Prove 3 Vectors Not Coplanar: Linear Algebra Exam

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In summary, Robphy suggests that x, y, z are coplanar if and only if their determinants are zero. This can be solved algebraically using vectors in the form (q\vec c-m\vec b) \times (m\vec a -p\vec c)
  • #1
kaalen
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I'm studying for linear algebra exam and don't know how to solve the following problem:
We have three arbitrary vectors a,b and c. Then we have another 3 vectors:
x = qc - mb
y = ma - pc
z = pb - qa
and I have to prove that these three vectors are not coplanar.

My idea is that I should use mixed vector product because if vectors are coplanar then they don't form parallelepiped so the result of a mixed vector product which expresses the volume of the parallelepiped is equal to 0.

The other possible option is to try to prove that one vector can be expressed with the other two... but this can be very complicated I think.

What do you think... am I on the right way?
 
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  • #2
kaalen said:
I'm studying for linear algebra exam and don't know how to solve the following problem:
We have three arbitrary vectors a,b and c. Then we have another 3 vectors:
x = qc - mb
y = ma - pc
z = pb - qa
and I have to prove that these three vectors are not coplanar.

My idea is that I should use mixed vector product because if vectors are coplanar then they don't form parallelepiped so the result of a mixed vector product which expresses the volume of the parallelepiped is equal to 0.

The other possible option is to try to prove that one vector can be expressed with the other two... but this can be very complicated I think.

What do you think... am I on the right way?

Have you actually tried either of these or are they just ideas? I would suggest taking the determinant of the matrix with these three vectors as rows (like you stated above) and see what you come up with.
 
  • #3
I'd vote: Mixed Vector Product.

my $0.02
 
  • #4
I would be inclined to say that the simplest way of deciding if x, y, z are coplaner is by determining if they are independent. If they are coplaner, then they span a 2 dimensional subspace (a plane) and so 3 vectors can't be independent. If they are coplaner they are dependent and vice-versa.

HOWEVER, the problem with that and either of the methods you mention is that you don't know anything about a,b,c! You say they are "three arbitrary vectors". If a,b, c themselves are coplanar, then x,y,z must be. If a,b,c are independent themselves, then x,y,z might be coplanar anyway. Unless-- hmm, yes, it is quite possible that the coefficients of a, b, c are chosen so that does work!
(And that almost gives away the answer!)
 
  • #5
Ok... obviously I made some mistake in the first attempt so my equation just kept getting longer and longer... but I started from the beginning and it was getting shorter and shorter until I got the result 0 :redface: I did it with mixed product and used a lot of basic rules about scalar and vector multiplication.

When I get to matrices I'll try this as well... but right now I don't have a clue how to solve this problem with matrix :uhh:

Thank you for your help and suggestions :smile:
 
  • #6
If you stare at the vectors given, you might see a pattern... from which you can see that [tex] p\vec x+q\vec y+m\vec z=\vec 0[/tex]. So, those vectors [tex]\vec x[/tex], [tex]\vec y[/tex], [tex]\vec z[/tex] are coplanar.

multiply each vector as shown...
px =p( qc - mb )
qy =q( ma - pc )
mz =m( pb - qa )
do the sum ... for example pqc + (-qpc)=0, etc...

In case the above was not obvious [It wasn't obvious to me, at first glance],
computing the mixed vector-product (i.e. [tex] (\vec x \times \vec y) \cdot \vec z [/tex]) is probably the best general method. However, if you do this with components in the determinant form, you probably have to introduce a rectangular basis to write, e.g., [tex]\vec a=a_1 \hat\imath +a_2\hat\jmath +a_3\hat k[/tex], etc... This looks quite tedious. If you do it algebraically, [tex] (\vec x \times \vec y) \cdot \vec z = ( (q\vec c-m\vec b) \times (m\vec a -p\vec c)) \cdot (p\vec b-q\vec a)[/tex], then distribute carefully..., it should be simpler.
 
  • #7
Aw! Robphy- that was just a skosh more than you really should have given in the homework threads! Leave something for kaalen to do!
 
  • #8
HallsofIvy said:
Aw! Robphy- that was just a skosh more than you really should have given in the homework threads! Leave something for kaalen to do!

Oops...
but kaalen did have the right idea in the original post... and did carry it out, as reported in #5.

I'll be quiet now.
 

1. How do you prove that three vectors are not coplanar?

To prove that three vectors are not 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. However, if the cross product is not equal to the third vector, then the three vectors are not coplanar.

2. What is the significance of proving that three vectors are not coplanar?

Proving that three vectors are not coplanar is important in linear algebra because it helps to determine the linear independence of the vectors. If the three vectors are not coplanar, then they are linearly independent, meaning that they cannot be represented as a linear combination of each other.

3. Can you prove that three vectors are not coplanar without using the cross product?

Yes, there are other methods to prove that three vectors are not coplanar. One method is by using determinants. If the determinant of the matrix formed by the three vectors is non-zero, then the vectors are not coplanar. Another method is by using scalar triple products. If the scalar triple product of the three vectors is non-zero, then the vectors are not coplanar.

4. What is the difference between coplanar and collinear vectors?

Coplanar vectors lie on the same plane, while collinear vectors lie on the same line. In other words, coplanar vectors can be in any orientation on the same plane, while collinear vectors must be in a straight line.

5. How can you use the concept of coplanarity in real-world applications?

Coplanarity is often used in engineering and physics to determine the stability and equilibrium of structures. For example, in bridge construction, the beams and cables must be placed in a way that they are coplanar to ensure the stability of the bridge. In physics, the concept of coplanarity is used to analyze the forces acting on an object in equilibrium.

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