Proving three vectors are coplanar

[FaraDazed]

Homework Statement

Prove that the vectors a=(3,2,-1), b=(5,-7,3), c=(11, -3, 1) are coplanar.

Homework Equations

not sure

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

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 :)

 

[Simon Bridge]

There is a way to check for yourself.

Any two vectors are coplanar.
The cross product of any two will be perpendicular to both.
If the third vector has a component in the direction of the cross product of the other two, then it is not coplanar with the other two.
The component of a vector along another vector is given by...

Also see: https://www.physicsforums.com/threads/proving-3-vectors-are-coplanar.155647/

Note: For matrixes in LaTeX:
http://latex.wikia.com/wiki/Matrix_environments

 

[FaraDazed]

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.

 

[Physgeek64]

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

 

[Ray Vickson]

Homework Statement

Prove that the vectors a=(3,2,-1), b=(5,-7,3), c=(11, -3, 1) are coplanar.

Homework Equations

not sure

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

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 :)

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
$$3 r + 5 s = 11\\<br /> 2r - 7 s = 3\\<br /> -r + 3s = 1$$
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 $\vec{b}$.

 

[FaraDazed]

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
$$3 r + 5 s = 11\\<br /> 2r - 7 s = 3\\<br /> -r + 3s = 1$$
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 $\vec{b}$.

Thank you, that has helped a lot :) . I agree about having to have a good understanding of what is happening, everything makes much more sense when it is understood what's going on. Like in this situation however, I had to hand in the coursework the following day so just knowing if the method and answer was correct was all I needed first, then I can get to grips with what's actually going on afterwards, like now :)