How Do You Calculate Specific Points in a Triangle Using Vectors?

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SUMMARY

The discussion centers on calculating specific points in triangle ABC using vectors, where vertices have position vectors a, b, and c. The correct formulas derived include: (a) P, the mid-point of AB, is given by ½(b + a); (b) Q, the point of trisection of AB closer to B, is ⅓(2b + a); and (c) R, the mid-point of the median CP, is ½(a + b + c). The participant's calculations align with the textbook answers, confirming their correctness and highlighting a misunderstanding in the textbook's explanation.

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Homework Statement


Consider the triangle ABC whose vertices have position vectors a, b, and c respectively.

http://img106.imageshack.us/img106/4240/vectorat6.png

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.


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(2b + a)
(c) ½(a + b + c)


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.
 
Last edited by a moderator:
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a) OB = OA + AB
b = a + AB
therefore,
AB = b - a, not b + a

OP = OA + ½(AB) = a + ½(b - a) = a + ½b - ½a
OP = ½(b + a)

b) OQ = OA + (2/3)AB (Q is 2/3 of way along AB)

c) the book is wrong, you are right!
 
Last edited:

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