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