# Thread: its a prove question ax(bxc)=....

1. ## its a prove question ax(bxc)=....

the question is a X (b X c)= (A.C)B - (A.B)C

THE DOT MEANS THOSE BIG DOTS BTW,
im not sure if you need what a b and c are but here they are,

a=9i-9j-4k
b=i-4j-2k
c=7i+j+9k

2. ## Re: its a prove question ax(bxc)=....

Originally Posted by newkidz
the question is a X (b X c)= (A.C)B - (A.B)C

If you are asked to prove, it is a messy, messy exercise in subscripts.

Let $\displaystyle A=<a_x,a_y,a_z>.~B=<b_x,b_y,b_z>.~C=<c_x,c_y,c_z>. ~$

Now do the cross-products , rearrange and factor to get $\displaystyle (A\cdot C)B-(A\cdot B)C$.

3. ## Re: its a prove question ax(bxc)=....

how do i cross product it 3 times,
i did b X c, and got i[ByCz-BzCy]-j[BxCz-BzCx]+k[BxCy-ByCx]
am i ment to cross product that against A=ax,ay,az.

also when i did it with my numbers i got -169i-125j-99k for Ax(Bxc)
but when you do (A.c) i get a single digit of 18 so not sure what to do

and lastly for the RHS there is no dot or multiply what does that mean, (a.c)b.

4. ## Re: its a prove question ax(bxc)=....

"A single digit of 18"? That's two digits! You mean a single NUMBER.

The cross product of two vectors is a vector and when you take the cross product of that with another vector, you get a vector again. That is, the left side of the equation is a vector and so the right side must also be a vector. Yes, the dot product of two vectors is a number and then (A.C)B means that you multiply that number by the vector B just as, for example, "2B". A number times a vector is a vector.

5. ## Re: its a prove question ax(bxc)=....

Originally Posted by newkidz
how do i cross product it 3 times,
i did b X c, and got i[ByCz-BzCy]-j[BxCz-BzCx]+k[BxCy-ByCx]
am i ment to cross product that against A=ax,ay,az.

I told you that this is a nightmare of subscripts. Using my suggested notation.

\displaystyle \begin{align*}(A\cdot C)B &=(a_xc_x+ a_yc_y+ a_zc_z)(b_xi+b_yj+b_zk)\\ &=(a_xb_xc_x+ a_yb_xc_y+ a_zb_xc_z)i \\&~+(a_xb_yc_x+ a_yb_yc_y+ a_zb_yc_z)j\\&~+(a_xb_zc_x+ a_yb_zc_y+ a_zb_zc_z)k \end{align*}.

-Dan

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# show a x (b x c) = (a.c)b - (a.b)c

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