Vector algebra problem "Show that $\vec a\times(\vec b\times \vec c)$ is perpendicular to '$\vec a$' and coplanar with '$\vec a$' and '$\vec b$' where $\vec a$,$\vec b$ and $\vec c$ are vectors."
I showed the first part but how shall I prove the "coplanar" part?
Thank you for the help.
 A: $$\vec {\mathbf a}\times(\vec {\mathbf b}\times \vec {\mathbf c})=(\vec {\mathbf a}\cdot\vec {\mathbf c})\vec {\mathbf b}-(\vec {\mathbf a}\cdot \vec {\mathbf b})\vec {\mathbf c}$$
Since dot product gives scaler quantity that's why $(\vec {\mathbf a}\cdot\vec {\mathbf c})$ and $(\vec {\mathbf a}\cdot\vec {\mathbf b})$ are scalers, so it will clearly give answer in this form $\;\;(k\vec {\mathbf b}-l\vec {\mathbf c})\;\;$ where k,l are scaler.and we know if two vectors are co planer so  there addition ,subtraction also in same plane (using triangle law in figure).

so 
$\;\;\vec {\mathbf a}\times(\vec {\mathbf b}\times \vec {\mathbf c})\;\;$ is co planer to $\;\;\vec {\mathbf b},\vec {\mathbf c}$.
this(edited link) will helpful.
A: The first part follows directly from the definition of the cross product. For the second part, recall that vectors are said to be coplanar if they all lie in the same plane. By the triple product expansion, we have
$$\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}.$$
Hence the vectors $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})$, $\mathbf{b}$, and $\mathbf{c}$ all lie in the plane spanned by $\mathbf{b}$ and $\mathbf{c}$, so they are coplanar.
