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Using Vectors to Prove Collinearity


Date: 06/05/99 at 10:11:10
From: william glenn
Subject: A killer vector problem

I'm one of those students who always wants (and usually gets) 100% on 
my math assignments. I love calculus, but my other class is killing me 
with this problem.

Three concurrent straight lines 0A, 0B, 0C are produced to D, E, and F 
respectively. Prove that the points of intersection of AB and DE, BC 
and EF, and CA and FD are collinear.

At first I thought it couldn't be right, but I've drawn several 
diagrams, and _every time_ it works. I tried to solve it with what I 
know about vector analysis, but I just keep going around in circles. I 
call the points of intersection P, Q, and R respectively, and I think 
that to solve it I have to express all the points in terms of a, b, 
and c, but there are unknowns everywhere.

P, Q and R are collinear if their vector joins are parallel and they 
have a point in common, right?

Please help me, doctor!
Thank you.


Date: 06/05/99 at 12:10:07
From: Doctor Anthony
Subject: Re: A killer vector problem

>Three concurrent straight lines 0A, 0B, 0C are produced to D, 
>E, and F respectively. Prove that the points of intersection of 
>AB and DE, BC and EF, and CA and FD are collinear.

This is Desargues's theorem. It will be more convenient to use A', B', 
and C' for the points D, E, and F so one can follow the proof more 
easily. Then if BC, B'C' meet at P; CA, C'A' meet at Q; and AB, A'B' 
meet at R then we must prove P, Q, and R are collinear.

Use the well-known result that if P, Q, and R are collinear then the 
two conditions that will be satisfied are 

      k1.p + k2.q + k3.r = 0
and         k1 + k2 + k3 = 0

where p, q, and r are position vectors respectively of P, Q and R, and 
k1, k2, and k3 are scalar constants.

Since OAA' are collinear we have (taking v as position vector of O)

      v + k1.a + k1'.a' = 0
       1 + k1 + k1'     = 0

Since OBB' are collinear we have

      v + k2.b + k2'.b' = 0
      1 + k2 + k2'      = 0

Since OCC' are collinear we have 

      v + k3.c + k3'.c' = 0
       1 + k3 + k3'     = 0

From the second and third pairs of equations

     k2.b - k3.c       k2'b' - k3'c'
    -------------  =  -------------- 
       k2 - k3           k2' - k3'

and these must both represent the point P where BC intersects B'C' = p

  and so  (k2-k3) x p = k2.b - k3.c

Similarly  (k3-k1) x q = k3.c - k1.a

  and      (k1-k2) x r = k1.a - k2.b

adding these three equations we get

  (k2-k3)p + (k3-k1)q + (k1-k2)r = 0   and of course
   (k2-k3) + (k3-k1) + (k1-k2)   = 0

and this is the condition for P, Q, and R to be collinear.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 06/06/99 at 10:24:26
From: William Glenn
Subject: Re: A killer vector problem

To Doctor Anthony (and also maybe Doctor Floor, who seems to know a 
lot about Desargues and Menelaus):

Dr. Anthony, thank you very much for the prompt reply. This is my 
first time here, so I don't know if that was lucky or this place is 
just very efficient!

I feel a bit of a fool to say this, but I tried for hours and couldn't 
understand your proof entirely...

I don't want to come across as ungrateful, thank you kindly for the 
time you took to reply.

please continue...

I searched for Desargues and I had no idea that this problem was 
actually so famous. I found all sorts of resources and diagrams, but 
no proofs. There were a few messages saying the proof needed to 
utilize the Menelaus theorem (I can understand Menelaus' theorem, but 
when I tried I couldn't apply it to Desargues'). There was even one 
message...

 http://mathforum.org/dr.math/problems/verhulst7.3.98.html   

that said that by just using the axioms on connection of points and 
intersection of lines, proving the theorem was impossible.

