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Question
Let `vec"a"` and `vec"b"` be the position vectors of the points A and B. Prove that the position vectors of the points which trisects the line segment AB are `(vec"a" + 2vec"b")/3` and `(vec"b" + 2vec"a")/3`
Solution
Let O be the origin and `vec"a"` and `vec"b"` are the position vectors of the points A and B respectively.
`vec"OA" = vec"a"` and `vec"OB" = vec"b"`
Let C and D be the points that trisect the line joining the points A and B.
∴ AC = CD = DB
C divides AB in the ratio 1 : 2 internally.
∴ `vec"OC" = (1 * vec"OB" + 2 * vec"OA")/(1 + 2)`
`vec"OC" = (vec"b" + 2vec"a")/3`
D diides AB n the ratio 2 : 1 internally.
∴ `vec"OD" = (2 * vec"OB" + 1 * vec"OA")/(2 + 1)`
`vec"OD" = (2vec"b" + vec"a")/3`
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