If $f(x,y,z)=(r \times A)\cdot (r\times B)$, where $r=(x,y,z)$, show that $\nabla f(x,y,z)=B\times (r\times A)+A\times (r\times B)$.

Question

If $f(x,y,z)=(r \times A)\cdot (r\times B)$, where $r=(x,y,z)$ and $A$ and $B$ are constant vectors, show that $\nabla f(x,y,z)=B\times (r\times A)+A\times (r\times B)$.

I'm a bit lost on this problem. I tried using cross product equalities to solve it, but I'm not getting a clean answer. I'm thinking that there's a simple way to solve it. I would greatly appreciate any solutions, suggestions, or hints.

Answer 1

We can use $(u\times v)\cdot w=u\cdot(v\times w)$ to prove this.

First, let's distinguish between the two $r$ vectors and write the expression $$ g(u,v)=(u\times A)\cdot(v\times B) =u\cdot(A\times(v\times B)) =v\cdot(B\times(u\times A)). $$ This makes $f(r)=g(r,r)$ where $r=(x,y,z)$, and differentiating in $r$ is the same as differentiating $g(u,v)$ in $u$ and in $v$ separately and inserting the value $r$ for both: $$ \nabla_r f(r)=\nabla_r(g(r,r)) =\left(\nabla_ug(u,v)+\nabla_v g(u,v)\right)|_{u=v=r}. $$ However, using the first expressions, we get $$ \nabla_u g(u,v)=A\times(v\times B),\quad \nabla_v g(u,v)=B\times(u\times A), $$ and so $$ \nabla_r f(r)=A\times(r\times B)+B\times(r\times A). $$

Answer 2

Taking into account that the cross product can be written as $$ a\times b=e_i\varepsilon_{ijk}a_jb_k, $$ we have \begin{align} \nabla f&=\nabla[(r\times A)\cdot(r\times B)]\\ &=[\nabla(r\times A)](r\times B)+(r\times A)[\nabla(r\times B)]\\ &=[e_i\partial_i\varepsilon_{jkl}x_kA_l](r\times B)_j+(r\times A)_j[e_i\partial_i\varepsilon_{jkl}x_kB_l]\\ &=[e_i\varepsilon_{jkl}\delta_{ik}A_l](r\times B)_j+(r\times A)_j[e_i\varepsilon_{jkl}\delta_{ik}B_l]\\ &=[e_i\varepsilon_{jil}A_l](r\times B)_j+(r\times A)_j[e_i\varepsilon_{jil}B_l]\\ &=e_i\varepsilon_{ilj}A_l(r\times B)_j+e_i\varepsilon_{ilj}B_l(r\times A)_j\\ &=A\times(r\times B)+B\times(r\times A). \end{align}

Note I used the summation convention, i.e. when an index is repeated, a sum over 1,2,3 is intended. Furthermore, in the cross product I used the Levi-Civita symbol $\varepsilon_{ijk}$, see Vector cross product, where $e_1, e_2, e_3$ are the versors $i, j, k$ of the axes.

Answer 3

From $$f({\bf x}):=({\bf x}\times {\bf a})\cdot({\bf x}\times{\bf b})$$ it follows from the bilinearity of the scalar and the the vector product that $$\eqalign{df({\bf x}).{\bf X}&=({\bf X}\times {\bf a})\cdot({\bf x}\times{\bf b})+({\bf x}\times {\bf a})\cdot({\bf X}\times{\bf b})\cr &=\epsilon({\bf X},{\bf a},{\bf x}\times{\bf b})+\epsilon({\bf X},{\bf b},{\bf x}\times{\bf a})\cr &={\bf X}\cdot\bigl({\bf a}\times({\bf x}\times{\bf b})+{\bf b}\times({\bf x}\times{\bf a})\bigr)\ ,\cr}$$ where $\epsilon(\ ,\ ,\ )$ denotes the triple vector product. The identity so obtained can be written as $$\nabla f({\bf x})={\bf a}\times({\bf x}\times{\bf b})+{\bf b}\times({\bf x}\times{\bf a})\ .$$