proof that a x (b x c) = b(a . c) - c(a . b)

KBriggs · Hodge Conjecture Offline Joined: 10 Jun 2009 Posts: 71

for vectors a,b,c in 3-space, Can someone show me how to prove that $a\ x (b\ x\ c) = b(a \cdotp c) - c(a \cdotp b)$ ?

I could do it the long and tedious way of expanding it with the determinant notation, but surely there is a simpler way?

This is an exerise posed by three of my textbooks, none of them show how to do it and it' driving me nuts since I can't figure it out.

**Posted:** Sun Sep 20, 2009 12:56 pm

Mathias_DK

**Posted:** Sun Sep 20, 2009 1:10 pm

KBriggs · Hodge Conjecture Offline Joined: 10 Jun 2009 Posts: 71

Sorry, $axb$ was supposed to be the cross product of a and b - I am not sure how to get the notation right in Latex. These will be vectors in three-space, since the cross product is only valid there, so if a = <a1, a2, a3> and b = <b1, b2, b3> then $a \cdotp b$ = a1*b1 + a2*b2 + a3*b3. The cross product of a and b can be expressed as the determinant of a 3x3 matrix whose top row is made up of the unit vectors i,j,k and the second row is the compnents of a and the third row the components of b.

**Posted:** Sun Sep 20, 2009 1:17 pm

t0rajir0u

$a \times (b \times c)$ is a vector perpendicular to both $a$ and $b \times c$ . Once you check that the RHS also has this property, the identity is proven up to a multiplicative factor, and all you have to do to figure out what that factor is is to plug in any particular choice of $a, b, c$ that makes both sides nonzero.

**Posted:** Sun Sep 20, 2009 1:30 pm

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suadnovic · New Member Offline Joined: 29 Oct 2009 Posts: 13

Hint:
Tray to prove this equality for koordinate vectors e1,e2,e3 (take special cases c=e1,e2,e3) and than use linearity for general case. But, this method is relatively long and "violent" (in proof You must add and subtract some naisly shosen members). Otherwise, You can find elementary proof of this formula in college books.

**Posted:** Mon Nov 02, 2009 8:41 am

J.Y.Choi

Cross and dot product of vector $\bold a, \bold b$ can be denoted by :
$(\bold a\times\bold b)_i=\varepsilon_{ijk}a_jb_k$ , $\bold a\cdot\bold b=\delta_{ij}a_ib_j$ . (Einstein notation)
Where $\varepsilon_{ijk}$ and $\delta_{ij}$ denotes Levi-Civita and Kronecker delta symbol respectively.
$(\bold a\times(\bold b\times\bold c))_i$

$=\varepsilon_{ijk}a_j(\bold b\times\bold c)_k=\varepsilon_{ijk}a_j(\varepsilon_{lmk}b_lc_m)$
$=\varepsilon_{ijk}\varepsilon_{lmk}a_jb_lc_m$ .
By using the identity $\varepsilon_{ijk}\varepsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$ , the form can be rewritten by :
$(\varepsilon_{ijk}\varepsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_jb_lc_m$
$=(b_l\delta_{il})(\delta_{jm}a_jc_m)-(c_m\delta_{im})(\delta_{jl}a_jb_l)$
$=b_i(\delta_{jm}a_jc_m)-c_i(\delta_{jl}a_jb_l)$ .
This means each componat of $\bold a\times(\bold b\times\bold c)$ and $\bold b(\bold a\cdot \bold c)-\bold c(\bold a\cdot\bold b)$ is equal.

**Posted:** Sat Nov 14, 2009 12:14 pm

suadnovic · New Member Offline Joined: 29 Oct 2009 Posts: 13

@J.Y.Choi
Very nice. Congratulations!

**Posted:** Mon Nov 23, 2009 2:55 pm