Cross Product?

sponsoredwalk

I asked this question on physicsforums a moment ago & was basically told I
need to understand exterior algebra to understand the following but that
can't be the case :pac: I mean, if that's true all of you must have taken this
on faith the first time you came across it & I doubt that!

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



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

[latex] \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}) [/latex]

[latex] 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})[/latex]

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

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

Serge Lang just defines [latex]\bold{a} \times \bold{b} \ = \ a_yb_z \ - \ a_z b_y \ ... [/latex] 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 [latex] \hat{i} \times \hat{i} \ = \ 0 [/latex] and [latex] \hat{i} \times \hat{j} \ = \ \hat{k} [/latex] 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 [latex] \hat{i} \times \hat{k} \ = \ - \hat{j} [/latex] etc...
so to get to that point you need to know why this happens in the first place...

thanks

equivariant

sponsoredwalk wrote: »

I asked this question on physicsforums a moment ago & was basically told I
need to understand exterior algebra to understand the following but that
can't be the case :pac: I mean, if that's true all of you must have taken this
on faith the first time you came across it & I doubt that!

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



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

[latex] \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}) [/latex]

[latex] 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})[/latex]

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

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

Serge Lang just defines [latex]\bold{a} \times \bold{b} \ = \ a_yb_z \ - \ a_z b_y \ ... [/latex] 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 [latex] \hat{i} \times \hat{i} \ = \ 0 [/latex] and [latex] \hat{i} \times \hat{j} \ = \ \hat{k} [/latex] 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 [latex] \hat{i} \times \hat{k} \ = \ - \hat{j} [/latex] etc...
so to get to that point you need to know why this happens in the first place...

thanks

Its tricky to answer this as it is difficult to know what level of explanation you are looking for. Anyway, here are a couple of things that are relevant

The cross product is a special case of some interesting constructions in mathematics.

1. A very important class of algebraic structures is the class of Lie algebras. A lie algebra is a vector space equipped with a binary operation satisfying certain axioms (see http://en.wikipedia.org/wiki/Lie_algebra). It turns out that these axioms correspond to the basic well known properties of the cross product, such as [latex] v \times v =0[/latex] and so on. So the cross product is a basic example of this important structure.

2. One can also think of the cross product [latex]u \times v[/latex] as the oriented area of the parallelogram spanned by the vectors u and v. The magnitude of [latex] u \times v[/latex] is the area of the parallelogram and the direction of [latex] u \times v[/latex] is the direction that is associated to that oriented parallelogram. Thus for example, this interpretation makes it clear that we should expect [latex] v \times v[/latex] to be 0.

antiselfdual

Hmm here's a simple explanation avoiding exterior algebra and relying on accepting a definition of the cross-product... So one way of defining the cross-product (wikipedia!) of vectors a and b is just as the vector perpendicular to a and b, with magnitude given by the area of the parallelogram they span, and direction given by the right-hand-rule... This latter implies that [latex]\vec{a} \times \vec{b} = - \vec{b} \times \vec{a}[/latex]. So [latex]\vec{a} \times \vec{a} = - \vec{a} \times \vec{a}[/latex] which can only be true if [latex]\vec{a} \times \vec{a} =0[/latex]. This explains your first question...

For the second, if [latex]\hat{i}[/latex] and [latex]\hat{j}[/latex] are unit vectors in the x and y directions then their cross-product is going to be a unit vector perpendicular to the x-y plane. So it has to be either [latex]\pm \hat{k}[/latex]. The convention of using the right-hand-rule to give the direction of the cross-product means that [latex]\hat{i} \times \hat{j} = \hat{k}[/latex].

Enkidu

equivariant has the right sort of idea, but I'll just add some more.

Essentially, the true explanation does some from Exterior algebra. Just to clear up a few things Exterior algebra is just a type of algebra and hence has nothing really to do with geometry like the cross-product. It would be more correct to say that the true explanation comes from differential forms and differential forms obey Exterior algebra. All the weirdness of the cross-product comes as an artefact of trying to shove something which is more natural in differential forms into the language of vectors. You'll encounter this in electromagnetism as well. For example the cross-product does not transform between coordinates correctly.

Anyway, I'll try an explanation that's accessible. The cross product essentially takes two vectors and spits out a vector whose magnitude is as great as the area contained between the two vectors, it also points perpendicular to the two vectors. Since there is no area swept out between the basis vector i and itself, the cross-product is zero.

