Do the BAC-CAB identity for triple vector product have some intepretation?

Question

As in the title, I was wondering if the formula: $$a\times (b\times c)=b(a\cdot c)-c(a \cdot b)$$ for $\mathbb R ^3$ cross product has some geometrical interpretation. I've recently seen a proof (from Vector Analysis - J.W. Gibbs) that's not at all difficult to understand, however I would hardly remember the steps of the proof (and I keep forgetting the correct order of A,B,C's), since it appears to me just as algebraic manipulation. So, why is this true? Or is it just an accident?

Answer 1

No, it's not an accident. The cross product is orthogonal to each factor, so the vector has to be orthogonal to $b\times c$, hence in the plane spanned by $b$ and $c$. But it also has to be orthogonal to $a$. So, writing $$a\times(b\times c) = xb + yc$$ and dotting with $a$, you get $x(b\cdot a) + y(c\cdot a)=0$. So the answer must be some scalar multiple of the correct formula. Now you only have to check that that scalar is $1$ by substituting $a=b$ and $a=c$. Better yet, let $a$ be a unit vector in the plane spanned by $b$ and $c$ that is orthogonal to $b$.

Answer 2

Well, we can prove the BAC-CAB identity using only geometric arguments as follows.

First consider a coordinate system where the $x$- and $y$-axes are skewed (non-orthogonal). We project the line $\mathrm{OA}$ orthogonally onto the axes, forming the quadrilateral $\square\mathrm{ABOC}$. We then construct the parallelogram $\square\mathrm{OFED}$ with sides $\mathrm{OF} = \mathrm{DE} = \mathrm{OC}$ and $\mathrm{OD} = \mathrm{FE} = \mathrm{BO}$ and the diagonal $\mathrm{OE} = \mathrm{BC}$. Then, by construction,

$$\begin{aligned} \overrightarrow{\mathrm{OE}} &= \mathrm{OF}\,\hat{x} + \mathrm{FE}\,\hat{y} \\ & = \mathrm{OC}\,\hat{x} + \mathrm{OB}\,\hat{y} \\ & = (\mathrm{OA} \cdot \cos{\angle \mathrm{AOC}})\,\hat{x} + (\mathrm{OA} \cdot \cos{\angle \mathrm{BOA}})\,\hat{y} \\ & = (\overrightarrow{\mathrm{OA}} \boldsymbol{\cdot} \hat{y})\,\hat{x} - (\overrightarrow{\mathrm{OA}} \boldsymbol{\cdot} \hat{x})\,\hat{y}, \end{aligned} \label{eq1} \tag{1}$$

where $\hat{x}$ and $\hat{y}$ are unit basis vectors. We note that $\square\mathrm{ABOC}$ is a cyclic quadrilateral since $\angle\mathrm{BAC} + \angle\mathrm{BOC} = \angle\mathrm{ABO} + \angle\mathrm{ACO} = 180^\circ$. It can therefore be inscribed in a circle and the intersecting chords theorem gives us $$\mathrm{AP} \cdot \mathrm{PO} = \mathrm{BP} \cdot \mathrm{PC}$$ such that $\Delta\mathrm{ABP} \sim \Delta\mathrm{COP}$ and $\Delta\mathrm{BOP} \sim \Delta\mathrm{ACP}$ (triangles are similar). From the law of sines we get that $$ \frac{\mathrm{OE}}{\sin \angle\mathrm{OFE}} = \frac{\mathrm{FE}}{\sin \angle\mathrm{EOF}} = \frac{\mathrm{BO}}{\sin \angle\mathrm{BCO}} = \frac{\mathrm{OA} \cdot \cos \angle\mathrm{BOA}}{\sin \angle\mathrm{BCO}} = \mathrm{OA}$$ as $\angle\mathrm{BOA} = \angle\mathrm{BCA} = 90^\circ - \angle\mathrm{BCO}$. Thus $$ \mathrm{OE} = \mathrm{OA} \cdot \sin \angle\mathrm{OFE} = \mathrm{OA} \cdot \sin \angle\mathrm{DOF}$$ which is equivalent to $$|\overrightarrow{\mathrm{OE}}| = |\overrightarrow{\mathrm{OA}}| \cdot \sin \angle\mathrm{DOF} = |\overrightarrow{\mathrm{OA}} \times (\hat{x} \times \hat{y})|. \label{eq2} \tag{2}$$ We can also see that $$ \angle\mathrm{AOE} = \angle\mathrm{AOC} + \angle\mathrm{COE} = \angle\mathrm{ABP} + \angle\mathrm{PBO} = 90^\circ. \label{eq3} \tag{3}$$

$(\ref{eq2})$ and $(\ref{eq3})$ combined thus proves that $$ \overrightarrow{\mathrm{OE}} = \overrightarrow{\mathrm{OA}} \times (\hat{x} \times \hat{y}) \label{eq4} \tag{4}$$ (correct magnitude and direction as given by the right-hand rule). We can now set $\vec{a} \equiv \overrightarrow{\mathrm{OA}}$ and combine eqs. $(\ref{eq1})$ and $(\ref{eq4})$;

$$ \vec{a} \times (\hat{x} \times \hat{y}) = (\vec{a} \boldsymbol{\cdot} \hat{y})\,\hat{x} - (\vec{a} \boldsymbol{\cdot} \hat{x})\,\hat{y}. $$

Finally, multiplying both sides with some scalars $b$ and $c$ gives us

$$ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \boldsymbol{\cdot} \vec{c})\,\vec{b} - (\vec{a} \boldsymbol{\cdot} \vec{b})\,\vec{c}. $$

Note that this also holds for any vector $\vec{a}$, even one which is not parallel to the $xy$-plane, as any component perpendicular to this plane is mapped to zero when taking the cross and dot products.

