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I'm stuck on one step of the proof for the identity: $$ \vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$$

So far, the proof follows as:

We know that $\vec{B}\times\vec{C}$ gives a vector perpendicular to both $\vec{B}$ & $\vec{C}$, and that $ \vec{A}\times(\vec{B}\times\vec{C})$ gives a vector perpendicular to both $\vec{A}$ & $(\vec{B}\times\vec{C})$. Therefore, the vector $\vec{A}\times(\vec{B}\times\vec{C})$ must lie in the plane containing both $\vec{B}$ & $\vec{C}$.

Provided $\vec{B}$ & $\vec{C}$ are not parallel (if they were, $\vec{A}\times(\vec{B}\times\vec{C}) = 0$ regardless), vectors $\vec{B}$ & $\vec{C}$ span the 2D plane containing them both.

Therefore, we can express any vector in the plane as a linear combination of both $\vec{B}$ and $\vec{C}$ and so we can write: $$\vec{A}\times(\vec{B}\times\vec{C}) = \alpha\vec{B} + \beta\vec{C} \tag{1}$$

Taking the scalar product of both sides with $\vec{A}$: $$\vec{A} \cdot (\vec{A}\times(\vec{B}\times\vec{C})) = \vec{A} \cdot (\alpha\vec{B} + \beta\vec{C}) = 0$$ So, $$\alpha(\vec{A} \cdot \vec{B}) + \beta(\vec{A} \cdot\vec{C}) = 0$$

Now writing, $$\lambda = \frac{\alpha}{\vec{A} \cdot\vec{C}} = -\frac{\beta}{\vec{A} \cdot \vec{B}}$$ and substituting $\alpha$ and $\beta$ back into (1) we get: $$ \vec{A}\times(\vec{B}\times\vec{C}) = \lambda(\vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B}))$$ I am able to show $\lambda = 1$ with particular choices of unit vectors for $\vec{A}, \vec{B}, \vec{C}$ but I am unable to prove that $\lambda$ is independent of the magnitude of vectors (i.e. $\lambda = 1$ for all choices of $\vec{A}, \vec{B}, \vec{C}$). This is the step that I am struggling with. Any suggestions?

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    $\begingroup$ If you want to finish your proof, then there some more care needs to be done. Split $\vec{C}$ into its orthogonal and parallel components with respect to the vector $\bf{x} = \vec{A}\times\vec{B}$ and then cross your equation $(1)$ from the left by $\vec{B}.$ After you reduce both sides as much as possible, you should get $\lambda = 1.$ $\endgroup$
    – dezdichado
    Jan 6 at 22:30
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Observe that both sides of the equation to be proved are linear in each of the three variables (when the other two are fixed).

Based on this, it's enough to prove the statement for all possible combinations of the standard basis, say.

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  • $\begingroup$ +1, and obviously one can omit the case $B=C$. $\endgroup$ Jan 6 at 14:15
  • $\begingroup$ so you are saying it's not clear at all at this point that the lambda is constant? i read this argument quite frequently and this step is always missing. i wonder if its actually obvious if regarded correctly. $\endgroup$
    – peter
    Jun 11 at 12:42
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(This is a long comment.) This is probably not the simplest way to derive the identity, but the identity follows quite quickly from a special case of Cauchy-Binet formula: \begin{aligned} \left(A\times(B\times C)\right)\cdot x &=\det(A,B\times C,x)\quad\text{(defintion of cross product)}\\ &=\det(x,A,B\times C)\\ &=(x\times A)\cdot(B\times C)\quad\text{(defintion of cross product)}\\ &=(x\cdot B)(A\cdot C)-(x\cdot C)(A\cdot B)\quad\text{(Cauchy-Binet formula)}\\ &=\left(B(A\cdot C)-C(A\cdot B)\right)\cdot x.\\ \end{aligned} Since $x$ is arbitrary, the result follows.

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