This was interesting, because when our teacher gave us the problem, 
she said that the proof could be done with what we had learned so far. 
Also, that it was tedious and about six pages long. She said that when 
they gave the same problem in an assignment six years ago, only one 
student got it right; and she is a quantum physicist or something now.

We have only just begun studying vectors in the "Matrices and Vectors" 
unit. We have not covered dot product and cross product yet, but we 
have done all sorts of addition and subtraction, and applications to 
forces, velocities, etc.

>Use the well-known result that if P, Q, and R are collinear 
>then the two conditions that will be satisfied are 
>
>      k1.p + k2.q + k3.r = 0
>and         k1 + k2 + k3 = 0
>
>where p, q, and r are position vectors respectively of P, Q and >R, 
and k1, k2, and k3 are scalar constants.

I wasn't aware of this result. Is there a proof? Could I please bother 
you for a little elaboration on the nature of this, because it would 
be nice to understand what I am doing. What determines the magnitude 
of each scalar constant?

>>P, Q and R are collinear if their vector joins are parallel 
>>and they have a point in common, right?

I searched for collinear results, I could not find this one - I just 
kept finding the determinant of a 3x3 containing the three points, 
with the 3rd column all ones. Apparently when that determinant is 0, 
the points are collinear, but I can't fathom that either.

>Since OAA' are collinear we have (taking v as position vector 
>of O)
>
>      v + k1.a + k1'.a' = 0
>       1 + k1 + k1'     = 0

Question: Why do we take v as position vector of O here? Couldn't we 
just leave it as O?

Also, how does v relate to 1 in the second term?


>From the second and third pairs of equations
>
>     k2.b - k3.c       k2'b' - k3'c'
>    -------------  =  -------------- 
>       k2 - k3           k2' - k3'
>
>and these must both represent the point P where BC intersects 
>B'C' = p

Algebraically, I understand what has happened here and that this is 
valid, but why do both these terms equal p?

Thank you for your time.

P.S. I think this place is great; everything here is so friendly :)


Date: 06/06/99 at 11:46:22
From: Doctor Anthony
Subject: Re: A killer vector problem

I have given a proof of the vector condition for three points to be 
collinear - see further down the page. If there are other 
difficulties, write back.

Doctor Anthony
----------------------------------------------------------------

To show that if P, Q, and R are collinear then the following pair of 
equations will be satisfied:

         k1.p + k2.q + k3.r = 0
   and         k1 + k2 + k3 = 0

Suppose that P, Q, and R are three collinear points with Q dividing PR 
in the ratio a:b

With origin O we can express the position vector of Q by going from O 
to P then a fraction a/(a+b) along the vector PR, that is along the 
vector (r-p)and so we have

                 q = p + a/(a+b) [r-p]

                     (a+b)p + a(r-p)
                 q = ---------------
                          a+b

            (a+b)q = ap + bp + ar - ap

            (a+b)q =  bp + ar

and so

              bp - (a+b)q + ar = 0

   or       k1.p + k2.q + k3.r = 0

also

        k1 + k2 + k3 = b -(a+b) + a = 0

Therefore we get

            k1.p + k2.q + k3.r = 0
    and           k1 + k2 + k3 = 0

I hope the rest of the proof was clear.  You might need to write down 
a few intermediate steps but it is all standard algebra.
----------------------------------------------------------------

>Question: Why do we take v as position vector of O here? 
>Couldn't we just leave it as O?

To make the equation fit the pattern that I gave at the start I have 
taken O as a point with position vector v. It doesn't matter if you 
prefer to call it 0.


>Also, how does v relate to 1 in the second term?

If you have an equation with 0 on the righthand side you can multiply 
or divide by ANY factor you like to make one of the coefficients equal 
to 1.


>>From the second and third pairs of equations
>>
>>     k2.b - k3.c       k2'b' - k3'c'
>>    -------------  =  -------------- 
>>       k2 - k3           k2' - k3'
>>
>>and these must both represent the point P where BC intersects 
>>B'C' = p
>
>Algebraically, I understand what has happened here and that 
>this is valid, but why do both these terms equal p?

