Finding the conditions for $\mathbf{A}×(\mathbf{B}×\mathbf{C}) = (\mathbf{A}×\mathbf{B})×\mathbf{C}$.

Question

I wanted to find the conditions in which $\mathbf{A}×(\mathbf{B}×\mathbf{C})$ is equal to $(\mathbf{A}×\mathbf{B})×\mathbf{C}$.

On solving $\mathbf{A}×(\mathbf{B}×\mathbf{C})-(\mathbf{A}×\mathbf{B})×\mathbf{C}=\mathbf{0}$, I got $$ \mathbf{A}(\mathbf{B}\cdot\mathbf{C})- \mathbf{C}(\mathbf{A}\cdot\mathbf{B})=\mathbf{0}\tag{1} $$ In the solution manual, it's given that this is possible if and only if either $\mathbf{A}$ is parallel to $\mathbf{C}$ or $\mathbf{B}$ is perpendicular to $\mathbf{A}$ and $\mathbf{C}$.

But as per what I have learnt, this means that $(1)$ will be $\mathbf{0}$ only when both terms are in it are $\mathbf{0}$. But here why don't we consider the equality of those two terms?

I am beginning vector analysis, So it will be really helpful if someone could help me out.

Answer 1

The Jacobi identity is $$ \mathbf{A}\times(\mathbf{B}\times\mathbf{C})+ \mathbf{B}\times(\mathbf{C}\times\mathbf{A})+ \mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{0} $$ Using anticommutativity, we get $$ \mathbf{A}\times(\mathbf{B}\times\mathbf{C})- (\mathbf{A}\times\mathbf{B})\times\mathbf{C}= -\mathbf{B}\times(\mathbf{C}\times\mathbf{A}) $$ If $\mathbf{A}$ is parallel to $\mathbf{C}$, then $\mathbf{C}\times\mathbf{A}=\mathbf{0}$. If $\mathbf{B}$ is perpendicular to both $\mathbf{A}$ and $\mathbf{C}$, then it is parallel to $\mathbf{C}\times\mathbf{A}$ and therefore $\mathbf{B}\times(\mathbf{C}\times\mathbf{A})=\mathbf{0}$.

What about the converse? Suppose the two triple products are equal and that $\mathbf{A}$ is not parallel to $\mathbf{C}$. Then, in order for $\mathbf{B}\times(\mathbf{C}\times\mathbf{A})=\mathbf{0}$ we need that $\mathbf{B}$ is parallel to $\mathbf{C}\times\mathbf{A}$.

Answer 2

If $A$ is parallel to $C$ then the two terms are equal, but aren't necessarily individually equal to zero.

If $A$ and $C$ are not parallel, then they are linearly independent, hence the coefficients of the linear combination must both be zero (which reduces to $B$ being perpendicular to both).