In another way you think of the cross product as giving information about the area swept between the vectors a and b. It's direction is determined by which area it is. The area swept between a and b is considered different than the area swept between b and a. It is an area with an orientation, if you like, the direction of the cross-product vector keeps track of the orientation and its magnitude keeps track of the size of the area. So the cross-product is a vector which records information about an area.

Differential forms basically says "Why bother?" and figures out a way of writing down the area directly, instead of recording it in a vector. However doing that requires more advanced mathematics so I'll leave it there.

sponsoredwalk

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 [latex] \hat{k} \times \hat{j} \ = \ - \hat{i} [/latex] which makes no sense to me.
My only problem here is understanding why [latex] \hat{i} \times \hat{j} \ = \ \hat{k} [/latex] 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 [latex] \hat{i} \cdot \hat{j} \ = \ 0 [/latex]
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 [latex] b \cdot h [/latex]
where b = ||a|| & h = ||b||sinθ, that's all fine.
My problem is that when computing the components you encounter having
to deal with [latex] \hat{k} \times \hat{j} \ = \ - \hat{i} [/latex] and all
of the fun variants, accepting it simply because it works isn't enough...

[latex] \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}) [/latex]

[latex] 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})[/latex]

[latex] 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)[/latex]

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

[latex] (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}[/latex]

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 [latex] \hat{k} \times \hat{j} \ = \ - \hat{i} [/latex]
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 throughout the internet, don't answer the question at all.

antiselfdual

What are you taking as the definition of the cross-product?

(Also [latex]\hat{i} \cdot \hat{j} = 0[/latex] has nothing to do with multiplying "numbers from different dimensions," that doesn't make sense, it's just saying they're orthogonal w.r.t inner product. If you take [latex]\hat{i} = (1,0,0)[/latex] and [latex]\hat{j} = (0,1,0)[/latex] then [latex]\hat{i} \cdot \hat{j} = 1\cdot 0 + 0 \cdot 1 + 0 \cdot 0= 0[/latex], only numbers from the same dimension as you put it are multiplied together.)

sponsoredwalk

Watch this video from 18:50 to 20:30, you'll see what I meant.

Well, one definition of the cross product is showing the determinant and
telling you what to do with it, and a second one is just giving you the answer
after calculating the determinant i.e. (a_y b_z - a_z b_y)i - .... etc... and a
third definition is the way I multiplied them out in my original post
a x b = (a_x i + a_y j + a_z k) x (b_x i + b_y j + b_z k)
- with the important point that [latex] \hat{i} \times \hat{j} \ = \ \hat{k} [/latex] etc...

Enkidu

sponsoredwalk wrote: »

It's a cheat because it uses [latex] \hat{k} \times \hat{j} \ = \ - \hat{i} [/latex] which makes no sense to me.
My only problem here is understanding why [latex] \hat{i} \times \hat{j} \ = \ \hat{k} [/latex] 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???

Well one of the ways of defining the cross product is that it is a vector completely orthogonal to the two original vectors. That's just part of its definition. The cross product is the vector operator which generates a vector perpendicular to the original ones with a magnitude equal to the area between the two vectors.

Now since we could define a lot of such operators why are we interested in this one? From a physical point of view it's because the cross-product is useful in describing electromagnetism, since it mathematically encodes the right-hand rule.

antiselfdual

That video freaks me out. However I think the clip from 17:53 on is more relevant to defining the cross-product: it is a vector perpendicular to the two original vectors, whose length is given by the area of the parallelogram formed by the two original vectors, and whose direction (because there are two possible directions which it could be perpendicular in) is given by the right-hand-rule. Obviously this is the sort of definition relevant for basic mechanics.

Now we come to [latex]\hat{i}, \hat{j}[/latex]. These are perpendicular unit vectors. If you imagine these in the Cartesian plane it should be clear that the parallelogram they form is a square of unit area. Their cross product is then a vector of unit length, perpendicular to the plane in which [latex]\hat{i}, \hat{j}[/latex] lie. The right-hand rule tells us that this cross-product corresponds to the unit vector [latex]\hat{k}[/latex] (as opposed to the other perpendicular possibility, [latex]-\hat{k}[/latex]). This is all just following from the basic definition of the cross-product in the first paragraph.

Obviously after you're familiar with this (and differential forms, etc) it is interesting to observe how the same structure can be expressed elegantly using exterior algebra etc. But I think the fundamentals of what you're asking can be answered using a simpler definition.

