Prove that four vectors are coplanar

I am given four vectors, A, B, C, and D, and need to show that they are coplanar.

I don't really know what to do. If it were three vectors, I know that I should scalar triple product them showing that the parallelepiped formed has zero volume, but when they add a fourth I'm not sure what to do.

Re: Prove that four vectors are coplanar

Would it work to show that A, B, and C are linearly dependent (and thus coplanar) and then do the same for B, C, and D?

Re: Prove that four vectors are coplanar

Are the vectors in $\displaystyle \mathbb{R}^3$? If so, you just need to show that the cross product of any two of those four vectors always results in a vector that is parallel (to any other choice of 2 vectors). That is, the set of all possible cross products should be vectors parallel to each other.

Your own idea above should work too, unless there's something counter-intuitive im not seeing.

Re: Prove that four vectors are coplanar

General approach that works for any number of vectors in any dimension of vector space:

Any set of vectors are coplanar if and only if the dimension of the subspace they span is less than or equal to 2.

For any list of vectors, the typical computational procedure of finding their linear span is by putting them in a matirx - each vector becomes a row - and then row reducing. (That's because the actions of row reducing a matrix are the same as the action of replacing a vecotr by some linear combination of it and other vectors in the linear span.) Once you've row reduced as far as possible, the non-zero rows remaining are necessarily linearly independent, and hence, when intepreted as vectors, are the basis of the linear span of the initial set of vectors. The number of basis vectors of the linear span is the dimension of the linear. If you're asking about coplanar, you're asking if the linear span has two or fewer basis vectors.

Thus, put your vectors into a matrix, and then row reduce as far as possible. If the number of non-zero rows is two or less, then your original vectors are coplanar, otherwise, they aren't coplanar.

Approach when your vectors are in $\displaystyle \mathbb{R}^{3}$:

Take the cross product of two of your vectors that produces a non-zero result (If that's impossible, then they're all coplanar - in fact, colinear). Call it N. Then all your vectors are coplanar if and only if all of them are in the plane perpendicular to the vector N (you already know the 1st two are in that plane). A vector is perpendicular to N if and only if its dot product with N is 0.

So in your case, with four vectors A, B, C, D in $\displaystyle \mathbb{R}^{3}$ they're all coplanar if and only if (assuming A and B are chosen so that AxB isn't 0):

<(A x B), C> = <(A x B), D> = 0. (There, <*,*> is the vector dot product, x is the vector cross product, and the N I refered to is AxB).