1. The problem statement, all variables and given/known data Consider the triangle ABC whose vertices have position vectors a, b, and c respectively. Find the position vector of (a) P, the mid-point of AB (b) Q, the point of trisection of AB, with Q closer to B. (c) R, the mid-point of the median CP. 2. Relevant equations None really. It just seems that I have a major conflict with the textbook answer and I'm not quite sure why. The textbook says: (a) ½(b + a) (b) ⅓(2b + a) (c) ½(a + b + c) 3. The attempt at a solution (a) I got this answer. Looking at the diagram, if we theoretically add OB and OA tip-to-tail, the half of the resultant vector should give us OP. (b) Similarly to (a), AB = b + a OQ = ⅓AB --> OQ = ⅓(b + a) I don't see why b is multiplied by 2. (c) First off: CP = CO + OP CP = -c + ½(b + a) Then, OR = OC + ½CP OR = c + ½(-c + ½(b + a)) OR = ½c + ¼b + ¼a Yeah, I don't see why my answers disagree with the textbook.