For cross product we have Lagrange's formula: $$\vec a \times (\vec b \times \vec c) = \vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b).$$

The formula can be verified by calculating in coordinates. I was thinking about a more geometrical approach and I have noticed this. If we denote $\vec v= \vec a \times (\vec b\times \vec c)$, then:

  • The vector $\vec v$ is perpendicular to $\vec b\times \vec c$, which means that is is a linear combination1 of $\vec b$ and $\vec c$. So we have $$\vec v= \beta\vec b+ \gamma \vec c$$ for some real numbers $\beta$ and $\gamma$.
  • This vector is also perpendicular to $\vec a$, which means that $$\vec a \cdot \vec v = 0 = \beta \vec a \cdot \vec b + \gamma \vec a \cdot \vec c.$$

The above equation suffices to determine the vector $\vec v$ up to a scalar multiple, since we know the ratio between $\beta$ and $\gamma$. In this way we get $$\vec a \times (\vec b \times \vec c) = K\left(\vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)\right) \tag{*}$$ where $K$ is some real number, still to be determined.

Is there some simple trick which helps to determine $K$ in $(*)$?

1To be more precise, this argument only works if $\vec b\times\vec c\ne\vec 0$, i.e., if $\vec b$ and $\vec c$ are linearly independent. But if $\vec b \times \vec c$ both sides in $(*)$ are equal to zero, to the equation $(*)$ holds in this case, too.

  • $\begingroup$ I was thinking about this formula just yesterday! Too funny. $\endgroup$ – Omnomnomnom Apr 6 '17 at 13:35
  • $\begingroup$ Also, neat derivation so far! I've never seen this argument. $\endgroup$ – Omnomnomnom Apr 6 '17 at 13:37
  • $\begingroup$ Another thought: perhaps it is sufficient to note that $$ \|a \times (b \times c)\| = \|a\|\,\|b\|\, \|c\| |\sin \theta_{a,(b\times c)} \sin \theta_{b,c}| $$ $\endgroup$ – Omnomnomnom May 1 '17 at 5:31

Here's an idea:

With the scalar triple product, we have $$ b \cdot v = b \cdot [a \times (b \times c)] = (b \times c) \cdot (b \times a) $$ Now, with the formula at the end of this section (which apparently can be seen as the three dimensional case of the Binet-Cauchy identity), we have $$ (b \times c) \cdot (b \times a) = (b \cdot b)(a \cdot c) - (b \cdot a)(c \cdot b) =\\ (b \cdot b) (a \cdot c) - (b \cdot c)(a \cdot b) $$ On the other hand, we have $$ b \cdot v = K b \cdot \left(\vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)\right) =\\ K[(b \cdot b)(a\cdot c) - (b \cdot c)(a \cdot b)] $$


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