I've always thought of the determinant as just being a trick to work out the cross-product, rather than a fundamental definition.

But seriously what's up with those people on the boat singing "99 Bottles of Beer on the Wall"???

sponsoredwalk

I really wanted to give you the video from around 17:30 on lol to get them
guys on the boat So funny! The voice-over guy always throws in little
digs like that The funniest one is when he's describing momentum using
snooker balls at a pool hall, he's just ripping it out of all the old guys drinking
and playing pool in the video "Chalky's Billiard Balls" or something...

I do view the determinant as a trick to get the right answer, that's why I
avoid it. I like the logic of this definition

a x b = (a_x i + a_y j + a_z k) x (b_x i + b_y j + b_z k)

the wiki does it nicely there but
it returns to the "definition" of i x j = k and all of that.

You have to understand where I'm coming from, this magic equation didn't
just appear out of thin air, defining such a complex arrangement of the
components

(a_y b_z - a_z b_y)i - (a_x b_z - a_z b_x)j + (a_x b_y - a_y b_x) k

as coming from a x b = (a_x i + a_y j + a_z k) x (b_x i + b_y j + b_z k)
and then telling us that the sine function is intimately related to this
takes a lot of foresight to find - or magically setting i x j = k
Furthermore, if you try to work backwards from ||a|| ||b|| sinθ
seeing as it's the only way to pre-conceive area requires a lot of forsight and
magically defining i x j = k
The only feasible way I can see the connection is by working backwards
this way but that still says nothing about i x j = k or anything and certainly
doesn't illustrate anything deep.

equivariant

sponsoredwalk wrote: »

I really wanted to give you the video from around 17:30 on lol to get them
guys on the boat So funny! The voice-over guy always throws in little
digs like that The funniest one is when he's describing momentum using
snooker balls at a pool hall, he's just ripping it out of all the old guys drinking
and playing pool in the video "Chalky's Billiard Balls" or something...

I do view the determinant as a trick to get the right answer, that's why I
avoid it. I like the logic of this definition

a x b = (a_x i + a_y j + a_z k) x (b_x i + b_y j + b_z k)

the wiki does it nicely there but
it returns to the "definition" of i x j = k and all of that.

You have to understand where I'm coming from, this magic equation didn't
just appear out of thin air, defining such a complex arrangement of the
components

(a_y b_z - a_z b_y)i - (a_x b_z - a_z b_x)j + (a_x b_y - a_y b_x) k

as coming from a x b = (a_x i + a_y j + a_z k) x (b_x i + b_y j + b_z k)
and then telling us that the sine function is intimately related to this
takes a lot of foresight to find - or magically setting i x j = k
Furthermore, if you try to work backwards from ||a|| ||b|| sinθ
seeing as it's the only way to pre-conceive area requires a lot of forsight and
magically defining i x j = k
The only feasible way I can see the connection is by working backwards
this way but that still says nothing about i x j = k or anything and certainly
doesn't illustrate anything deep.

I'm not sure what kind of explanation you are asking for? On one hand you seem unsatisfied with just assuming that i x j = k etc as a starting point. But on the other hand, you don't seem satisfied with people using words like exterior algebra or Lie algebra. Basic Lie theory (or the basic notions of exterior algbera) is not incredibly difficult, so maybe read a little of that and you will find a satisfactory explanation.

sponsoredwalk

I knew it!!!! I KNEW IT!!!!! :D:D

I knew there was a subtle logic to all of this :cool:

Given two vectors
[latex] \overline{x} \ = \ x_1 \hat{i} \ + \ x_2 \hat{j} \ + \ x_3 \hat{k} [/latex]
and
[latex] \overline{y} \ = \ y_1 \hat{i} \ + \ y_2 \ + \ y_3 \hat{k} [/latex]
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
[latex] \overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k} [/latex]
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

[latex] ^{ \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} [/latex]

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!

[latex] ^{ 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} [/latex]

[latex] (x_1y_3 \ - \ x_3y_1)w_1 \ + \ (x_2y_3 \ - \ x_3y_2)w_2 \ = \ 0 [/latex]

Obviously
[latex] w_1 \ = \ (x_2y_3 \ - \ x_3y_2) [/latex]
[latex] w_2 \ = \ -(x_1y_3 \ - \ x_3y_1) \ = \ (x_3y_1 \ - \ x_1y_3)[/latex]

will give us zero here!

