Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are `vec(a) , vec(b) , vec(c ) and (vec(a) + vec(b) + vec(c ))/(4) `
respectively, then the position vector of the orthocentre of this triangle is

Options

A. `vec(0)`
B. `-(vec(a)+vec(b)+vec(c))/(2)`
C. `vec(a) + vec(b)+vec( c) `
D. `(vec(a)+vec(b)+vec(c))/(2)`

Answer

D

Explanation

Let G be the centroid of `Delta ABC `. Then, the position vector of G is `(vec(a)+vec(b)+vec(c ))/(3).` Let the position vector of orthocentre O be ` vec(alpha)` .

We know that O, G and P are collinear and G divides OP in the ratio `2:1`.
`therefore (2((vec(a)+vec(b)+vec(c))/(4))+1*vec(alpha))/(2+1)=(vec(a)+vec(b)+vec(c))/(3)`
` rArr (vec(a)+vec(b)+vec(c ) )/(2)+vec(alpha)=vec(a)+vec(b)+vec(c ) rArr vec(alpha)=(vec(a)+vec(b)+vec(c ))/(2)`