# Necessary and sufficient condition for zero scalar triple product

I want to show that the vector triple product of

$[x-y, b, a\times b]$, where $a$ and $b$ are not parallel is NOT equal to $0$ iff $[x-y, a, b]=0$.

Attempt: As $a$ and $b$ not parallel, $a \times b$ is not equal to $0$, the cross product of this with $b$ will produce some (non-zero) scalar multiple of $a$. Now this is scalar product-ed with $x-y$, which is not equal to $0$ as long a isn't perpendicular to $x-y$ but I can't seem to formulate we must have $[x-y,a,b]=0$.

This is motivated from finding the necessary and sufficient condition for two lines to intersect.

## 1 Answer

Hence you're attempting to prove two implications: $$\left[c,a,b\right] = 0 \implies \left[c,b,a\times b\right] \neq 0$$ and $$\left[c,a,b\right] \neq 0 \implies \left[c,b,a\times b\right] = 0$$ where $c = x - y$.

Immediately we have that if $c = 0$, the first implication fails, hence I'm assuming we have the additional restriction that $x \neq y$.

Even so, both implications fail to hold. As a counterexample for the first implication, consider $$a = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 1 \end{array}} \right],\;\;b=c = \left[ {\begin{array}{*{20}{c}} 0 \\ 1 \\ 0 \end{array}} \right]$$ where $\left[ {c,a,b} \right] = \left[ {c,b,a \times b} \right] = 0$.

As a counterexample for the second implication, consider $$a = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 1 \end{array}} \right],\; b = \left[ {\begin{array}{*{20}{c}} 0 \\ 1 \\ 0 \end{array}} \right],\; c = \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right]$$ where $\left[ {c,a,b} \right] = - 1 \ne 0$ and $\left[ {c,b,a \times b} \right] = 1 \ne 0$.

If you're instead considering the vector triple product (which you mention once at the beginning of the preamble), the first implication holds, but the second does not.