# Proving three vectors are coplanar

1. Oct 19, 2014

1. The problem statement, all variables and given/known data
Prove that the vectors a=(3,2,-1), b=(5,-7,3), c=(11, -3, 1) are coplanar.

2. Relevant equations
not sure

3. The attempt at a solution
First time I have ever came across the term coplanar, I know what it means but did not know any tests for it. So I did a little research and found that one way is to test if they are linearly independent and that if that are not linearly independent then that means they must be coplanar.

So where I found this out, it had a few example of testing for linearly independency using the determent of a matrix formed by the three vectors (as column vectors) and I think I have done it correct as my answer comes out to zero. But as I am extremely new to this (the past half hour or so) I wanted a second opinion.

$Det \begin{vmatrix} 3 & 5 & 11 \\ 2 & -7 & 3 \\ -1 & 3 & 1 \end{vmatrix} = 3(-7+9)-5(2-3)+11(6-7) = 6+5+(-11)=0$

And therefore as the determent is zero, that means they are NOT linearly independent, and hence means that they are coplanar. At least this is what I have gathered from a little bit of research.

Appreciate it if someone could double check the thinking behind the method. Thanks :)

Last edited: Oct 19, 2014
2. Oct 19, 2014

### Simon Bridge

3. Oct 19, 2014

Oops, I meant independent not dependendent, so I have gone back and corrected that. Plus added in the matrix, (thanks Simon for the link) .

Thanks for the link to the other thread too, I did come across it before but did not read all of it.

4. Oct 21, 2014

### Physgeek64

you have to show that one vector is a scalar multiple of the other and that they share a common point. xx

5. Oct 21, 2014

### Ray Vickson

Yes, your reasoning is OK. However, it would be better if you really understood what is happening, rather than just plugging in formulas you do not fully grasp. The point is that if the three vectors are coplanar, then (unless they all point along a single line), one of them will be a linear combination of the other two. In other words, there should be numbers r and s giving $\vec{c} = r \vec{a} + s \vec{b}$. In detail, these say
$$3 r + 5 s = 11\\ 2r - 7 s = 3\\ -r + 3s = 1$$
You can solve for r and s from the first two equations, then see if the resulting solution also satisfies the third equation; if it does, you are done, as you will have found a way to express $\vec{c}$ as a linear combination of $\vec{a}$ and $\vec{b}$.

6. Oct 21, 2014