Why is $(\vec{b} \times \vec{a}) \times \vec{b}$ non zero? If I use the rule of vector triple product, it becomes:
$$ \vec{b} \times \vec{a} \times \vec{b} = \vec{a} ( |\vec{b}|^2) - \vec{b}(\vec{b} \cdot \vec{a})$$
which is generally non-zero, but suppose I use properties of cross product:
$$ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}$$
Hence,
$$ \vec{b} \times \vec{a} \times \vec{b} = - \vec{a} \times \vec{b} \times \vec{b} =  \vec{a} \times (\vec{b} \times \vec{b})=0$$
What did I do wrong?
 A: $\vec S= \vec P \times (\vec Q  \times \vec R)= (\vec P. \vec R) \vec Q-(\vec P. \vec Q) \vec R$
So $(\vec b \times a)\times \vec b=-\vec b\times(\vec b \times \vec a)=-[(\vec b. \vec a) \vec b- (\vec b . \vec b) \vec a] \ne 0$
A: Removing the parenthesis makes you perform invalid transformations, because the cross product is not associative.
In fact $$\vec a\times\vec b\times\vec c$$ has no defined meaning.

Unless $\vec a,\vec b$ are parallel, $\vec b\times\vec a$ is orthogonal to $\vec b$, hence not parallel, and $(\vec b\times\vec a)\times\vec b$ is nonzero.
A: If $\mathbf{x}$ and $\mathbf{y}$ are perpendicular, then $\|\mathbf{x} \times \mathbf{y}\| = \|\mathbf{x}\|\;\|\mathbf{y}\|$.  So in that case, if the two factors are nonzero, so is the cross product.  But $\mathbf{b} \times \mathbf{a}$ is perpendicular to $\mathbf{b}$, so we can conclude: if $\mathbf{b} \times \mathbf{a}$  is nonzero, then so is $(\mathbf{b}\times\mathbf{a})\times \mathbf{b}$.
Note, however, that if $\mathbf{a}$ and $\mathbf{b}$ are parallel, then
$\mathbf{b} \times \mathbf{a} = \mathbf{0}$ and your conclusion fails.
