If $\vec a, \vec b$ are the position vectors of A, B respectively and C is a point on AB produced such that AC = 3 AB, then the position vector of C is

$3 \vec a-2\vec b$ $3 \vec b-2\vec a$ $3 \vec b+2\vec a$ $2\vec a-3\vec b$

$3 \vec b-2\vec a$

Let the position vector for C be $\vec c$ Clearly, B divides AC internally in the ratio 1 : 2. $∴\vec b=\frac{2\vec a+1.\vec c}{2+1}⇒\vec c=3\vec b-2\vec a$