# Cross Product?

1. Aug 24, 2010

We all know and love the determinant form of the cross product:

$$\bold{a}\times\bold{b} = \left|\begin{array}{ccc} \bold{i} & \bold{j} & \bold{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z\end{array}\right|$$

and can calculate this by going through the motions but when I try to think of why this is the case I get stuck.

$$\bold{a} \times \bold{b} \ = \ (a_x \hat{i} \ + \ a_y \hat{j} \ + \ a_z \hat{k}) \times (b_x \hat{i} \ + \ b_y \hat{j} \ + \ b_z \hat{k})$$

$$a_x b_x (\hat{i} \times \hat{i}) \ + \ a_x b_y (\hat{i} \times \hat{j}) \ + \ a_x b_z (\hat{i} \times \hat{k}) \ + \ a_y b_x (\hat{j} \times \hat{i}) \ + \ a_y b_y (\hat{j} \times \hat{j}) \ + \ a_y b_z (\hat{j} \times \hat{k}) \ + \ a_z b_x (\hat{k} \times \hat{i}) \ + \ a_z b_y (\hat{k} \times \hat{j}) \ + \ a_z b_z (\hat{k} \times \hat{k})$$

I have it memorized how to calculate all this fine, but I don't know why $$\hat{i} \times \hat{i} \ = \ 0$$ and $$\hat{i} \times \hat{j} \ = \ \hat{k}$$

Thinking about the inner product I can understand why $$\hat{i} \cdot \hat{j} \ = \ 0$$ because they represent different dimensions but the cross product makes no sense.

Serge Lang just defines $$\bold{a} \times \bold{b}$$ as aybz - azby etc... but that is hardly satisfying.
Furthermore, Marsden & Tromba axiomatically define it as following from the determinant
but again that is not satisfying at all, I don't understand why the determinant answers this question.

I can see reason in using the determinant as a shortcut way of calculating the cross
product as just following from calculating the components the way I have it laid out
above so if I can get a reason why $$\hat{i} \times \hat{i} \ = \ 0$$ and $$\hat{i} \times \hat{j} \ = \ \hat{k}$$ then it all makes sense!
Invoking the sine function isn't justification enough for me because as I know it
the sine function arises from squaring the magnitude of the cross product ||a x b||² and that involves squaring the components.
The minus signs in the component form arise from calculating $$\hat{i} \times \hat{k} \ = \ - \hat{j}$$ etc...
so to get to that point you need to know why this happens...

thanks

Last edited: Aug 24, 2010
2. Aug 24, 2010

### phyzguy

It was Hermann Grassmann that first realized that the fully anti-symmetric product (also called exterior product or wedge product) of two vectors has special properties (see http://en.wikipedia.org/wiki/Exterior_algebra). Note that in general, the exterior product of two vectors is not another vector, but is a different geometric object referred to as a bi-vector (see http://en.wikipedia.org/wiki/Bivector). It is only in three dimensions that the space of bi-vectors is dual to the space of vectors, so that we can identify the three basis bi-vectors with the three basis vectors and define a vector cross-product. In 4D space-time, for example, this procedure does not work, since there are four basis vectors and six basis bi-vectors. The book "Geometric Algebra for Physicists" by Doran and Lasenby is an excellent introduction to all of this.

Last edited by a moderator: Apr 25, 2017
3. Aug 24, 2010

So what you're saying is that I have to take it on faith that $$\hat{i} \times \hat{j} \ = \ \hat{k}$$ until I am advanced enough to study exterior algebra????

4. Aug 24, 2010

### phyzguy

It's basically a definition, so it has to be accepted as is. We have learned that if you define the cross product in this way, then it has certain useful properties. You're not going to be able to 'derive' or 'prove' that it is true, because it is a definition.

5. Aug 24, 2010

### gomunkul51

it is a definition, learn it and use it wisely ;)

P.S: of course if it makes sense than it came form somewhere, so ask your teacher for a direction for a satisfactory answer.
and read a theoretical algebra math book.

Last edited: Aug 24, 2010
6. Aug 24, 2010

That's horrible!!! Seriously there has to be a better explanation than that!!!!!!!!

