Vector Calculus - Curl of Vector

Question

I'm asked to prove the following identity, using index notation:

$(\nabla\times A)\times A=A \cdot\nabla A - \nabla(A \cdot A)$

However, when I work it out, I find that the actual solution should be:

$(\nabla\times A)\times A=A \cdot\nabla A - \frac{1}{2}\nabla(A \cdot A)$

Am I missing something, or is the book wrong?

Answer 1

Expanding everything out componentwise, I get

$(\nabla\times A)\times A$ $$ \begin{align} &{\lower{1pt}{\Large(}}(\partial_1,\partial_2,\partial_3)\times(A_1,A_2,A_3){\lower{1pt}{\Large)}}\times(A_1,A_2,A_3)\\ &=(\partial_2A_3-\partial_3A_2,\partial_3A_1-\partial_1A_3,\partial_1A_2-\partial_2A_1)\times(A_1,A_2,A_3)\\ &=(A_3\partial_3A_1+A_2\partial_2A_1-A_3\partial_1A_3-A_2\partial_1A_2,\\ &\hphantom{=(}A_1\partial_1A_2+A_3\partial_3A_2-A_1\partial_2A_1-A_3\partial_2A_3,\\ &\hphantom{=(}A_2\partial_2A_3+A_1\partial_1A_3-A_2\partial_3A_2-A_1\partial_3A_1) \end{align} $$ $(A\cdot\nabla)A$ $$ \begin{align} &(A_1\partial_1+A_2\partial_2+A_3\partial_3)(A_1,A_2,A_3)\\ &=(A_1\partial_1A_1+A_2\partial_2A_1+A_3\partial_3A_1,\\ &\hphantom{=(}A_1\partial_1A_2+A_2\partial_2A_2+A_3\partial_3A_2,\\ &\hphantom{=(}A_1\partial_1A_3+A_2\partial_2A_3+A_3\partial_3A_3) \end{align} $$ $\frac12\nabla(A\cdot A)$ $$ \begin{align} &\tfrac12(\partial_1,\partial_2,\partial_3)(A_1A_1+A_2A_2+A_3A_3)\\ &=(A_1\partial_1 A_1+A_2\partial_1 A_2+A_3\partial_1 A_3,\\ &\hphantom{=(}A_1\partial_2 A_1+A_2\partial_2 A_2+A_3\partial_2 A_3,\\ &\hphantom{=(}A_1\partial_3 A_1+A_2\partial_3 A_2+A_3\partial_3 A_3) \end{align} $$ And from these, I get $$ (\nabla\times A)\times A=(A\cdot\nabla)A-\tfrac12\nabla(A\cdot A) $$