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How would I prove the following statement:

$\vec{A}\cdot(\vec{B}\times \vec{C})=0 \Leftrightarrow \vec{A}, \vec{B}, \vec{C} \text{ is linearly dependent.}$

The inverse is quite straightforward. However, going from left to right is kind of difficult. I just can't find out how to start the proof.

I am trying to show for a general coordinate system, so calculation using elements is not what I need.

biology12323
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    Since the statement to prove is coordinate-independent, it actually suffices to prove the statement in any one coordinate system you choose. – Travis Willse Sep 23 '19 at 05:31
  • Also, it would be useful for you to list which facts about the cross and dot products you have available. The left-hand side is $\det\pmatrix{{\bf A} & {\bf B} & {\bf C}}$, from which both directions follow from the fact that a square matrix has determinant $0$ iff its columns are linearly dependent. – Travis Willse Sep 23 '19 at 05:34
  • That was a very helpful insight. Then, I suppose I can just prove in Cartesian coordinate system. So the left-hand side is equal to det(A B C). How would we prove det(A B C)=0 implies linear dependence? – biology12323 Sep 23 '19 at 05:52

3 Answers3

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If $B$ and $C$ are linearly dependent, then we are done.

If $B$ and $C$ are linearly independent, then they span a two-dimensional plane in $\Bbb{R}^3$. Note that $B \times C$ is normal to this plane. If $A \cdot (B \times C) = 0$, then $A$ is perpendicular to this normal vector, meaning that it must be contained in this plane. In other words, $A \in \operatorname{span}(B, C)$.

In either case, if $A \cdot (B \times C) = 0$, we have linear dependence of $A, B, C$.

Theo Bendit
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  • So linear dependence is another name for coplanarity? If sum of unit vectors of three vectors is not zero does it mean they are not coplanar and vice-versa? Since $$\vec A.(\vec B\times \vec C)=0\implies ABC[\hat A.(\hat B\times \hat C)]=0$$ – Aurelius Jan 22 '24 at 20:49
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    @Aurelius Coplanarity and linear independence are indeed connected, but they are certainly not the same thing. Coplanarity (of four or more points) is actually a form of affine dependence (see here and here), but affine dependence is more general than coplanarity. It also covers colinearity, or whenever there are more points than necessary to define the affine hull (the smallest translate of a linear subspace containing the points). – Theo Bendit Jan 24 '24 at 18:01
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    @Aurelius Three vectors $u, v, w$ will be linearly dependent if and only if the four points $0, u, v, w$ are coplanar. Two vectors $v, w$ will be linearly dependent if and only if the three points $0, v, w$ are colinear. This is a special case of the more general connection: $v_1, \ldots, v_n$ is linearly dependent if and only if $0, v_1, \ldots, v_n$ is affine dependent. – Theo Bendit Jan 24 '24 at 18:04
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    Regarding your final question, no, it's easy to find three unit vectors that don't sum to $0$ that are still linearly dependent. For example, if $\hat{A} = \hat{B} = \hat{C}$, all of which are unit vectors, then their sum has length $3$, not $0$. – Theo Bendit Jan 24 '24 at 18:06
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$َA \cdot B \times C$ is the volume of a parallelepiped whose edges are formed by three vectors. The only way the volume can be zero is if all three vectors are co-planar and every three vectors on a plane are linearly independent.

Andrew Chin
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M. Foster
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This triple product of the three vectors is basically the determinant formed by the three given vectors taken as the columns of the matrix associated with this determinant. Now, think about a standard property of the determinant function: If any column is a linear combination of the other columns, then the determinant is zero. Conversely, the determinant zero means the given matrix is not invertible that means the rank of the associated matrix is not 3. That also means that the given three vectors don't span the whole space R^3. Hence, these three vectors span a two or one-dimensional subspace of R^3. Both cases imply that the given three vectors are not linearly independent. Please think about it.

Matha Mota
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