JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
If the points P and Q are respectively the circumcenter and the orthocentre of a △ ABC, then PA⃗+PB⃗+PC⃗ is equal to :
- (a) QP⃗
- (b) PQ⃗
- (c) 2PQ⃗
- (d) 2QP⃗
Answer and solution
Answer (b) PQ⃗
Solution
Maths sir · 4 steps
Put the circumcentre at the origin, and the orthocentre turns out to be just a⃗ + b⃗ + c⃗.
- Origin at P. Take the circumcentre P as the origin and let A, B, C have position vectors a⃗, b⃗, c⃗. Then PA⃗ + PB⃗ + PC⃗ = a⃗ + b⃗ + c⃗. Since P is the circumcentre, |a⃗| = |b⃗| = |c⃗| = R.
- Test a point. Let H have position vector h⃗ = a⃗ + b⃗ + c⃗. Then AH⃗ = h⃗ - a⃗ = b⃗ + c⃗. Dot it with CB⃗ = b⃗ - c⃗:
(b⃗ + c⃗)·(b⃗ - c⃗) = |b⃗|² - |c⃗|² = R² - R² = 0
- H is orthocentre. So AH is perpendicular to BC. In the same way, BH is perpendicular to CA and CH is perpendicular to AB. H lies on all three altitudes, so H is the orthocentre Q, and PQ⃗ = a⃗ + b⃗ + c⃗.
- So. PA⃗ + PB⃗ + PC⃗ = PQ⃗, which is option (b).
The idea to keep. With the circumcentre as origin, the orthocentre is a⃗ + b⃗ + c⃗. Keep the twin result too: from the orthocentre, QA⃗ + QB⃗ + QC⃗ = 2QP⃗.
Answer from the NTA final answer key. Solution by PrepWiser.