Thread: Vector laws involving dot/cross product

1. Vector laws involving dot/cross product

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!

2. Re: Vector laws involving dot/cross product

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.

3. Re: Vector laws involving dot/cross product

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.

4. Re: Vector laws involving dot/cross product

Why didn't your tutor do it for you?

5. Re: Vector laws involving dot/cross product

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

6. Re: Vector laws involving dot/cross product

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 $\displaystyle A\times C=-(C\times A).$

7. Re: Vector laws involving dot/cross product

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.

8. Re: Vector laws involving dot/cross product

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.

,

,

,

,

,

,

,

,

,

,

,

,

,

,

show that (axb)x(axc)= (axb)ca vectors

Click on a term to search for related topics.