So to answer the question, I guess the rule can be said to hold due to a (rather complex) geometrical relation between the two quadrilaterals $\square\mathrm{ABOC}$ and $\square\mathrm{OFED}$ formed as above. The last multiplication step can be thought of as scaling the relevant sides of these equally.

Answer 3

Trigonometric proof of vector triple product expansion

The vector identity \begin{equation}\label{e1}\tag{1} \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{a}{\cdot}\mathbf{c}~\mathbf{b} - \mathbf{a}{\cdot}\mathbf{b}~\mathbf{c} \end{equation} is related to the trigonometric identity \begin{equation}\label{e2}\tag{2} \sin(\beta-\gamma) = \cos\gamma\,\sin\beta - \cos\beta\,\sin\gamma \,, \end{equation} which I have arranged so as to emphasize the connection.

Proof of (\ref{e1})

If  $\mathbf{b}$ and $\mathbf{c}$ are linearly dependent, then either $\mathbf{b}$ is a scalar multiple (possibly zero) of $\mathbf{c}$, or vice versa; and in either case it is trivial to verify that both sides of (\ref{e1}) are zero.

If  $\mathbf{b}$ and $\mathbf{c}$ are linearly independent, then the left side of (\ref{e1}) is unchanged if $\mathbf{a}$ is replaced by its projection on the plane normal to $\mathbf{b}{\times}\mathbf{c}$, i.e. on the plane of  $\mathbf{b}$ and $\mathbf{c}$; and the same replacement leaves the right side of (\ref{e1}) unchanged, because any component of $\mathbf{a}$ normal to the said plane is normal to both $\mathbf{b}$ and $\mathbf{c}$ and therefore makes no contribution to the dot-products.

So, to complete the proof of (\ref{e1}), we need only prove it for the special case in which $\mathbf{a}$ is in the plane of  $\mathbf{b}$ and $\mathbf{c}$. In this case let $\mathbf{i},\mathbf{j},\mathbf{k}$ be mutually perpendicular unit vectors with  $\mathbf{i}\times\mathbf{j}=\mathbf{k}\,,$ and let them be oriented so that the plane of  $\mathbf{i}$ and $\mathbf{j}$ is parallel to the plane of  $\mathbf{b}$ and $\mathbf{c}\,$ while $\mathbf{a}$ (if nonzero) is in the $\mathbf{i}$ direction. Let $\mathbf{a},\mathbf{b},\mathbf{c}\,$ have magnitudes $a,b,c$. Let  $\mathbf{b}$ and $\mathbf{c}$ make angles $\beta$ and $\gamma$ (respectively) with $\mathbf{i}$ (measured clockwise while looking in the $\mathbf{k}$ direction). Then, by definition, $$\mathbf{b}\times\mathbf{c} = bc\sin(\gamma-\beta)\,\mathbf{k} \,,$$ so that \begin{equation}\label{e3}\tag{3} \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = abc\sin(\gamma-\beta)\,(-\mathbf{j}) = abc\sin(\beta-\gamma)\,\mathbf{j} \,. \end{equation} But  $\mathbf{a}=a\mathbf{i}\,,$ and  $\mathbf{b}=b\mathbf{i}\cos\beta+b\mathbf{j}\sin\beta\,,$ and  $\mathbf{c}=c\mathbf{i}\cos\gamma+c\mathbf{j}\sin\gamma\,,$ so that  $\mathbf{a}{\cdot}\mathbf{b}=ab\cos\beta\,$ and  $\mathbf{a}{\cdot}\mathbf{c}=ac\cos\gamma\,,$ whence $$\mathbf{a}{\cdot}\mathbf{c}~\mathbf{b} - \mathbf{a}{\cdot}\mathbf{b}~\mathbf{c} = ac\cos\gamma\,(b\mathbf{i}\cos\beta+b\mathbf{j}\sin\beta) - ab\cos\beta\,(c\mathbf{i}\cos\gamma+c\mathbf{j}\sin\gamma) \,.$$ The terms in $\mathbf{i}$ cancel, leaving $$\mathbf{a}{\cdot}\mathbf{c}~\mathbf{b} - \mathbf{a}{\cdot}\mathbf{b}~\mathbf{c} = abc(\cos\gamma\,\sin\beta - \cos\beta\,\sin\gamma)\,\mathbf{j} \,,$$ which, by identity (\ref{e2}), matches the right side of (\ref{e3}), completing the proof.

Notes

• The above proof uses conveniently chosen basis vectors—so conveniently chosen that it does not need to invoke the distributive law for the cross-product (but does invoke the distributive law for the dot-product).
• I don't think "bac-cab"; I think "outer dot-product first"—which also works for $$(\mathbf{a}\times\mathbf{b})\times\mathbf{c} = \mathbf{a}{\cdot}\mathbf{c}~\mathbf{b} - \mathbf{b}{\cdot}\mathbf{c}~\mathbf{a} \,.$$