OAB is a given triangle such that `vec(OA)=vec(a), vec(OB)=vec(b)`. Also C is a point on `vec(AB)` such that `vec(AB)=2vec(BC)`. What is `vec(AC)` equal to ?

Options

A. `1/2(vec(b)-vec(a))`
B. `1/2(vec(b)+vec(a))`
C. `3/2(vec(a)-vec(b))`
D. `3/2(vec(b)-vec(a))`

Answer

A

Explanation

In `DeltaOAB`,
Let `vec(OA)=vec(a)`
`vec(OB)=vec(b) `
`vec(OA)+vec(AB)=vec(OB)`
`rArr vec(AB)=vec(OB)-vec(OA)`
`=vec(b)-vec(a)`
`vec(AB)=2vec(BC)`
`rArr C ` is mid point of `vec(AB)`
`:. vec(AC) = (1)/(2) vec(AB)=(1)/(2)(vec(b)-vec(a))`