The proof of Triple Vector Products

THe question is to show the following: $$\vec{a}\times (\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$$ $$(\vec{a}\times \vec{b})\times\vec{c}=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{b}\cdot\vec{c})\vec{a}$$ In fact, I am able to finish the proof using high school method, which is calculate $$\vec{b}\times\vec{c}$$ and then cross with $$\vec{a}$$. However, I would like to use another more mathematical way to prove this triple vector product.

For the first one, $$\vec{b}\times\vec{c}$$ is a perpendicular vector towards b and c. Then this vector is cross with a. Then, the final results $$\vec{a}\times (\vec{b}\times\vec{c})$$ is a vector lies on a plane where b and c do also. Hence, it is a linear combination of b and c. $$\vec{a}\times (\vec{b}\times\vec{c})=x\vec{b}+y\vec{c}$$ Take a dot product with $$\vec{a}$$to both side, L.H.S becomes $$0$$. $$0=x(\vec{a}\cdot\vec{b})+y(\vec{a}\cdot\vec{c})$$ Then...how an I go further?? I have read a similar proof, but I don't understand the step after I got $$0=x(\vec{a}\cdot\vec{b})+y(\vec{a}\cdot\vec{c})$$. The reference link is http://www.fen.bilkent.edu.tr/~ercelebi/Ax(BxC).pdf

Thank you.

First, note that we are dealing with polynomial functions here so we can freely impose nonvanishing of certain expressions (as long as there is some $$\vec{a},\vec{b},\vec{c}$$ that achieves this) and then use a continuity argument to pass to the limiting case. You have already use this when you say you can write the triple product as a linear combination of $$\vec{b}$$ and $$\vec{c}$$ (you have thrown away cases where $$\vec{b}\parallel\vec{c}$$ only to regain them at the limit).

To start with, assume $$\vec{a}\cdot\vec{b},\vec{a}\cdot\vec{c}$$ are not simultaneously zero. Then the solutions $$(x,y)$$ to $$0=(\vec{a}\cdot\vec{b})x+(\vec{a}\cdot\vec{c})y$$ lies on the line perpendicular to $$(\vec{a}\cdot\vec{b},\vec{a}\cdot\vec{c})$$ and hence you can introduce a constant of proportionality $$\lambda$$ (a priori could depend on $$\vec{a},\vec{b},\vec{c}$$) so that $$(x,y)=\lambda(\vec{a}\cdot\vec{c},-\vec{a}\cdot\vec{b})$$ and what's more, this constant $$\lambda$$ can be chosen to be independent of $$\vec{a},\vec{b},\vec{c}$$. Your reference just assert this and went on to calculate $$\lambda$$ using specific case. But here is a simpler way to justify part of it that would suffice for the derivation.

We can further restrict our attention to the case $$\vec{a}\times(\vec{b}\times\vec{c})\neq\vec{0}$$. Then it doesn't hurt to replace $$\vec{a}$$ by its orthogonal projection along $$\vec{b}\times\vec{c}$$ to the plane containing $$\vec{b}$$ and $$\vec{c}$$ (because it wouldn't change the vector triple product nor $$\lambda$$). Hence we just need to show that $$\lambda$$ is the same for $$\vec{b}\times(\vec{b}\times\vec{c})$$ and $$\vec{c}\times(\vec{b}\times\vec{c})$$. Taking dot product with $$\vec{c}$$ in the $$\vec{b}\times(\vec{b}\times\vec{c})$$ gives $$[\vec{c},\vec{b},\vec{b}\times\vec{c}]= \lambda(\vec{c}\cdot\vec{b})(\vec{b}\cdot\vec{c})-\lambda(\vec{c}\cdot\vec{c})(\vec{b}\cdot\vec{b})$$ but the LHS is, by cyclic permutation of the scalar triple product \begin{align*} \vec{c}\cdot(\vec{b}\times(\vec{b}\times\vec{c})) &=(\vec{b}\times\vec{c})\cdot(\vec{c}\times\vec{b})\\ &=-|\vec{b}\times\vec{c}|^2\\ &=-b^2c^2\sin^2\theta\\ &=(bc\cos\theta)^2-b^2c^2\\ &=(\vec{b}\cdot\vec{c})^2-(\vec{b}\cdot\vec{b})(\vec{c}\cdot\vec{c}) \end{align*} So $$\lambda=1$$ in this case. Similarly, taking scalar product with $$\vec{b}$$ for $$\vec{c}\times(\vec{b}\times\vec{c})$$, we get $$\lambda=1$$ too in this case. So we can take $$\lambda=1$$ for all $$\vec{a},\vec{b},\vec{c}$$.

