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September 25th, 2013, 01:25 AM   #1
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2-D vector proofing

A, B and C are the mid-points of PQ, QR, RQ respectively. G is the point of intersection of AR, BP and CQ. Given that the position vectors of P, Q and R relative to an origin O, (which is not shown in the diagram), are p, q and r respectively, prove that p + q + r = 3g, where g is the position vector of G relative to the origin O.
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 September 25th, 2013, 02:51 PM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: 2-D vector proofing You will have to keep the sketch in your view while examining this edited proof. Comment if any statement I made needs better explanation. Proof starts here--- Claim:$\overline{PG} + \overline{QG} + \overline{RG}= 0$ Proof of claim By completing the parallelogram in the image with base $\overline{RQ}$ and slant height $\overline{PQ}$, we observe that $\overline{RQ} + \overline{PQ}= 2\overline{CQ} \cdots (1)$. But also observe that $\overline{GQ}= 2\overline{CG}$. This is because each of the segments $\overline{PB}, \overline{QC}, \overline{RA}$ divide the triangle $\triangle PQR$ into two halves equal in area. It is then easy to deduce that each of the six triangular segments in the diagram have the same area; and then that the area of $\triangle PGR$ is one-third that of $\triangle PQR$. Hence, $\overline{CQ}= \frac{3}{2}\overline{GQ} \ \Rightarrow \ \overline{RQ} + \overline{PQ} = 3\overline{GQ}$ by (1) $\overline{RQ} + \overline{PQ}= 3\overline{GQ} \ \Leftrightarrow \ \overline{RQ} + 2\overline{QG} = \overline{GQ} + \overline{QP} = \overline{GP} \cdots (2)$ But also, $\overline{RQ}= \overline{RG} + \overline{GQ} \ \Leftrightarrow \ \overline{RQ} + 2\overline{QG} = \overline{RG} + \overline{QG} \cdots (3)$ By validity of (2) and (3) we get that $\overline{RG} + \overline{QG}= \overline{GP} \ \Leftrightarrow \ \overline{PG} + \overline{QG} + \overline{RG} = 0$, concluding the proof of our claim. The following relations hold: $p + \overline{PG}= g \ ; \ r + \overline{RG} = g \ ; \ q + \overline{QG} = g .$ Adding them up, we get that $p + q + r + \overline{PG} + \overline{QG} + \overline{RG}= 3g \ \Leftrightarrow \ p+q+r=3g.$

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