Is there a simpler method for proving the cross product vector theorem?

[planauts]

Not sure if this is the correct section. I apologize if it's not.

Homework Statement

For any vectors $\vec{a}, \vec{b}, \vec{c}$ show that:
$(\vec{a} \times \vec{b} ) \times \vec{c}$
lies in the plane of $\vec{a}$ and $\vec{b}$

Homework Equations

The Attempt at a Solution

I assigned $\vec{a} = (e,f,g), \vec{b} = (x,y,z), \vec{c} = (s,t,u)$

then I used the cross product formula and got:

(u(gx-ez)-t(ey-fx), s(ey-fx)-u(fz-gy), t(fz-gy)-s(gx-ez))

which expanded comes to:

(gux-ezu-tey+tfx, sey-sfx-fuz+yug, zft-tgy-sgx+sez)

I'm not sure if that helps...
Thanks.

 

[vela]

Hint: If a vector lies in the plane of A and B, you can express it as a linear combination of the two vectors. So see if you can write your result in the form

(something)x(e, f, g) + (something)x(x, y, z)

 

[Dick]

Try using what you should know about the geometry of the cross product. axb is perpendicular to a and b, right? So it's a normal to the plane of a and b. Anything that is perpendicular to that normal is in the plane of a and b. Is (axb)xc perpendicular to axb?

 

[planauts]

Try using what you should know about the geometry of the cross product. axb is perpendicular to a and b, right? So it's a normal to the plane of a and b. Anything that is perpendicular to that normal is in the plane of a and b. Is (axb)xc perpendicular to axb?

Yes, that makes sense. I get it :D But is it possible to prove it algebraically?

Hint: If a vector lies in the plane of A and B, you can express it as a linear combination of the two vectors. So see if you can write your result in the form

(something)x(e, f, g) + (something)x(x, y, z)

I don't understand :S Could you explain it? Do you mean:
(something)*(e, f, g) + (something)*(x, y, z)

:S

 

[Dick]

I don't understand :S Could you explain it? Do you mean:
(something)*(e, f, g) + (something)*(x, y, z)

:S

Sure, that's what vela meant. And you can find those two somethings with a little work. Write (axb)xc=i*a+j*b and try to find i and j.

 

[planauts]

Sure, that's what vela meant. And you can find those two somethings with a little work. Write (axb)xc=i*a+j*b and try to find i and j.

Ah... I see...
So it would be like:

(ae+bx, af+by, ag+bz) where a and b are the somethings... right?

The goal is to get something like this:
(gux-ezu-tey+tfx, sey-sfx-fuz+yug, zft-tgy-sgx+sez)

right.

the constants s, t, u could be formed by multiplying the a and b "something" right?

Am I on the right track?

-----

EDIT:
Is that the long way of doing it? Is there an easier way? I feel that there is an easier way of solving it algebraically...

 

[Dick]

Ah... I see...
So it would be like:

(ae+bx, af+by, ag+bz) where a and b are the somethings... right?

The goal is to get something like this:
(gux-ezu-tey+tfx, sey-sfx-fuz+yug, zft-tgy-sgx+sez)

right.

the constants s, t, u could be formed by multiplying the a and b "something" right?

Am I on the right track?

-----

EDIT:
Is that the long way of doing it? Is there an easier way? I feel that there is an easier way of solving it algebraically...

You already used 'a' and 'b' for the vectors, so I like (axb)xc=i*a+j*b better. But, yes, that's the general idea. That's three linear equations in the two unknown constants i and j. You can solve it systematically. There's a lot of letters flying around but everything is a constant except for i and j. I like the geometrical approach better, but you can pull it off algebraically.