I've put up with hearing that $$\sum_{i=1}^{n} i^2 \ = \ \frac{n(2n + 1)(n + 1)}{6}$$ is definition too before :tongue: I'll find the answer!

7. Aug 24, 2010

### D H

Staff Emeritus
Love? Some of us hate that determinant form because it is an abuse of notation.

If you read the history of the vector notation used in physics it is pretty clear that it originated from the quaternions, not Grassmann's work on linear algebra. Physicists began using vector notation in the late 1800s. They didn't come across Grassmann's work for another 30 years.

Take, for example, the use of i, j, and k to represent the unit vectors in three space. That comes straight out of Hamilton's use of i, j, and k as the three distinct roots of -1 in the quaternions. Gibbs and Heaviside didn't particularly like the complexity of the quaternions and developed vector analysis in its stead.

The principal equation of Hamilton's quaternions is

$$i^2=j^2=k^2 = ijk = -1$$

From this the following relations are easily derived:

\aligned ij &= k \\ ji &= -k \\ jk &= i \\ kj &= -i \\ ki &= j \\ ik &= -j \endaligned

Making those quaternionic roots of -1 into unit vectors leads to the dot and cross products:

\aligned \hat{\mathbf i} \cdot \hat{\mathbf i} &= \hat{\mathbf j} \cdot \hat{\mathbf j} = \hat{\mathbf k} \cdot \hat{\mathbf k} = 1 \\ \hat{\mathbf i} \cdot \hat{\mathbf j} &= \hat{\mathbf i} \cdot \hat{\mathbf k} = \hat{\mathbf j} \cdot \hat{\mathbf k} = 0 \\ \hat{\mathbf i} \times \hat{\mathbf i} &= \hat{\mathbf j} \times \hat{\mathbf j} = \hat{\mathbf k} \times \hat{\mathbf k} = \boldsymbol 0 \\ \hat{\mathbf i} \times \hat{\mathbf j} &= \hat{\mathbf k} \\ \hat{\mathbf j} \times \hat{\mathbf k} &= \hat{\mathbf i} \\ \hat{\mathbf k} \times \hat{\mathbf i} &= \hat{\mathbf j} \endaligned

8. Aug 24, 2010

### D H

Staff Emeritus
A lot of things are defined the way they are because those definitions are useful.

Some people like the determinant form of the cross product because it's easy for them to remember. If you don't like that expression, there are others. For example
• The magnitude of the cross product of $\mathbf a$ and $\mathbf b$ is $ab\sin\theta$. The resultant vector is normal to both $\mathbf a$ and $\mathbf b$ and its direction is given by the right hand rule.
• The cross product of a vector with itself is always zero. The cross product of one of the unit vectors $\hat i$, $\hat j$, or $hat k$ with another one of those unit vectors is $\pm1$ times the omitted unit vector. Using +1 for a right-handed sequence, -1 for a left-handed sequence.
• Alternatively, use +1 if the sequence is an even permutation of (i,j,k), -1 if the sequence is an odd permutation.

That's easily derived. Use recursion, for example.

9. Aug 24, 2010

I went back and checked Gibb's book (the one his student wrote!)
and he just defines the cross product in the exact same manner any other
book would do. I got Heaviside's book and peeked at it & it's about 10
pages of explanation and he never writes a x b he defines it a bit
differently so I'm going to read that now in the hopes he'll give some
deeper explanation but I'd just like to point out that viewing the cross
product as ||a|| ||b|| sinθ is a cheat insofar as I know this.

It's a cheat because it uses $\hat{k} \times \hat{j} \ = \ - \hat{i}$ which makes no sense to me.
My only problem here is understanding why $\hat{i} \times \hat{j} \ = \ \hat{k}$ and all of the variations etc...
Why does "crossing" two unit vectors result in the third unit vector in a
different dimension? Surely teachers get stopped in a linear algebra class
by students calling them out???
As I said, it makes sense for the inner product when you have $\hat{i} \cdot \hat{j} \ = \ 0$ because you're multiplying numbers
from different dimensions and that makes no sense but the cross
product has no physical explanation like that, insofar as I can see

I know how it computes the area of a parallelogram & can derive the
sinθ expression amidst all the confusion of the components fine.
I'm even more convinced when I look at a picture of 2 vectors and
realise that the area contained in them is just $b \cdot h$
where b = ||a|| & h = ||b||sinθ, that's all fine.
My problem is that when computing the components you encounter having
to deal with $\hat{k} \times \hat{j} \ = \ - \hat{i}$ and all
of the fun variants, accepting it simply because it works isn't enough...

