Basically that just says that A\cdot (B\times C)=0 (or any reordering thereof). BXC is a vector that is perpendicular to B and C. If A is coplanar with B and C, then it can be expressed as a linear combination of the two, i.e. A=bB+cC where b and c are real numbers. In that case, then it's obvious that A dotted into this vector which is perpendicular to both B and C would be 0.
Another way to think about it is to note that the above triple product has a value which is the volumn of the parallelepiped defined by A, B and C. If A, B and C are coplanar, then the parallelepiped has 0 volume.
#3
Jim Kata
197
10
What Matterwave said is correct it follows from a theorem that says the determinant of a matrix is non zero if and only if the vectors which make it up are all linearly independent. So if the vectors are coplanar obviously one the vectors is linearly dependent so the determinant of the matrix they form must be zero.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?