Let `G` be the centroid of triangle`\ A B Cdot` If ` vec A B= vec a ,\ vec A C= vec b ,` then the bisector ` vec A G ,` in terms of ` vec a\ a n d\ vec b` is `2/3( vec a+ vec b)` b. `1/6( vec a+ vec b)` c. `1/3( vec a+ vec b)` d. `1/2( vec a+ vec b)1`

Options

A. `(2)/(3)(veca+vecb)`
B. `(1)/(6)(veca+vecb)`
C. `(1)/(3)(veca+vecb)`
D. `(1)/(2)(veca+vecb)`

Answer

C

Explanation

Let A be the origin. Then, `A vec B = vec a , A vec C = vec b `
implies that the position vectors of B and C are `vecb and vec c ` respectively. Let AD be the median and G be the centroid.
Then,

P.V of D `= (veca + vec b)/(2)`, P. V. of G `= (veca + vec b)/(3)`
` therefore A vec G = (veca + vec b)/(3)`