$\overline{a} \times \overline{b} \ = \ (a_x \hat{i} \ + \ a_y \hat{j} \ + \ a_z \hat{k}) \times (b_x \hat{i} \ + \ b_y \hat{j} \ + \ b_z \hat{k})$

$a_x b_x (\hat{i} \times \hat{i}) \ + \ a_x b_y (\hat{i} \times \hat{j}) \ + \ a_x b_z (\hat{i} \times \hat{k}) \ + \ a_y b_x (\hat{j} \times \hat{i}) \ + \ a_y b_y (\hat{j} \times \hat{j}) \ + \ a_y b_z (\hat{j} \times \hat{k}) \ + \ a_z b_x (\hat{k} \times \hat{i}) \ + \ a_z b_y (\hat{k} \times \hat{j}) \ + \ a_z b_z (\hat{k} \times \hat{k})$

$a_x b_x (0) \ + \ a_x b_y (\hat{k}) \ + \ a_x b_z (- \hat{j}) \ + \ a_y b_x (- \hat{k}) \ + \ a_y b_y (0) \ + \ a_y b_z (\hat{i}) \ + \ a_z b_x (\hat{j}) \ + \ a_z b_y (- \hat{i}) \ + \ a_z b_z (0)$

You see all of that, yes it's all understandable as rote memorization

$(a_y b_z - \ a_z b_y) \hat{i} + \ (a_z b_x \ - \ a_x b_z) \hat{j} \ + \ (a_x b_y \ - \ a_y b_x) \hat{k} \ = \ (a_y b_z - \ a_z b_y) \hat{i} \ - \ (a_x b_z \ - \ a_z b_x) \hat{j} \ + \ (a_x b_y \ - \ a_y b_x )\hat{k}$

Using all of this we can then derive the relationship that the sin of the angle between the
vectors has with respect to the overall formula by squaring and sifting through it all but
we had to know that $\hat{k} \times \hat{j} \ = \ - \hat{i}$ in order to get
here in the first place!

Do people really have to take it on faith that these things work until they
can understand the deep reasons why it works when studying exterior
& Lie algebra? The wiki link, and all of the explanations people have given
in online posts, don't answer the question at all.

Also, "love" was just a joke

10. Aug 24, 2010

### D H

Staff Emeritus
Unless you delve into either the quaternionic or Lie algebra derivations, you will have to take the fact that the cross product is antisymmetric ($\mathbf b \times \mathbf a = -\mathbf a \times \mathbf b$) as definitional. Even if you knew those deeper meanings it is still good to remember that the cross product is antisymmetric right off the bat. For example, that the cross product is antisymmetric immediately means that the cross product of a vector with itself is zero. You don't have to commit this latter fact to memory. It is immediately derivable from the antisymmetric nature of the cross product.

There are lots of things you have to take as fact that later on you will find can be derived. A good example: Newton's third law. It follows from conservation of linear momentum and angular momentum, and those in turn derive from symmetries of space thanks to Noether's theorems. Until you know that stuff you just have to take Newton's third law as fact.

11. Aug 24, 2010

I knew it!!!! I KNEW IT!!!!!

I knew there was a subtle logic to all of this

Given two vectors
$\overline{x} \ = \ x_1 \hat{i} \ + \ x_2 \hat{j} \ + \ x_3 \hat{k}$
and
$\overline{y} \ = \ y_1 \hat{i} \ + \ y_2 \ + \ y_3 \hat{k}$
we want to find a vector that is orthogonal to both x & y.
In order to do this we'll need to find a vector
$\overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$
such that;

w • x = 0
w • y = 0

Beautiful! Not only does this get rid of the i's and j's but it makes beautiful
sense falling out of the beautiful dot product

$^{ \overline{w} \cdot \overline{x} \ = \ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ \overline{w} \cdot \overline{y} \ = \ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0}$

I really should have though this through & recognised this! I feel terrible
for missing it This is a serious lesson! However, the next part wouldn't
have registered for me, not yet anyway!