The lefthand side MUST represent some point on BC, and the righthand 
side MUST represent some point on B'C'. Since the two sides are equal 
the point must be where BC and B'C' intersect. That is the point P 
(position vector p)

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 06/07/99 at 10:29:00
From: William.
Subject: Re: A killer vector problem

>I have given a proof of the vector condition for three points 
>to be collinear - see further down the page. If there are other 
>difficulties, write back.

Yes that proof made sense, thank you!


>I hope the rest of the proof was clear.  You might need to 
>write down a few intermediate steps but it is all standard 
>algebra.

Yes, that's all clear now.  :)


>>Question: Why do we take v as position vector of O here? 
>>Couldn't we just leave it as O?
>
>To make the equation fit the pattern that I gave at the start I 
>have taken O as a point with position vector v. It doesn't 
>matter if you prefer to call it 0.

Okay, not confused here anymore.


>>Also, how does v relate to 1 in the second term?
>
>If you have an equation with 0 on the right hand side you can 
>multiply or divide by ANY factor you like to make one of the 
>coefficients equal to 1.

Yes.

So, the constant in front of the null vector could just as easily be, 
say, seven - and that wouldn't produce different results? Dividing 
through by seven would change the other two constants, is that okay? 
It seems that using the null vector limits the freedom of the other 
two coefficients. Are you just allowed to decide that in this case to 
make it one? Couldn't the null vector coefficient validly be 0, would 
that mess anything up?

>The left hand side MUST represent some point on BC, the right 
>hand side MUST represent some point on B'C'. Since the two 
>sides are equal the point must be where BC and B'C' intersect, >that 
is the point P (position vector p)

In this linear combination, why must each side represent a point on BC 
or B'C'? It _sounds_ reasonable enough, but this is the only step 
which still seems trivial to me, now.

Is it a problem that each side only has two variable terms? I mean, 
two variable points will always be collinear, just as three variable 
points will always be coplanar.

Thanks for your time
  -wim


Date: 06/07/99 at 11:45:46
From: Doctor Anthony
Subject: Re: A killer vector problem

>So, the constant in front of the null vector could just as 
>easily be, say, seven - and that wouldn't produce different 
>results? Dividing through by seven would change the other two 
>constants, is that okay? It seems that using the null vector 
>limits the freedom of the other two coefficients. Are you just 
>allowed to decide that in this case to make it one? Couldn't 
>the null vector coefficient validly be 0, would that mess 
>anything up?

Normally you can't make one of the coefficients zero because there is 
no factor that will make a coefficient go to zero except zero and that 
would knock out all the other terms as well.


>In this linear combination, why must each side represent a 
>point on BC or B'C'? It _sounds_ reasonable enough, but this is 
>the only step which still seems trivial to me, now.
>
>Is it a problem that each side only has two variable terms? I 
>mean, two variable points will always be collinear, just as 
>three variable points will always be coplanar.

b is position vector of B and c is position vector of C. So any point 
on BC will be of the form:

         b + k(c-b)   = b(1-k) + kc

Note that the sum of the coefficients is 1. Provided the sum of the 
coefficients is 1, then ANY expression of the form  p.b + q.c  where 
p+q = 1 is a point on BC.

Now look at the expression that you asked about:

     k2.b - k3.c      k2'b' - k3'c'
    ------------- =  --------------
       k2 - k3          k2' - k3'

   
The sum of the coefficients on the lefthand side is

       k2       -k3       k2-k3
     -----  + -------  = ------- =  1
     k2-k3     k2-k3      k2-k3

so the point in space represented by the lefthand side of the equation 
is a point on the line BC. Similarly, the sum of the coefficients on 
the righthand side is 1, so the righthand side represents a point on 
B'C'.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Linear Algebra

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