Can someone help with this question?

Let $\times$ denote the cross product of vectors in $\mathbb{R}^3$.

Find $u$, $v$, $w$ such that $u \times (v \times w)$ is different from $(u \times v) \times w$?

I don't know how to start, Thanks!

  • $\begingroup$ See Jacob identity $\endgroup$ May 16, 2017 at 12:28
  • 5
    $\begingroup$ Hint: for any vector $v$ we have $v\times v = 0$. $\endgroup$
    – florence
    May 16, 2017 at 12:29
  • 1
    $\begingroup$ Just start by trying random vectors. I bet there's a (counter)example using various $\pm e_i$ for your vectors. $\endgroup$
    – pjs36
    May 16, 2017 at 12:30
  • $\begingroup$ I've edited the Question. Please review to check that I did not unintentionally change your meaning (you can rollback the edit if I did). $\endgroup$
    – hardmath
    May 16, 2017 at 14:25

2 Answers 2


Let $i,j,$ and $k$ be the standard basis vectors in $\mathbb{R}^3$. Then $$i\times (i\times j) = i\times k = -j$$ $$(i\times i)\times j = 0\times j = 0$$


Jacob Identity

$$a\times (b\times c)+b\times(c \times a) +c\times(a \times b)=0$$

$$a\times(b\times c)=-c\times (a \times b) $$ iff$$b\times(c\times a)=0$$

Now note $-c\times (a \times b)=(a \times b)\times c$.

then $$a\times(b\times c)=(a \times b) \times c$$

So we mustn't have $$b\times (c \times a) =0$$

Or as you wrote all vectors which do not satisfy $v\times (w \times u)=0$.

  • $\begingroup$ What part dont you get (or should I repeat the same thing and hope it works) $\endgroup$ May 16, 2017 at 12:56
  • $\begingroup$ You'll have to remain concrete in your objection to help me understand where a problem is. My answer rigorously defines the set which was asked for? $\endgroup$ May 16, 2017 at 13:05
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    $\begingroup$ $$\{ \{u,v,w\} \ : u,v,w \in \Bbb R^3 \ \cap \ v\times (w \times u) \neq 0 \}$$ $\endgroup$ May 16, 2017 at 13:08
  • $\begingroup$ @I guess I didnt try and find a single one which maybe was the question to just find one such example? Is that your issue you couldn't find words for? $\endgroup$ May 16, 2017 at 13:14
  • $\begingroup$ Well there is $c\times (a\times b)$ type. $\endgroup$ May 16, 2017 at 13:23

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