$^{ x_1y_3w_1 \ + \ x_2y_3w_2 \ + \ x_3y_3w_3 \ = \ 0}_{ y_1x_3w_1 \ + \ y_2x_3w_2 \ + \ y_3x_3w_3 \ = \ 0}$

$(x_1y_3 \ - \ x_3y_1)w_1 \ + \ (x_2y_3 \ - \ x_3y_2)w_2 \ = \ 0$

Obviously
$w_1 \ = \ (x_2y_3 \ - \ x_3y_2)$
$w_2 \ = \ -(x_1y_3 \ - \ x_3y_1) \ = \ (x_3y_1 \ - \ x_1y_3)$

will give us zero here!

If we now take

$x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0$

from

$^{ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0}$

and solve for $x_3w_3$ we get;

$x_3w_3 \ = \ - x_1w_1 - x_2w_2 \ = \ - x_1(x_2y_3 \ - \ x_3y_2) - x_2(x_3y_1 \ - \ x_1y_3) \ = \ - x_1x_2y_3 + x_1x_3y_2 \ - \ x_2x_3y_1 + \ x_1x_2y_3$

$x_3w_3 \ = \ x_1x_3y_2 \ - \ x_2x_3y_1$

$w_3 \ = \ x_1y_2 \ - \ x_2y_1$

This is the solution to just $w_3$ obviously the same chain of
logic is used to deduce $w_1$ & $w_2$.
This way we've found a vector that is perpendicular to both of these
vectors.

When we've computed all of

$\overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$

we can take it's magnitude & seeing as we've got $w_1$, $w_2$ & $w_3$ in the $x_1y_2 \ - \ x_2y_1$
form we're going to end up getting the sine function out of it
when we do the dirty work. Note that we can just randomly define
a x b as a symbol that will encode this as falling out of the dirty work above!

All is right with the world again

the source of my happiness!!!

12. Aug 25, 2010

I'm having trouble with a minus sign

Given two vectors
$\overline{x} \ = \ x_1 \hat{i} \ + \ x_2 \hat{j} \ + \ x_3 \hat{k}$
and
$\overline{y} \ = \ y_1 \hat{i} \ + \ y_2 \ + \ y_3 \hat{k}$
we want to find a vector that is orthogonal to both x & y.
In order to do this we'll need to find a vector
$\overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$
such that;

w • x = 0
w • y = 0

$^{ \overline{w} \cdot \overline{x} \ = \ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ \overline{w} \cdot \overline{y} \ = \ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0}$

This is the central equation to return to. We'll find $w_1$ first.

$^{ x_1y_1w_1 \ + \ x_2y_1w_2 \ + \ x_3y_1w_3 \ = \ 0}_{ y_1x_1w_1 \ + \ y_2x_1w_2 \ + \ y_3x_1w_3 \ = \ 0}$

$(x_2y_1 \ - \y_2x_1)w_2 \ + \ (x_3y_1 \ - \ x_1y_3)w_3 \ = \ 0$

$w_2 \ = \ (x_3y_1 \ - \ x_1y_3)$
$w_3 \ = \ - (x_2y_1 \ - \y_2x_1) \ = \ (y_2x_1 \ - \ x_2y_1)$

Solve for $w_1$ now.

$x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 \Rightarrow \ x_1w_1 \ = \ - \ x_2w_2 \ - \ x_3w_3$

$x_1w_1 \ = \ - \ x_2(x_3y_1 \ - \ x_1y_3) \ - \ x_3(y_2x_1 \ - \ x_2y_1)$

$x_1w_1 \ = x_1x_2y_3 \ - \ x_2x_3y_1 \ + \ x_2x_3y_1 \ - \ x_1x_3y_2$

$x_1w_1 \ = x_1x_2y_3 \ - \ x_1x_3y_2$

$w_1 \ = (x_2y_3 \ - \ x_3y_2)$

So now we have $w_1$ from
$\overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$

We'll now work on $w_2$

$^{ x_1y_2w_1 \ + \ x_2y_2w_2 \ + \ x_3y_2w_3 \ = \ 0}_{ y_1x_2w_1 \ + \ y_2x_2w_2 \ + \ y_3x_2w_3 \ = \ 0}$