Carrying on from where you left of, \begin{align} y = -\lambda (\vec{a} \cdot \vec{b} ) &\Rightarrow x = \lambda (\vec{a} \cdot \vec{c} ) \\ \vec{a} \times ( \vec{b} \times \vec{c} ) &= \lambda [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} ] \\ \end{align} Comparing the magnitude of the vectors, gives: \begin{align} ||\vec{a} \times ( \vec{b} \times \vec{c} )||^2 &= ||\lambda [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} ]||^2 \\ &= \lambda^2 [ (\vec{a} \cdot \vec{c})^2 ||\vec{b} ||^2 + (\vec{a} \cdot \vec{b})^2 ||\vec{c} ||^2 - 2(\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c} )] \\ LHS &= ||\vec{a} \times ( \vec{b} \times \vec{c} )||^2 \\ &= ||\vec{a} ||^2 ||\vec{b} \times \vec{c} ||^2 - [\vec{a} \ \vec{b} \ \vec{c} ]^2 \\ &= ||\vec{a} ||^2 ( ||\vec{b} ||^2||\vec{c} ||^2 - ( \vec{b} \cdot \vec{c})^2) - [ \ ||\vec{a} ||^2 ||\vec{b} ||^2||\vec{c} ||^2 + 2(\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c} ) - ||\vec{a}||^2(\vec{b}\cdot\vec{c})^2 - ||\vec{b}||^2(\vec{a}\cdot\vec{c})^2 - ||\vec{c}||^2(\vec{a}\cdot\vec{b})^2] \\ &= (\vec{a} \cdot \vec{c})^2 ||\vec{b} ||^2 + (\vec{a} \cdot \vec{b})^2 ||\vec{c} ||^2 - 2(\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c} ) \\ &\Rightarrow \lambda^2 = 1 \Rightarrow \lambda = \pm1 \\ \Rightarrow \vec{a} \times ( \vec{b} \times \vec{c} ) &= \pm [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} ] \end{align} Determining the sign: \begin{align} \vec{a} \times ( \vec{b} \times \vec{c} ) &= \pm [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} ] \\ \Rightarrow [\vec{a} \times ( \vec{b} \times \vec{c} )] \times \vec{c} &= \pm (\vec{a} \cdot \vec{c}) [\vec{b} \times \vec{c}] \end{align} Using right-hand thumb rule for cross product, it is easy to see that the direction of the LHS and RHS vectors matches with the $$+$$ sign. \begin{align} \therefore \vec{a} \times ( \vec{b} \times \vec{c} ) &= [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} ] \\ (\vec{a} \times \vec{b}) \times \vec{c} &= - [\vec{c} \times ( \vec{a} \times \vec{b} )], \text{ thus solvable.}\\ \end{align}

Note: In the proof of $$\lambda = \pm1$$, I have used the following properties:

(1) $$\vec{p} = \alpha\vec{x} + \beta \vec{y} \Rightarrow ||\vec{p} ||^2 = |\alpha|^2||\vec{x}||^2 + |\beta|^2 ||\vec{y}||^2 + 2\alpha \beta (\vec{x} \cdot \vec{y})$$.

(2) Lagrange's Identity: $$||\vec{x} \times \vec{y}||^2 + ||\vec{x} \cdot \vec{y}||^2 = ||\vec{x}||^2 ||\vec{y}||^2$$

(3) $$[\vec{a} \ \vec{b} \ \vec{c} ]^2 = ||\vec{a} \cdot( \vec{b} \times \vec{c}) ||^2 = \ ||\vec{a} ||^2 ||\vec{b} ||^2||\vec{c} ||^2 + 2(\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c} ) - ||\vec{a}||^2(\vec{b}\cdot\vec{c})^2 - ||\vec{b}||^2(\vec{a}\cdot\vec{c})^2 - ||\vec{c}||^2(\vec{a}\cdot\vec{b})^2$$

• Sorry, I would like to know how to come out with the first sentence: $$y=\lambda(\vec{a}\cdot\vec{b}) \quad \text{and} x=\lambda(\vec{a}\cdot\vec{c})$$ – Jason Ng Sep 23 '18 at 5:10
• I made a typo: I meant $y = -\lambda (\vec{a}\cdot \vec{b} )$. Edited now. Obviously this dismisses the case when $\vec{a}\cdot \vec{b} = \vec{a}\cdot \vec{c} = 0$, but this case is trivial. – Kaind Sep 23 '18 at 10:29