Proving three vectors are coplanar

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In summary, to prove that the vectors a=(3,2,-1), b=(5,-7,3), c=(11, -3, 1) are coplanar, we can use the determinant of a matrix formed by the three vectors. If the determinant is equal to zero, then the vectors are not linearly independent and therefore must be coplanar. Another way to prove coplanarity is to check if one vector is a linear combination of the other two.
  • #1
FaraDazed
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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.
[itex]
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
[/itex]

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 :)
 
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  • #2
  • #3
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
you have to show that one vector is a scalar multiple of the other and that they share a common point. xx
 
  • #5
FaraDazed said:

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.
[itex]
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
[/itex]

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
[tex] 3 r + 5 s = 11\\
2r - 7 s = 3\\
-r + 3s = 1
[/tex]
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
Ray Vickson said:
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
[tex] 3 r + 5 s = 11\\
2r - 7 s = 3\\
-r + 3s = 1
[/tex]
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 :)
 

1. How do you determine if three vectors are coplanar?

To determine if three vectors are coplanar, you can use the cross product. If the cross product of any two of the vectors is equal to the third vector, then the three vectors are coplanar.

2. What is the mathematical definition of coplanar vectors?

Coplanar vectors are vectors that lie in the same plane. This means that they can be represented using the same set of coordinates in a two-dimensional plane.

3. Can three non-coplanar vectors be parallel?

No, three non-coplanar vectors cannot be parallel. If three vectors are parallel, they must lie in the same plane, making them coplanar.

4. How many vectors are needed to determine a plane in three-dimensional space?

At least three non-collinear vectors are needed to determine a unique plane in three-dimensional space. This means that the vectors cannot lie on the same line.

5. Can three vectors be coplanar in four-dimensional space?

No, three vectors cannot be coplanar in four-dimensional space. In four-dimensional space, three vectors must be linearly dependent, meaning that one vector can be expressed as a linear combination of the other two. This is not possible if the three vectors are coplanar.

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