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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