Home
Class 12
MATHS
Orthocenter of an equilateral triangle A...

Orthocenter of an equilateral triangle ABC is the origin O. If `vec(OA)=veca, vec(OB)=vecb, vec(OC)=vecc`, then `vec(AB)+2vec(BC)+3vec(CA)=`

A

3c

B

3a

C

0

D

3b

Text Solution

Verified by Experts

The correct Answer is:
B

For an equilateral triangle, centroid is the samme as orthocentre
`therefore(OA+OB+OC)/(3)=0`
`thereforeOA+OB+OC=0`
Now, `AB+2BC+3CA`
`=OB-OA+2OC-2OB+3OA-3OC`
`=-OB+2OA-OC`
`=(OB+OA+OC)+3OA=3OA=3a`
Promotional Banner

Similar Questions

Explore conceptually related problems

If |vec(a)| = 3, |vec(b) = 4 and |vec(a) + vec(b)| = 1 , then |vec(a) - vec(b)| =

Let O, O' and G be the circumcentre, orthocentre and centroid of a Delta ABC and S be any point in the plane of the triangle. Statement -1: vec(O'A) + vec(O'B) + vec(O'C)=2vec(O'O) Statement -2: vec(SA) + vec(SB) + vec(SC) = 3 vec(SG)

Statement 1 : In DeltaABC , vec(AB) + vec(BC) + vec(CA) = 0 Statement 2 : If vec(OA) = veca, vec(OB) = vecb , then vec(AB) = veca + vecb

Prove that 3vec(OD)+vec(DA)+vec(DB)+vec(DC) is equal to vec(OA)+vec(OB)+vec(OC) .

[vec a + vec b vec b + vec c vec c + vec a] =

ABCDE is a pentagon. Prove that the resultant of forces vec (AB), vec(AE), vec(BC), vec(DC), vec(ED) and vec(AC) is 3vec(AC) .

Prove that [vec(a)+vec(b),vec(b)+vec( c) , vec( c) +vec(a)] =2 [vec(a),vec(b),vec(c )] .

If A, B and C are the vertices of a triangle ABC, then what is the value of vec(AB) + vec(BC) + vec(CA) ?

Prove that [vec(a),vec(b),vec( c) +vec(d)]=[vec(a),vec(b),vec( c)] +[vec(a),vec(b),vec(d)] .