# Cross-product Vector Proof

1. Oct 6, 2005

### zanazzi78

Ive been given a question on a problem sheet that is supposed to be "just for Fun"

The question is;

show that:
$${\hat a}\times({\hat b}\times{\hat c})= {\hat b}({\hat a}.{\hat c})-{\hat c}({\hat a}.{\hat b})$$

Im not really sure where to start so any hints would be greatly appreciated.

2. Oct 6, 2005

### robphy

This is the so-called "bac-cab" rule.
The most efficient way is with the tensorial epsilon-delta identities. This may be too advanced.

However, you could do it using brute-force. Write each vector in terms of components. Then, use the properties of the cross- and dot-products.

I'm sure there has to be a neat geometrical proof somewhere.

3. Oct 6, 2005

### zanazzi78

I have a feeling there is a neat proof, my prof said that it could be done in a matter of lines (but that might be an exageration on his part!) he said you could do it the long way, refering to expanding the vectors using the;
$$\hat a \times \hat b = (a_{2}b_{3} - a_{3}b_{2}\hat i ) ... etc.$$
but i can`t get started!

4. Oct 6, 2005

### Staff: Mentor

$${\hat a}\,\dot\,{\hat b}\,=\,a_x\,\dot\,b_x\,+\,a_y\,\dot\,b_y\,+\,a_z\,\dot\,b_z$$

and

$${\hat a}\times{\hat b}\,=\,(a_y\,b_z - a_z\,b_y){\hat x}\,+\,(a_z\,b_x - a_x\,b_z){\hat y}\,+\,(a_x\,b_y - a_y\,b_x){\hat z}$$

Then rearrange terms - this is the long way. Instead of x, y, z, one could use i, j, k.

It would be shorter to use $\delta_{ij}$ and $\epsilon_{ijk}$ notation.

5. Oct 6, 2005

### big man

yeah I only know how to do it the long way by using the components of each vector...wouldn't mind finding out how to do it in a shorter manner though.

6. Oct 6, 2005

### robphy

Do the long way first... and do it correctly. You'll hopefully see the various patterns involved.

Then, learn the other methods, which encodes and summarizes those patterns.

7. Oct 7, 2005

### zanazzi78

well it loks like i have no choice but to go down the long road. Hopefully on the way i figure out how to simplify the proof.

Thanx for you help

8. Oct 7, 2005

### CarlB

Re:
$${\hat a}\times({\hat b}\times{\hat c})= {\hat b}({\hat a}.{\hat c})-{\hat c}({\hat a}.{\hat b})$$

Both the LHS and the RHS are vectors. That's a good start. It's easy to see that if you dot $\hat{a}$ with the RHS you will get zero. Is that also the case with the LHS? Yes, because the vector $\hat{a} \times \hat{v}$ is perpendicular to both $\hat{a}$ and $\hat{v}$. Thus you have two vectors that both are annihilated when dotted with $\hat{a}$.

A similar argument will show that both sides are perpendicular to
$\hat{b}\times\hat{c}$

If
$\hat{a}$
and
$\hat{b}\times\hat{c}$
are not parallel or antiparallel, that will get you that the two sides are parallel or antiparallel.

It remains to determine the scaling. Hmmmmmm.

Carl

And hey, what happened to LaTex?

Last edited: Oct 7, 2005
9. Oct 7, 2005

### Physics Monkey

These are always to fun to try to get as elegantly as you can.

Start with $$v = a \times (b \times c)$$. Clearly v is a vector and there are only three vectors in the problem so it must be representable as $$v = k_a a + k_b b + k_c c$$. Now we know from the properties of the cross product that v is perpendicular to a so we must have $$k_a = 0$$. Progress. Now let's use the linearity of v in its arguments. If I change a by a factor of $$\lambda$$ then v should change by a factor $$\lambda$$ which means $$k_{c,b}$$ have to be linear in a. A similar argument for b and c means implies that $$k_b$$ is independent of b and $$k_c$$ is independent of c. We are almost home free now. Now we know that since $$k_b$$ is a scalar that is linear in linear in a and c it must be proportional to $$a\cdot c$$ since there are no other scalars available that fit the bill. The coeffecient of proportionality must be independent of a and c. A similar argument holds for $$k_c$$. So where are we now? We have v in the following form $$v = \alpha (a \cdot c) b + \beta (a \cdot b) c$$. Since we know that v is antisymmetric under exchange of b and c, we must have $$\alpha = -\beta$$. So now we have $$v = \alpha ( (a\cdot c) b - (a\cdot b) c)$$ and we can evaluate this constant using a simple choice for a, b, and c. We easily find $$\alpha =1$$ using for instance a = j, b = i, c = j. And there you have it, no crazy Levi-Civita symbols (which are the easiest way) and no rote calculation, just some vector space algebra and geometry.

Last edited: Oct 7, 2005
10. Oct 7, 2005

### robphy

11. May 27, 2008

### a.a

Long WAy

I've been trying to do this proof the long way for almost 3 hrs now... and I've been working on it since two days ago.

my two sides look nothing alike at this point...

12. Mar 19, 2011

### Nes_physics

thanks for Physics Monkey's proof of the property of cross product:
ax(bxc) = (a.c)b - (a.b)c
like your arguments, I have another way to show
v= k((a.c)b-(a.b)c)

One have v, b, c are coplanar because they are all perpendicular to vector (bxc),
thus, v= xb+ yc where the set (x,y) is unique,
Since a and v are orthogonal, the dot product: a.v=0
therefore, x(a.b) + y(a.c) = 0 (1)
from (1), choose x = k a.c and y = -k a.c
so v= k((a.c)b-(a.b)c)

After that, apply your above method to prove k=1
^^

Last edited: Mar 19, 2011