Please prove the following laws


a.(bxc)= b.(cxa)=c.(bxa)
ax(bxc)=b(a.c) -c(a.b)
(axb).(cxd)= (a.c)(b.d) -(a.d)(b.c)
ax(bxc) +bx(cxa) + cx(axb) = 0
(axb)x(cxd) = b(a.(cxd))-a.(b.(cxd))
= c(a.(bxd)) - d(a.(bxc)
(axb).((bxc)x(cxa))= (a.(bxc))^2
thank you!

Try expanding out each vector in components, A = Ax(x) + Ay(y) + Az(z), and reorganizing to get your desired result. For example,

A.(B X C) = A.[(ByCz - BzCy)(x) + (BzCx - BxCz)(y) + (BxCy - ByCx)(z)]
= AxByCz - AxBzCy + AyBzCx - AyBxCz + AzBxCy - AzByCx
= Bx(CyAz - CzAy) + By(CzAx - CxAz) + Bz(CxAy - CyAx)
= B.(C X A)

For the components of a cross product it's helpful to remember the following. Consider the cross product Q = R X S, which gives a vector Q. The x-component Q will involve cross terms of the non-x components of R and S, ie. RySz and RzSy. The positive term will be cyclic in (x,y,z) and the negative will be anticyclic. So, for example, in (RySz - RzSy)(x), the first term RySz(x) is cyclic (y,z,x), while the second term is anticyclic (z,y,x). You can verify this in the cross products above.

Hello,
Thanks for your reply! I tried expanding out each vector component and was told by my tutor that it was a waste of time and there were laws or rules that would make it much simpler.

Why didn't your tutor do it for you?

She showed me how to one however I recently tried the other ones and i am still confused and was hoping seeing how the other ones were solved would help me understand how they were done. Any help would be appreciated

Originally Posted by Trianagt

She showed me how to one however I recently tried the other ones and i am still confused and was hoping seeing how the other ones were solved would help me understand how they were done. Any help would be appreciated

This only a guess as to what your tutor may have meant.
$\displaystyle A\cdot(B\times C)=\left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{ {b_1}}&{{b_2}}&{{b_3}}\\{{c_1}} &{{c_2}}&{{c_3}}\end{array}} \right|$
Now by manipulating the rows of the determinate #1 is easy to see.
Recall that $\displaystyle A\times C=-(C\times A).$

Well I can't immediately think of how to prove the first two without expanding components. However, these first two identities can be used to prove all the rest.

Originally Posted by JohnDMalcolm

Well I can't immediately think of how to prove the first two without expanding components. However, these first two identities can be used to prove all the rest.

If you look at reply #6, you will see that #1 is quite easily done without expansion.
However, I agree that #2 must be done with expansion. It is a nightmare of an exercise in subscripts.