Help proving that $\vec{a}\times(\vec{a}\times(\vec{a}\times\vec{b}))\cdot\vec{c} = -\|\vec{a}\|^2\vec{a}\cdot\vec{b}\times\vec{c}$

Question

This is problem 13 from Chapter 13, Section 14 of Apostol's Calculus Vol 1.

I need to prove or disprove the formula $\vec{a}\times(\vec{a}\times(\vec{a}\times\vec{b}))\cdot\vec{c} = -\|\vec{a}\|^2\vec{a}\cdot\vec{b}\times\vec{c}$.

This is what I have so far:

We have $\vec{a}\times(\vec{a}\times(\vec{a}\times\vec{b})) = k\vec{a}\times\vec{b}$ for some $k\in\mathbb{R}$.Hence $\vec{a}\times(\vec{a}\times(\vec{a}\times\vec{b}))\cdot\vec{c} = k\vec{a}\cdot\vec{b}\times\vec{c} $ and thus we need to determine if $k=-\|\vec{a}\|^2$. Using the identities $\vec{x}\cdot\vec{y} = \|\vec{x}\|\,\|\vec{y}\|\,\cos\theta$, where $\theta$ is the angle between $x$ and $y$, and $\|\vec{x}\times\vec{y}\|=\|\vec{x}\|\,\|\vec{y}\|\sin\theta$, we get $$ \|\vec{a}\|^3\,\|\vec{b}\|\,\|\vec{c}\|\,\cos\theta_1\sin\theta_3\sin\theta_4\sin\theta_5 = k\|\vec{a}\|\,\|\vec{b}\|\,\|\vec{c}\|\,\cos\theta_2\sin\theta_4 \ . $$ The angles correspond to: \begin{align*} &\theta_1 \ : \quad k\vec{a}\times\vec{b} \ \text{and} \ \vec{c} \\ &\theta_2 \ : \quad \vec{a} \ \text{and} \ \vec{b}\times\vec{c} \\ &\theta_3 \ : \quad \vec{a} \ \text{and} \ \vec{a}\times(\vec{a}\times\vec{b}) \\ &\theta_4 \ : \quad \vec{a} \ \text{and} \ \vec{b} \\ &\theta_5 \ : \quad \vec{a} \ \text{and} \ \vec{a}\times\vec{b}. \end{align*}

If $\vec{a},\vec{b},\vec{c}$ are dependent, then the equality we wish to prove becomes $0=0$ and is thus true. Assume that $\vec{a},\vec{b},\vec{c}$ are independent.Then we can simplify the previous equation to $$ \|\vec{a}\|^2\cos\theta_1\sin\theta_3\sin\theta_5 = k\cos\theta_2 \ . $$ Notice that $\theta_3=\theta_5=\frac{\pi}{2}$. Therefore $\|\vec{a}\|^2\cos\theta_1 = k\cos\theta_2.$

This is where I am stuck. Any hints or suggestions? Am I making this a lot harder than it needs to be?

Answer 1

If you know the triple product identity $$ u \times (v \times w) = (u \cdot w)v - (u \cdot v)w, $$ and the cyclic identity $a \times b \cdot c = b \times c \cdot a$, you can simply write $$ a \times (a \times (a \times b)) \cdot c = a \times ((a \cdot b)a - (a \cdot a)b) \cdot c = \cdots $$

Answer 2

There is a trick to proving complex vector identities such as these. It is called a proof by linearity. The guts of it is that each of the vectors in the identity can be written as a sum of standard basis vectors with coefficients. If you do this for the intended identity, it will split into a sum of simpler identities, which we can then prove separately.

Essentially, you need to prove the identity when $\vec{a}$ and $\vec{c}$ are each basis vectors. You need to prove it for any choice of basis vectors for each to be, but there is further symmetry that lowers the number of cases that must be checked. Then general case $\vec{a}$ and $\vec{a}$ can be reformed using linearity of the cross product.

Note the norm is not linear but the dot product is bilinear.

Sorry, I have just realized the two formulae are not linear in each variable. The one on the left is ternary in $\vec{a}$ for example. You can still use this method, but you have to actually prove some more complicated identity, which I imagine is $\vec{a_1} \times(\vec{a_2} \times (\vec{a_3} \times \vec{b})) \cdot \vec{c} = - (\vec{a_1} \cdot \vec{a_2})\vec{a_3} \cdot (\vec{b} \times \vec{c})$ This is linear in each variable so can be simplified and proved using linearity. If it is true.

The formalisation is that each expression acts as a tensor onto each component of the resulting vector. Since a tensor is uniquely determined by its action on a basis, you only need to prove both expressions are equal on a basis input.

Answer 3

This is much easier with clifford algebra and geometric products.

The identity would read,

$$(a \wedge [a \cdot \{a \wedge b\} ]) \wedge c= a^2 (a \wedge b \wedge c)$$

This is facilitated by considering the geometric product $aaabc$ and looking at the trivector component of this product. Grouping the product $(aa)(abc)$, we have

$$\langle aaabc \rangle_3 = (aa) \langle abc \rangle_3 = a^2 (a \wedge b \wedge c)$$

Alternatively, we can group the product $aaabc = (a[a\{ab\}])c$, yielding

$$\langle aaabc \rangle_3 = (a \cdot b) (a \wedge a \wedge c) + (a \wedge (a \cdot (a \wedge b))) \wedge c + (a \cdot [a \wedge a \wedge b]) \wedge c$$

The first and third terms are zero, as $a \wedge a = 0$ always for any vector $a$.

The unifying properties of the geoemtric product makes a very tedious problem extremely simple. It also shows that $c$ is arbitrary here and could be canceled, giving the related identity

$$a \wedge [a \cdot (a \wedge b)] = a^2 (a \wedge b)$$

which is obscured in the original problem due to the usual form of the triple scalar product (however, you could get it if you permuted $a, b, c$ on the right).

Answer 4

Using $$(u×v)·w=\det([u,v,w])=\det([v,w,u])=v·(w\times u),$$ one can transform the expression to a scalar product of two cross products, which after Cauchy-Binet is $$(a×b)·(c×d)=[a·c][b·d]-[a·d][b·c]$$. With the anti-symmetry of the cross product one then gets \begin{align} a×(a×(a×b))·c&=-(a×(a×b))·(a×c) \\&=-\|a\|^2[(a×b)·c]+[a·c][(a×b)·a] \\&=-\|a\|^2[a·(b×c)] \end{align}