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Vector Proof
Date: 01/17/99 at 00:42:34
From: Olivia
Subject: Vector proofs
I've been having some difficulty with solving vector proofs. I kind of
know how to do them but don't know the proper way to write the
solution. I was wondering if you could please give me some tips or
help me out with the questions. It would be much appreciated.
1. Given: P, Q, R, and S any 4 non-collinear points
A and B the midpoints of PR and QS respectively
Prove: vector PQ + vector RS = 2(vector AB)
2. Given parallelogram PQRS with A dividing vector SR in the ratio
3:4, B and C the points of trisection of vector QR, and T the
point of intersection of vector SB, show that BT:TS = 4:9
_____________________ S
\ . \ 3
\ . / \ A
\ / . \
\ / . \ 4
\/_______.___________\
Q B C R
There is supposed to be a line from B to S and another from Q to A.
Date: 01/17/99 at 16:37:38
From: Doctor Anthony
Subject: Re: Vector proofs
Question 1:
Take origin at A, and let b, p, q, r, s be the position vectors of B,
P, Q, R, S relative to the origin A.
We have r = -p and b = (q+s)/2
Then:
PQ = q-p RS = s-r
PQ + RS = q-p + s-r
= q+s - p + p since r = -p
= q+s
= 2b
= 2 AB as required.
Question 2:
Take origin at R and let s, q, be the vectors RS ans RQ respectively.
Any point on the line QA is q + k(4s/7 - q) = q(1-k) + s(4k/7) where k
is a scalar.
Any point on the line SB is s + k'(2q/3 - s) = q(2k'/3) + s(1-k')
where k' is a scalar.
Where these lines intersect at T we can equate coefficients of q and s
for the two lines.
We get the two equations:
Coeffs of q 1-k = 2k'/3 (1)
Coeffs of s 4k/7 = 1-k' (2)
From (1) k' = 3(1-k)/2 and putting this in (2) we get:
4k/7 = 1 - 3(1-k)/2
8k = 14 - 21(1-k)
8k = 14 - 21 + 21k
7 = 13k
k = 7/13
From this:
k' = 3(1-k)/2 = 3(6/13)/2
k' = 9/13
So ST/SB = 9/13, and therefore ST/TB = 9/4 or BT:TS = 4:9, as required.
- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
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