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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.