= Solution
Write all <vectors> in Cartesian components. The first component of the nested <cross product> is
$$
\begin{aligned}
[\mathbf a\times(\mathbf b\times\mathbf c)]_x
&=a_y(b_xc_y-b_yc_x)-a_z(b_zc_x-b_xc_z)\\
&=b_x(a_yc_y+a_zc_z)-c_x(a_yb_y+a_zb_z)\\
&=b_x(\mathbf a\cdot\mathbf c)-c_x(\mathbf a\cdot\mathbf b).
\end{aligned}
$$
The missing $a_x$ terms cancel as $-b_xa_xc_x+c_xa_xb_x=0$. Computing the $y$ and $z$ components by cyclic permutation gives the same form, proving the <vector triple product identity>
$$
\boxed{\mathbf a\times(\mathbf b\times\mathbf c)
=(\mathbf a\cdot\mathbf c)\mathbf b-(\mathbf a\cdot\mathbf b)\mathbf c.}
$$
Apply this expression to the three cyclic terms. Their sum is
$$
(\mathbf a\cdot\mathbf c)\mathbf b-(\mathbf a\cdot\mathbf b)\mathbf c
+(\mathbf b\cdot\mathbf a)\mathbf c-(\mathbf b\cdot\mathbf c)\mathbf a
+(\mathbf c\cdot\mathbf b)\mathbf a-(\mathbf c\cdot\mathbf a)\mathbf b=0.
$$
Each coefficient cancels because the <dot product> is symmetric. This is the <Jacobi identity> for the <cross product>.
Back to article page
