# Proof of $BAC-CAB$ identity missing step

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?

• 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 $\lambda = 1.$ Jan 6 at 22:30

• +1, and obviously one can omit the case $B=C$. Jan 6 at 14:15
(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.