$(x_1y_2 \ - \ y_1x_2)w_1 \ + \ (x_3y_2 \ - \ y_3x_2)w_3 \ = \ 0$

$w_1 \ = \ (x_3y_2 \ - \ y_3x_2)$

$w_3 \ = \ - \ (x_1y_2 \ - \ y_1x_2) \ = \ (y_1x_2 \ - \ x_1y_2)$

Now solve for $w_2$

$x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 \Rightarrow \ x_2w_2 \ = \ - \ x_1w_1 \ - \ x_3w_3$

$x_2w_2 \ = \ - \ x_1(x_3y_2 \ - \ y_3x_2) \ - \ x_3(y_1x_2 \ - \ x_1y_2)$

$x_2w_2 \ = \ x_1x_2y_3 \ - \ x_1x_3y_2 \ + \ x_1x_3y_2 \ - \ x_2x_3y_1$

$x_2w_2 \ = \ x_1x_2y_3 \ - \ x_2x_3y_1$

$w_2 \ = \ x_1y_3 \ - \ x_3y_1$

We now have

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ w_3 \hat{k}$

This is obviously incorrect - for some reason I can't place.
There is a minus sign that's off & I can't see how that happens...

We'll now get $w_3$

$^{ x_1y_3w_1 \ + \ x_2y_3w_2 \ + \ x_3y_3w_3 \ = \ 0}_{ y_1x_3w_1 \ + \ y_2x_3w_2 \ + \ y_3x_3w_3 \ = \ 0}$

$(x_1y_3 \ - \ x_3y_1)w_1 \ + \ (x_2y_3 \ - \ x_3y_2)w_2 \ = \ 0$

Obviously
$w_1 \ = \ (x_2y_3 \ - \ x_3y_2)$
$w_2 \ = \ -(x_1y_3 \ - \ x_3y_1) \ = \ (x_3y_1 \ - \ x_1y_3)$

will give us zero here!

If we now take

$x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0$

from

$^{ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0}$

and solve for $x_3w_3$ we get;

$x_3w_3 \ = \ - x_1w_1 - x_2w_2 \ = \ - x_1(x_2y_3 \ - \ x_3y_2) - x_2(x_3y_1 \ - \ x_1y_3) \ = \ - x_1x_2y_3 + x_1x_3y_2 \ - \ x_2x_3y_1 + \ x_1x_2y_3$

$x_3w_3 \ = \ x_1x_3y_2 \ - \ x_2x_3y_1$

$w_3 \ = \ x_1y_2 \ - \ x_2y_1$

We now have

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ w_3 \hat{k}$

$\overline{a} \times \overline{b} \ = \ \overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ (x_1y_2 \ - \ x_2y_1) \hat{k}$

13. Aug 25, 2010

I'm having trouble with a minus sign

Given two vectors
$\overline{x} \ = \ x_1 \hat{i} \ + \ x_2 \hat{j} \ + \ x_3 \hat{k}$
and
$\overline{y} \ = \ y_1 \hat{i} \ + \ y_2 \ + \ y_3 \hat{k}$
we want to find a vector that is orthogonal to both x & y.
In order to do this we'll need to find a vector
$\overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$
such that;

w • x = 0
w • y = 0

$^{ \overline{w} \cdot \overline{x} \ = \ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ \overline{w} \cdot \overline{y} \ = \ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0}$

This is the central equation to return to. We'll find $w_1$ first.

$^{ x_1y_1w_1 \ + \ x_2y_1w_2 \ + \ x_3y_1w_3 \ = \ 0}_{ y_1x_1w_1 \ + \ y_2x_1w_2 \ + \ y_3x_1w_3 \ = \ 0}$

$(x_2y_1 \ - \y_2x_1)w_2 \ + \ (x_3y_1 \ - \ x_1y_3)w_3 \ = \ 0$

$w_2 \ = \ (x_3y_1 \ - \ x_1y_3)$
$w_3 \ = \ - (x_2y_1 \ - \y_2x_1) \ = \ (y_2x_1 \ - \ x_2y_1)$

Solve for $w_1$ now.

