# What is the relationship between $(u\times v)\times w$ and $u\times(v\times w)$?

Given three vectors $u$, $v$, and $w$, $(u\times v)\times w\neq u\times(v\times w)$. This has been a stated fact in my recent class. But what is the ultimate relationship between them? I would presume that one is a scalar multiple of the other?

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–  vadim123 May 27 '13 at 22:43
Also this: en.wikipedia.org/wiki/… –  vadim123 May 27 '13 at 22:45

What you see here by the inequality of the left-hand side and the right-hand side is that the cross product (in this case, the vector triple product) is not associative $(\dagger)$:

$$(\vec u\times \vec v)\times \vec w\neq \vec u\times(\vec v\times \vec w)$$

Using boldface to represent the vectors:

$$\mathbf{u}\times (\mathbf{v}\times \mathbf{w}) = (\mathbf{u}\cdot\mathbf{w})\mathbf{v} - (\mathbf{u}\cdot\mathbf{v})\mathbf{w}\tag{1}$$

$$(\mathbf{u}\times \mathbf{v})\times \mathbf{w} = -\mathbf{w}\times(\mathbf{u}\times \mathbf{v}) = -(\mathbf{w}\cdot\mathbf{v})\mathbf{u} + (\mathbf{w}\cdot\mathbf{u})\mathbf{v}\tag{2}$$

See Wikipedia for more about triple products of vectors

$(\dagger)$ "In mathematics the Jacobi identity is a property that a binary operation [like the triple vector cross product] can satisfy which determines how the order of evaluation behaves for the given operation. Unlike for associative operations, order of evaluation is significant for operations satisfying Jacobi identity. It is named after the German mathematician Carl Gustav Jakob Jacobi." (Wikipedia).

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Forever the teacher! :-) +1 –  Amzoti May 28 '13 at 0:38

As others have mentioned, the ultimate relationship between $(u\times v)\times w$ and $u\times(v\times w)$ is given by the Jacobi identity: $u \times (v \times w) + w \times (u \times v) + v \times (w \times u) = 0$.

Hence, $(u\times v)\times w = u\times(v\times w)$ iff $v \times (w \times u) = 0$, and this happens iff $u$ and $w$ are parallel or $v$ is perpendicular to both $u$ and $w$.

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Here I give two pure geometrical proof of the Jacobi identity based on the right hand rule of the cross product. You could draw a picture yourself to see the geometrical relation of them more clearly.

Without loss of generality, consider the normalized version where $\|\u\| = \|\v\| = \|\w\| = 1$.

