If D, E and F be the middle points of the sides BC,CA and AB of the `DeltaABC`, then `AD+BE+CF` is

Options

A. a zero vector
B. a unit vector
C. 0
D. none of these

Answer

A

Explanation

`AD=OD-OA`
`=(b+c)/(2)-a=(b+c-2a)/(2)` [where, O is the origin for reference]
Similarly, `BE=OE-OB=(c+a)/(2)-b`
`=(c+a-2b)/(2)`
and `CF=(a+b-2b)/(2)`

Now, `AD+BE+CF=(b+c-2a)/(2)+(c+a-2b)/(2)+(a+b-2c)/(2)=0`