# Colinear points in a vectorplane

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1. Oct 8, 2016

### Marcus95

1. The problem statement, all variables and given/known data
If a, b and c are coplanar vectors related by λabc=0, where the constants are non-zero, show that the condition for the points with position vectors αa, βb and γc to be collinear is:

λ/α + μ/β + ν/γ = 0

2. Relevant equations
Dot product
Cross product
Tripple product
Vector equation of a line

3. The attempt at a solution
I am fairly new to vectors, so I don't really know where to begin.
Firstly, we know that for co-planar vectors, the tripple product is zero: [a,b,c]=0
Then, for the points so be colinear, the "slope" unit vector between them must be equal. Thus if we assume that point βb is in the middle:

ac)/|αac| = ab)/|αab|

From here what I can do is to expand the expressions in terms of the vector components, but this doesn't really bring me anywhere. I guess there should be an elegant solution without having to use components of vectors. Any suggestions where to begin?

Many thanks!

2. Oct 8, 2016

### andrewkirk

There are many ways to do it. An easy one given the options offered (although not the most intuitive one) is that two vectors are parallel iff their cross product is zero, where we also include vectors with exactly opposite directions as parallel.

Using what's above, can you write the formulas for two vectors that must be parallel for the three points to be collinear?

3. Oct 9, 2016

### Marcus95

Sure! We must have (αac) ⋅ (αab) = 0
If I evaluate it, I am left with: (αa)^2 - αβa*b - αγa*c + βγb*c = 0
from here, I don't know how to continue. I have tried to substitute c from the equation of the plane, which leaves me with:
(α^2ν+αγλ)a^2 + (αγμ-αβν-βγλ)a*b - βγμ b^2 = 0
but I don't know if this brings me any further.

Last edited: Oct 9, 2016
4. Oct 9, 2016

### andrewkirk

That is a dot product, not a cross product.

Re-write it as a cross product and expand the expression using the distributive law.

After that you can use the first equation in the OP to replace one of the vectors, say c, by an expression involving the other two, and things should then start to cancel out to give you the result you seek.

On reflection, I think it may have been easier to do this using the vector equation of a line instead. But one can never have too much practice with cross products (at least I can't) so let's do it that way first.

5. Oct 10, 2016

### Marcus95

Thank you very much, I got it!
However, would it be educational for me to try the solution with vector equation of lines as well? How should I start that approach?