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"`.

Answer 1

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".`