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**1. Homework Statement**

Consider the triangle ABC whose vertices have position vectors

**a**,

**b**, and

**c**respectively.

http://img106.imageshack.us/img106/4240/vectorat6.png [Broken]

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. Homework 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(2

**b**+

**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 = ½

**c**+ ¼

**b**+ ¼

**a**

Yeah, I don't see why my answers disagree with the textbook.

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