Prove 3 Vectors Not Coplanar: Linear Algebra Exam

[kaalen]

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?

 

[amcavoy]

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.

 

[robphy]

I'd vote: Mixed Vector Product.

my $0.02

 

[HallsofIvy]

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!)

 

[kaalen]

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 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

Thank you for your help and suggestions

 

[robphy]

If you stare at the vectors given, you might see a pattern... from which you can see that p\vec x+q\vec y+m\vec z=\vec 0. So, those vectors \vec x, \vec y, \vec z 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. (\vec x \times \vec y) \cdot \vec z) 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., \vec a=a_1 \hat\imath +a_2\hat\jmath +a_3\hat k, etc... This looks quite tedious. If you do it algebraically, (\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), then distribute carefully..., it should be simpler.

 

[HallsofIvy]

Aw! Robphy- that was just a skosh more than you really should have given in the homework threads! Leave something for kaalen to do!

 

[robphy]

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.