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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/   
    
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Linear Algebra

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