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

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?

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 $b$.

• I'm confused. How do you know a priori that the scalar multiple doesn't depend on the vectors $a$ and $b$? I.e., if you substitute $a=b$ and $a=c$, how do you know that the scalar multiplier doesn't change? Dec 9, 2016 at 2:36
• @Matthew: Note that the coefficient $x$ has to be bilinear in $a$ and $c$ and the coefficient $y$ has to be bilinear in $a$ and $b$. Dec 9, 2016 at 3:08
• Thank you very much for your help. I think I understand now. Dec 9, 2016 at 6:58
• @ASlowLearner, you're trying to solve fir $x$ and $y$. Work it out step by step with pencil and paper. Apr 16, 2018 at 2:32
• There are zillions of linear algebra books out there. I don't know what's best for your interests and background. For just the basic notions of linearity, properties of dot and cross product, you might check out some of the earlier lectures in my YouTube lectures (linked in my profile). Apr 18, 2021 at 2:23

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.

## Trigonometric proof of vector triple product expansion

The vector identity $$$$\label{e1}\tag{1} \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{a}{\cdot}\mathbf{c}~\mathbf{b} - \mathbf{a}{\cdot}\mathbf{b}~\mathbf{c}$$$$ is related to the trigonometric identity $$$$\label{e2}\tag{2} \sin(\beta-\gamma) = \cos\gamma\,\sin\beta - \cos\beta\,\sin\gamma \,,$$$$ 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 $$$$\label{e3}\tag{3} \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = abc\sin(\gamma-\beta)\,(-\mathbf{j}) = abc\sin(\beta-\gamma)\,\mathbf{j} \,.$$$$ But  $$\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 $$(\mathbf{a}\times\mathbf{b})\times\mathbf{c} = \mathbf{a}{\cdot}\mathbf{c}~\mathbf{b} - \mathbf{b}{\cdot}\mathbf{c}~\mathbf{a} \,.$$
• At the time of this comment, internal links work correctly in Firefox but not in Chrome. Jan 6 at 0:18