$x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 \Rightarrow \ x_1w_1 \ = \ - \ x_2w_2 \ - \ x_3w_3$

$x_1w_1 \ = \ - \ x_2(x_3y_1 \ - \ x_1y_3) \ - \ x_3(y_2x_1 \ - \ x_2y_1)$

$x_1w_1 \ = x_1x_2y_3 \ - \ x_2x_3y_1 \ + \ x_2x_3y_1 \ - \ x_1x_3y_2$

$x_1w_1 \ = x_1x_2y_3 \ - \ x_1x_3y_2$

$w_1 \ = (x_2y_3 \ - \ x_3y_2)$

So now we have $w_1$ from
$\overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$

We'll now work on $w_2$

$^{ x_1y_2w_1 \ + \ x_2y_2w_2 \ + \ x_3y_2w_3 \ = \ 0}_{ y_1x_2w_1 \ + \ y_2x_2w_2 \ + \ y_3x_2w_3 \ = \ 0}$

$(x_1y_2 \ - \ y_1x_2)w_1 \ + \ (x_3y_2 \ - \ y_3x_2)w_3 \ = \ 0$

$w_1 \ = \ (x_3y_2 \ - \ y_3x_2)$

$w_3 \ = \ - \ (x_1y_2 \ - \ y_1x_2) \ = \ (y_1x_2 \ - \ x_1y_2)$

Now solve for $w_2$

$x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 \Rightarrow \ x_2w_2 \ = \ - \ x_1w_1 \ - \ x_3w_3$

$x_2w_2 \ = \ - \ x_1(x_3y_2 \ - \ y_3x_2) \ - \ x_3(y_1x_2 \ - \ x_1y_2)$

$x_2w_2 \ = \ x_1x_2y_3 \ - \ x_1x_3y_2 \ + \ x_1x_3y_2 \ - \ x_2x_3y_1$

$x_2w_2 \ = \ x_1x_2y_3 \ - \ x_2x_3y_1$

$w_2 \ = \ x_1y_3 \ - \ x_3y_1$

We now have

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k}$

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ w_3 \hat{k}$

This is obviously incorrect - for some reason I can't place.
There is a minus sign that's off & I can't see how that happens...

We'll now get $w_3$

$^{ x_1y_3w_1 \ + \ x_2y_3w_2 \ + \ x_3y_3w_3 \ = \ 0}_{ y_1x_3w_1 \ + \ y_2x_3w_2 \ + \ y_3x_3w_3 \ = \ 0}$

$(x_1y_3 \ - \ x_3y_1)w_1 \ + \ (x_2y_3 \ - \ x_3y_2)w_2 \ = \ 0$

Obviously
$w_1 \ = \ (x_2y_3 \ - \ x_3y_2)$
$w_2 \ = \ -(x_1y_3 \ - \ x_3y_1) \ = \ (x_3y_1 \ - \ x_1y_3)$

will give us zero here!

If we now take

$x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0$

from

$^{ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0}$

and solve for $x_3w_3$ we get;

$x_3w_3 \ = \ - x_1w_1 - x_2w_2 \ = \ - x_1(x_2y_3 \ - \ x_3y_2) - x_2(x_3y_1 \ - \ x_1y_3) \ = \ - x_1x_2y_3 + x_1x_3y_2 \ - \ x_2x_3y_1 + \ x_1x_2y_3$

$x_3w_3 \ = \ x_1x_3y_2 \ - \ x_2x_3y_1$

$w_3 \ = \ x_1y_2 \ - \ x_2y_1$

We now have

$\overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ w_3 \hat{k}$

$\overline{a} \times \overline{b} \ = \ \overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ (x_1y_2 \ - \ x_2y_1) \hat{k}$

14. Aug 25, 2010

### ehild

The cross product is connected to such physical quantities as torque or area of a parallelogram which depend on the perpendicular component of a vector to an other one. The magnitude of the cross product of the vectors a and b is the product of their magnitudes and the sine of the enclosed angle. Because of the sine, the cross product of parallel vectors is zero. The product itself is a vector that is perpendicular to both vectors, and points in that direction from where the rotation of the first vector, a into the second one, b, with angle less than 180°, looks anti-clockwise.
Or: if we imagine that we turn a screwdriver in the same way as we turned a into b, the product vector would point in the direction of motion of the screw.
If the unit vectors along the x, y, z axes are i, j, k, respectively, ixj=k, but jxi=-k, as the direction of rotation of j into i is clockwise . Similarly, jxk=i and kxi=j. You need cyclic permutations of i, j, k to get positive sign of the product.
Knowing the cross-product of the unit vectors, you can calculate the cross product of any two vectors from their components.

ehild

15. Aug 25, 2010

Yes I know, I was very unhappy accepting the "permutative" nature of i, j & k until I
found out the reason why you do that with the unit vectors, I've just got a problem with the
minus sign in the above derivation if you'd be able to spot the problem.

