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A book I'm reading contains the following (paraphrased) \begin{equation} (a \times b) \times c = (a \cdot c)b - (b \cdot c)a \end{equation} This is supposed to follow from: \begin{equation} (a \times b) \cdot (c \times d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c) \end{equation} Where in both equations $a, b, c, d$ are all vectors. The question is how to prove this?

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  • $\begingroup$ The first equation doesn't seem right. Take a=i, b=j, c=k, for example. $\endgroup$
    – user84413
    Mar 31, 2015 at 21:45
  • $\begingroup$ You're right. corrected. $\endgroup$ Mar 31, 2015 at 21:49
  • $\begingroup$ How does the book justify the 2nd equation? (This seems like a good way to verify the 1st equation.) $\endgroup$
    – user84413
    Mar 31, 2015 at 22:10
  • $\begingroup$ The 2nd equation is introduced (before the 1st) without any proof unfortunately. $\endgroup$ Mar 31, 2015 at 22:45
  • $\begingroup$ Thanks for your reply; it's too bad they don't prove it. $\endgroup$
    – user84413
    Mar 31, 2015 at 23:01

2 Answers 2

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Chappers' proof is the way that was most likely intended (using the identity given). However, here is another approach (if only to show how much work is saved by using the identity given)

Since $(a\times b)\times a$ is perpendicular to $a$, $$ ((a\times b)\times a)\cdot a=0\tag{1} $$ Since $|a\times b|=|a|\,|b|\,|\sin(\theta)|$ and $a\cdot b=|a|\,|b|\cos(\theta)$, we have $$ |a\times b|^2+|a\cdot b|^2=|a|^2|b|^2\tag{2} $$ Since $(a\times b)\cdot c=(b\times c)\cdot a$, we can use $(2)$ to get $$ \begin{align} (\color{#C00000}{(a\times b)}\times\color{#00A000}{a})\cdot\color{#0000F0}{b} &=(\color{#00A000}{a}\times \color{#0000F0}{b})\cdot\color{#C00000}{(a\times b)}\\ &=(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)\tag{3} \end{align} $$ $(a\times b)\times a$ is perpendicular to $a\times b$ which is perpendicular to both $a$ and $b$. Therefore, $(a\times b)\times a$ is in the plane of $a$ and $b$. Thus, we can verify the following equation by taking dot products with $a$ and $b$ and using $(1)$ and $(3)$ $$ (a\times b)\times a=(a\cdot a)b-(a\cdot b)a\tag{4} $$ Swapping $a$ and $b$ and negating $(4)$ gives $$ (a\times b)\times b=(a\cdot b)b-(b\cdot b)a\tag{5} $$ Suppose that $c$ is in the plane of $a$ and $b$. Writing $c=xa+yb$, we can take the dot product of this equation with $a$ and $b$ to get two equations to solve for $x$ and $y$. This gives $$ c=\tfrac{(a\cdot c)(b\cdot b)-(b\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}a+\tfrac{(a\cdot a)(b\cdot c)-(a\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}b\tag{6} $$ Furthermore, since $c$ only appears on the right side of $(6)$ in a dot product with $a$ or $b$, any component of $c$ perpendicular to the plane of $a$ and $b$ will not change the right side. Thus, the right side of $(6)$ is a formula for the perpendicular projection of $c$ onto the plane of $a$ and $b$.

Using $(4)$, $(5)$, and $(6)$, we have $$ \begin{align} (a\times b)\times c &=\tfrac{(a\cdot c)(b\cdot b)-(b\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}((a\cdot a)b-(a\cdot b)a)\\ &+\tfrac{(a\cdot a)(b\cdot c)-(a\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}((a\cdot b)b-(b\cdot b)a)\\[4pt] &=(a\cdot c)b-(b\cdot c)a\tag{7} \end{align} $$

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  • $\begingroup$ Thanks for this answer. I know the identity that is given in the problem (the 2nd equation) can be verified by "brute force", but is there a simple way to show this? $\endgroup$
    – user84413
    Apr 1, 2015 at 16:27
  • $\begingroup$ @user84413: of course, it follows by taking the dot product of $(7)$ with $d$, but as for a simple proof without $(7)$ or breaking things down into $\vec{i},\vec{j},\vec{k}$ components (I assume that is what you mean by "brute force"), I have not encountered one. $\endgroup$
    – robjohn
    Apr 1, 2015 at 16:52
  • $\begingroup$ Thank you for your reply, and that is what I meant by "brute force". $\endgroup$
    – user84413
    Apr 1, 2015 at 18:46
  • $\begingroup$ In answer of robjohn in above: math.stackexchange.com/a/1215955/1265767, I don't really know why: $$ (a\times b)\times b=(a\cdot b)b-(b\cdot b)a\tag{5} $$ Suppose that $c$ is in the plane of $a$ and $b$. Writing $c=xa+yb$, we can take the dot product of this equation with $a$ and $b$ to get two equations to solve for $x$ and $y$. This gives $$ c=\tfrac{(a\cdot c)(b\cdot b)-(b\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}a+\tfrac{(a\cdot a)(b\cdot c)-(a\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}b\tag{6} $$ Can Anyone give me more detail? $\endgroup$
    – Tien
    Dec 11, 2023 at 0:16
  • $\begingroup$ Sorry, i want to comment answer of robjohn to ask answer of him but need 50 repulation, i don't know how to get link of this ans $\endgroup$
    – Tien
    Dec 11, 2023 at 1:24
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If you know the identity $$x \cdot (y \times z) = (x \times y) \cdot z$$ for arbitrary vectors $x,y,z$, then putting $x=a \times b$ and $y=c$, $z=d$, you have $$ (a \times b) \cdot (c \times d) = ((a \times b) \times c) \cdot d $$ so you can arrange your second equation so that everything is dotted with $d$: $$ ((a \times b) \times c) \cdot d = ((a \cdot c)b-(b \cdot c)a) \cdot d $$ and $d$ is arbitrary, so the vectors dotted on both sides must be the same.

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  • $\begingroup$ Your answer seems reasonable, although it doesn't seem to follow from that identity like the book stated, so I'm a bit hesistant to accept. $\endgroup$ Mar 31, 2015 at 22:47
  • $\begingroup$ @user3042966 I think this follows from $((a\times b)\times c)\cdot d=(a \times b)\cdot(c\times d)$ and then the identity given in the book. $\endgroup$
    – user84413
    Mar 31, 2015 at 23:04
  • $\begingroup$ @user3042966 I've rewritten it a bit. Does it satisfy you, or is there still something unclear? $\endgroup$
    – Chappers
    Mar 31, 2015 at 23:10
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    $\begingroup$ @user3042966: It does follow from the identity. We can add one step to make this clearer: $$ \begin{align} ((a\times b)\times c)\cdot d &=(c\times d)\cdot(a\times b)\\ &=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)\\ &=((a\cdot c)b-(b\cdot c)a)\cdot d \end{align} $$ $\endgroup$
    – robjohn
    Apr 1, 2015 at 14:57
  • $\begingroup$ Thank you. I have to say, neither of the two answers is how I would prove this: the way I was taught is to use summation convention, and use that $ (a \times b)_i = \epsilon_{ijk} a_j b_k $, where $\epsilon_{ijk}$ is the Levi-Civita symbol. (You can check this is equivalent to the geometric definition using @robjohn's argument). You then have the identity $\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$, which basically gives you everything. (This may be rather an Applied Maths way of working, but it's fine in flat 3D space.) $\endgroup$
    – Chappers
    Apr 1, 2015 at 19:16

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