# If ABC is a triangle whose orthocentre is P and the circumcentre is Q, prove that PAPBPCPQPA¯+PB¯+PC¯=2PQ¯. - Mathematics and Statistics

Sum

If ABC is a triangle whose orthocentre is P and the circumcentre is Q, prove that bar"PA" + bar"PB" + bar"PC" = 2bar"PQ".

#### Solution

Let G be the centroid of the Δ ABC.

Let A, B, C, G, Q have position vectors bar"a",bar"b",bar"c",bar"g",bar"q" w.r.t. P. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.

∴ bar"g" = (1. bar"p" + 2bar"q")/(1 + 2) = (2 bar"q")/3            .........[∵ bar"p" = bar0]

∴ 3bar"g" = 2bar"q"

∴ (3(bar"a" + bar"b" + bar"c"))/3 = 2bar"q"

∴ bar"a" + bar"b" + bar"c" = 2bar"q"

∴ bar"PA" + bar"PB" + bar"PC" = 2bar"PQ"

Concept: Vectors and Their Types
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