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Question
If P is orthocentre, Q is the circumcentre and G is the centroid of a triangle ABC, then prove that `bar"QP" = 3bar"QG"`.
Solution
Let `bar"p" and bar"g"` be the position vectors of P and G w.r.t. the circumcentre Q.
i.e. `bar"QR" = bar"p" and bar"QG" = bar"g"`
We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.
∴ by section formula for internal division,
`bar"g" = (1.bar"p" + 2bar"q")/(1 + 2) = bar"p"/3` .....`[∵ bar"q" = 0]`
∴ `bar"p" = 3bar"g"`
∴ `bar"QP" = 3bar"QG".`
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