vec"GA" + vec"GB" + vec"GC" = vec0`
Solution
Let ABC be the triangle with centroid G.
Le `vec"a", vec"b"` and `vec"c"` be the position vectors of the vertices A, B and C respectively with respect to the origin O.
Then `vec"OA" = vec"a"`
`vec"OB" = vec"b"`
`vec"OC" = vec"c"`
`vec"OG" = (vec"a" + vec"b" + vec"c")/3`
`vec"GA" + vec"GB" + vec"GC" = vec"OA" - vec"OG" + vec"OB" - vec"OG" + vec"OC" - vec"OG"`
= `vec"OA" + vec"OB" + vec"OC" - 3vec"OG"`
= `vec"a" + vec"b" + vec"c" - 3 xx (vec"a" + vec"b" + vec"c")/3`
= `vec"a" + vec"b" + vec"c" - (vec"a" + vec"b" + vec"c")`
`vec"GA" + vec"GB" + vec"GC" = vec"0"`
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