Geometrical proof 1: $\u\times \v$ is a vector perpendicular to the plane $S_{\u\v}$ spanned by $\u$ and $\v$, i.e., parallel to the unit normal vector $n$ of $S_{\u\v}$ pointing the same direction. We assume the angle between $\w$ and this plane is acute. Now cross product with $\w$: $$(\u\times \v)\times \w = \sin\alpha \,\n \times \w = \sin\alpha\,\n \times (\mathrm{Proj}_{S_{\u\v}} \w),\tag{1}$$ where $\alpha$ is the angle between $\u$ and $\v$. The projection $\w_{\perp}$ can be decomposed into two parts if $\u$ and $\v$ are not parallel: $$\w_{\perp}:= \mathrm{Proj}_{S_{\u\v}} \w = a \u_{\perp} + b\v,$$ where $$\u_{\perp} = \u - (\u\cdot \v)\v = \u - \cos\alpha \,\v\tag{2}$$ is perpendicular to $\v$. Dot product both sides with $\u_{\perp}$ and $\v$: $$\w_{\perp} \cdot \u_{\perp} = a \sin^2 \alpha \quad \text{and}\quad \w_{\perp} \cdot \v = b,$$ and $$\w_{\perp} \cdot \u_{\perp} = \|\w_{\perp}\| \|\u_{\perp}\|\cos(\alpha_u+\frac{\pi}{2}-\alpha) = \sin\theta \sin\alpha \cos(\frac{\pi}{2}-\alpha_v), \quad \text{and}\quad \w_{\perp} \cdot \v=\|\w_{\perp}\| \|\v\| \cos\alpha_v = \sin\theta \cos\alpha_v,$$ where $\theta$ is the angle between $\w$ and $\n$, $\alpha_u$ is the angle between $\u$ and $\w_{\perp}$, and $\alpha_v$ is the angle between $\v$ and $\w_{\perp}$, such that $$\alpha_u + \alpha_v = \alpha.\tag{3}$$ Now $$\w_{\perp}=\frac{\sin\theta\sin(\alpha_v)}{\sin\alpha}\u_{\perp} + \sin\theta \cos\alpha_v\,\v ,$$ Notice $\n$ cross product with a vector on the plane $S_{\u\v}$ is rotating the vector by $\pi/2$ counterclockwisely on the plane with respect to the normal $\n$: $$\n\times \u_{\perp} = \|\u_{\perp}\|\v = \sin\alpha\,\v$$ and $$\n\times \v= -\u_{\perp}/\|\u_{\perp}\| = -\u_{\perp}/\sin\alpha$$ Finally by (2) and (3), (1) becomes \begin{aligned} (\u\times \v)\times \w =&\sin\alpha\sin\theta\sin\alpha_v\,\v - \sin\theta \cos\alpha_v\, \u + \sin\theta \cos\alpha_v\cos\alpha \,\v \\ =&-\sin\theta \cos\alpha_v\, \u+ \sin\theta \cos(\alpha-\alpha_v)\,\v = -\sin\theta \cos\alpha_v\, \u + \sin\theta \cos\alpha_u \,\v.\tag{4} \end{aligned} For $\theta$ is the angle between $\w$ and the normal of $\u\v$-plane, drawing a perpendicular line from the tip of $\w$ to $\v$ we can see $$\sin\theta \cos\alpha_v = \sin\theta\, \w_{\perp}\cdot \v = (\w\cdot \v),$$ similarly $\sin\theta \cos\alpha_u = (\w\cdot \u)$, (4) is $$(\u\times \v)\times \w = (\w\cdot \u)\v - (\w\cdot \v)\u.$$

Geometrical proof 2: Let $$\s = (\u\times \v)\times \w,$$ It is easily to see from above argument, $\s$ lies in the plane of $\u\v$, while isperpendicular to $\w$. Hence we let again: $$\s = a\u+b\v,$$ Dot product with $\w$ we see that: $$\s\cdot \w = 0 = a(\u\cdot \w)+b(\v\cdot \w),$$ Hence $$a = c(\v\cdot \w) \quad \text{and}\quad b = -c(\u\cdot \w),$$ hence: $$\s = c(\v\cdot \w)\u-c(\u\cdot \w)\v \tag{5}$$ Now cross product with $\u$ on both sides of (5), left hand side is: \begin{aligned} \s\times \u &=\big( (\u\times \v)\times \w \big)\times \u = (\sin\alpha\,\n\times \w)\times \u \\ &=\sin\alpha( \n\times\w_{\perp})\times \u \end{aligned} From above we know that $\n\times\w_{\perp}$ is the counterclockwise rotation of $\w_{\perp}$ by $\pi/2$ on this plane with respect the normal $\n$, so the angle between $\n\times\w_{\perp}$ and $\u$ is $\pi/2 + \alpha_u$ (the definition see above (3)), and $$\s\times \u = -\sin\alpha \sin\theta \sin(\frac{\pi}{2} + \alpha_u)\n = -\sin\alpha \sin\theta \cos(\alpha_u)\,\n.$$ Right hand side of (5) is: $$-c(\u\cdot \w)\v \times \u = c(\u\cdot \w)\sin\alpha\,\n = c\sin\theta \cos(\alpha_u)\sin\alpha \,\n.$$ Therefore $c = -1$, and (5) is: $$(\u\times \v)\times \w = (\w\cdot \u)\v - (\w\cdot \v)\u.$$

Now for any three vectors in general we can use the formula for the normalized vectors and $$\left(\frac{\u}{\|\u\|}\times \frac{\v}{\|\v\|}\right)\times \frac{\w}{\|\w\|} = \left(\frac{\w}{\|\w\|} \cdot \frac{\u}{\|\u\|}\right)\frac{\v}{\|\v\|} - \left(\frac{\w}{\|\w\|} \cdot \frac{\v}{\|\v\|}\right)\frac{\u}{\|\u\|},$$ cancelling their norms would give the result.

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