I'm starting to think it's just inbuilt into one of the components, namely it falls out of the
fact that a component changes sign due to the orthogonality of the new vector but it should
fall out of the calculation here I thought

16. Aug 25, 2010

### ehild

I can not (or I am just lazy) to follow your derivation. It is not enough to find a vector which is perpendicular to both original ones. And why to start with general vectors? If you fix the cross product of the unit vectors i, j, k, you will find the product of any two vectors.

So let be two vectors

$$\vec {a}=a_x\vec {i }+a_y \vec{j}+a_z\vec{k}$$

$$\vec {b}=b_x\vec {i }+b_y \vec{j}+b_z\vec{k}$$

$$\vec {a}\times\vec {b}=(a_x\vec {i }+a_y \vec{j}+a_z\vec{k})\times (b_x\vec {i }+b_y \vec{j}+b_z\vec{k})=a_x b_x\vec {i }\times\vec {i }+a_x b_y \vec {i }\times\vec {j }+a_x b_z\vec {i }\times\vec {k }+a_y b_x\vec {j }\times\vec {i }+a_y b_y\vec {j }\times\vec {j }+a_y b_z\vec {j }\times\vec {k }+a_z b_x\vec {k}\times\vec {i }+a_z b_y\vec {k}\times\vec {j }+a_z b_z\vec {k }\times\vec {k }$$

Knowing that

$$\vec {i}\times\vec {i}=\vec {j}\times\vec {j}=\vec {k}\times\vec {k}=0$$

$$\vec {i}\times\vec {j}=\vec{k},\vec {j}\times\vec {k}=\vec{i},\vec {k}\times\vec {i}=\vec{j}$$

$$\vec {j}\times\vec {i}=-\vec {k}, \vec {k}\times\vec {j}=-\vec {i}, \vec {k}\times\vec {i}=-\vec {j},$$

$$\vec {a}\times\vec {b}=a_x b_y \vec {k }-a_x b_z\vec {j }-a_y b_x\vec {k}+a_y b_z\vec {i }+a_z b_x\vec {j}-a_z b_y\vec {i}=(a_y b_z-a_z b_y)\vec {i }+(a_z b_x-a_x b_z)\vec {j}+(a_x b_y -a_y b_x)\vec {k }$$

ehild

Last edited: Aug 25, 2010
17. Aug 25, 2010

When you say it's not enough to just find a vector perpendicular to the other two you're
telling me not to listen to, as of now, two books that have told me this is the correct derivation.

here is the second source of this
derivation. Furthermore the teacher in this pdf calls what you've given me "a convenient mnemonic" for getting the right answer.

Just copying the book because it says $$\hat{i} \times \hat{j} \ = \ \hat{k}$$ is
in no way satisfying, offers no motivation & doesn't explain anything.

This derivation explains everything & my only problem is understanding the minus sign in
the derivation above.

18. Aug 25, 2010

Well I went back and just reversed my choices for $w_1$ & $w_3$ when trying to find the solution to $w_2$ and got the right answer

I just found that pdf 5 minutes ago and if you look at page 3 he says

"5. We are left with one true ambiguity in the de¯nition, and that is which sign to take. In our
development we chose z3 = x1y2 ¡ x2y1, but we could have of course chosen z3 = x2y1 ¡ x1y2.
In this case, the entire mathematical community agrees with the choice we have made."

Lol! Arbitrary choices!!!

I didn't read page 3 until a second ago

He says in the pdf that we're just choosing the values as we please here
in order to make the equation work & then defining a x b as that.

So, this is the reason why $\hat{i} \times \hat{j}$ is defined as $\hat{k}$