Proof of a Relation Between Cross Product and Dot Product

This comes from an extra-credit question in my math class. 

Prove: \( \vec a \times (\vec b \times \vec c) = (\vec a * \vec c)\vec b - (\vec a * \vec b)\vec c \)

I can prove it alternatively by manipulating the components algebraically with determinant calculations, but such an approach doesn't make any intuitive sense, and after all, it's boring.

Proof:

Let \( \vec a = |a|e_a , \vec b = |b|e_b , \vec c = |c|e_c, \) where e is a unit vector of length 1.

\(\vec a \times (\vec b \times \vec c) = |a||b||c|(e_a \times ( e_b \times e_c))\)

\((\vec a * \vec c)\vec b - (\vec a * \vec b)\vec c=|a||b||c|((e_a * e_c)e_b - (e_a * e_b)e_c)\)

Therefore, we need to show \(e_a \times ( e_b \times e_c)=(e_a * e_c)e_b - (e_a * e_b)e_c\).

Define \(e_{\perp b}=\frac{c-(c*e_b)e_b}{|c-(c*e_b)e_b|}\), \(e_{b\times c} \) as \( \frac{b \times c}{|b \times c|}\).

Then \(e_b\), \(e_{\perp b}\), and \(e_{b\times c}\) forms an orthonormal basis in \(R^3\).

The basis vectors are shown in red

Let \(S\) denote the orthogonal basis matrix \( [e_b\ \  e_{\perp b}\  \ e_{b\times c}] \), then

\(e_a = S a'\), where \(a'=(x,y,z)\)

\(e_b = S b'\), where \(b'=(1,0,0)\)

\(e_c = S c'\), where \(c'=(\cos{\theta_{cb}},\sin{\theta_{cb}},0)\)

         \(  a' \times (b' \times c') =a' \times ((1,0,0) \times ( \cos{ \theta_{cb} },\sin{ \theta_{cb} },0) \)

                                  $   = a' \times (0,0,\sin{\theta_{cb}})$
                                  $   =(x,y,z) \times (0,0,\sin{\theta_{cb}})$
                                  $   =(y\sin{\theta_{cb}},-x\sin{\theta_{cb}},0)$
                                  $   =(y\sin{\theta_{cb}},0,0)-(0,x\sin{\theta_{cb}},0)$
                                  $   =(x\cos{\theta_{cb}}+y\sin{\theta_{cb}},0,0)-(x\cos{\theta_{cb}},x\sin{\theta_{cb}},0)$
                                  $   =(a'*c',0,0)-(x\cos{\theta_{cb}},x\sin{\theta_{cb}},0)$
                                  $   =(a'*c')(1,0,0)-(x\cos{\theta_{cb}},x\sin{\theta_{cb}},0)$
                                  $   =(a'*c')b-(x\cos{\theta_{cb}},x\sin{\theta_{cb}},0)$
                                  $   =(a'*c')b-x(\cos{\theta_{cb}},\sin{\theta_{cb}},0)$
                                  $   =(a'*c')b-xc'$
                                  $   =(a'*c')b-(x,y,z)*(1,0,0)c'$
                                  $   =(a'*c')b-(a'*b')c'$

With this reference, we have the following: For an orthogonal matrix $A$, \[A\vec x \times A\vec y = \det (A)*A(x \times y),\ \det (A)=\pm 1 \]\[A\vec x * A\vec y=x*y\]

Since $S$ is orthogonal, it applies to $S$.

    $ e_a \times ( e_b \times e_c) = e_a \times ( S b' \times S c') $

                               $=e_a \times(\det (S) *S(b' \times c'))$

                               $=\det (S) * (e_a \times S(b' \times c'))$

                               $=\det (S) * (Sa' \times S(b' \times c'))$

                               $=(\det (S))^2 * S(a' \times (b' \times c'))$

                               $=1*S(a' \times (b' \times c'))$

                               $=S(a' \times (b' \times c'))$

   $(e_a * e_c)e_b - (e_a * e_b)e_c=(Sa' * Sc')e_b - (Sa' *Sb')e_c$

                                                $=(a' * c')e_b - (a' *b')e_c$

                                                $=(a' * c')Sb' - (a' *b')Sc'$

                                                $=S((a'*c')b-(a'*b')c')$

    Because $ a' \times (b' \times c')=(a'*c')b-(a'*b')c'$ 

    Therefore $ S(a' \times (b' \times c'))=S((a'*c')b-(a'*b')c')$

    Therefore $ e_a \times ( e_b \times e_c)=(e_a * e_c)e_b - (e_a * e_b)e_c$

Therefore, \[\vec a \times (\vec b \times \vec c) = (\vec a * \vec c)\vec b - (\vec a * \vec b)\vec c \]
$ \Box$