If we now take

[latex] x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 [/latex]

from

[latex] ^{ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0} [/latex]

and solve for [latex] x_3w_3[/latex] we get;

[latex] 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 [/latex]

[latex] x_3w_3 \ = \ x_1x_3y_2 \ - \ x_2x_3y_1 [/latex]

[latex] w_3 \ = \ x_1y_2 \ - \ x_2y_1 [/latex]

:D:D

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

When we've computed all of

[latex] \overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k} [/latex]

we can take it's magnitude & seeing as we've got [latex]w_1[/latex], [latex]w_2[/latex] & [latex]w_3[/latex] in the [latex]x_1y_2 \ - \ x_2y_1 [/latex] 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!

:D:D

All is right with the world again :cool:


the source of my happiness!!!

sponsoredwalk

I'm having trouble with a minus sign

Given two vectors
[latex] \overline{x} \ = \ x_1 \hat{i} \ + \ x_2 \hat{j} \ + \ x_3 \hat{k} [/latex]
and
[latex] \overline{y} \ = \ y_1 \hat{i} \ + \ y_2 \ + \ y_3 \hat{k} [/latex]
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
[latex] \overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k} [/latex]
such that;

w • x = 0
w • y = 0



[latex] ^{ \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} [/latex]

This is the central equation to return to. We'll find [latex] w_1[/latex] first.

[latex] ^{ 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} [/latex]

[latex] (x_2y_1 \ - \y_2x_1)w_2 \ + \ (x_3y_1 \ - \ x_1y_3)w_3 \ = \ 0[/latex]

[latex]w_2 \ = \ (x_3y_1 \ - \ x_1y_3) [/latex]
[latex]w_3 \ = \ - (x_2y_1 \ - \y_2x_1) \ = \ (y_2x_1 \ - \ x_2y_1) [/latex]

Solve for [latex] w_1 [/latex] now.

[latex] x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 \Rightarrow \ x_1w_1 \ = \ - \ x_2w_2 \ - \ x_3w_3 [/latex]

[latex]x_1w_1 \ = \ - \ x_2(x_3y_1 \ - \ x_1y_3) \ - \ x_3(y_2x_1 \ - \ x_2y_1) [/latex]

[latex]x_1w_1 \ = x_1x_2y_3 \ - \ x_2x_3y_1 \ + \ x_2x_3y_1 \ - \ x_1x_3y_2[/latex]

[latex]x_1w_1 \ = x_1x_2y_3 \ - \ x_1x_3y_2[/latex]

[latex]w_1 \ = (x_2y_3 \ - \ x_3y_2)[/latex]

So now we have [latex] w_1[/latex] from
[latex] \overline{w} \ = \ w_1 \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k} [/latex]

[latex] \overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k} [/latex]

We'll now work on [latex] w_2[/latex]

[latex] ^{ 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} [/latex]

[latex] (x_1y_2 \ - \ y_1x_2)w_1 \ + \ (x_3y_2 \ - \ y_3x_2)w_3 \ = \ 0[/latex]

[latex] w_1 \ = \ (x_3y_2 \ - \ y_3x_2) [/latex]

[latex] w_3 \ = \ - \ (x_1y_2 \ - \ y_1x_2) \ = \ (y_1x_2 \ - \ x_1y_2) [/latex]

Now solve for [latex] w_2[/latex]

[latex] x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 \Rightarrow \ x_2w_2 \ = \ - \ x_1w_1 \ - \ x_3w_3 [/latex]

[latex] x_2w_2 \ = \ - \ x_1(x_3y_2 \ - \ y_3x_2) \ - \ x_3(y_1x_2 \ - \ x_1y_2) [/latex]

[latex] x_2w_2 \ = \ x_1x_2y_3 \ - \ x_1x_3y_2 \ + \ x_1x_3y_2 \ - \ x_2x_3y_1 [/latex]

[latex] x_2w_2 \ = \ x_1x_2y_3 \ - \ x_2x_3y_1 [/latex]

[latex] w_2 \ = \ x_1y_3 \ - \ x_3y_1 [/latex]

We now have

[latex] \overline{w_2} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ w_2 \hat{j} \ + \ w_3 \hat{k} [/latex]

[latex] \overline{w_2} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ w_3 \hat{k} [/latex]

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 [latex] w_3 [/latex]


[latex] ^{ 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} [/latex]

[latex] (x_1y_3 \ - \ x_3y_1)w_1 \ + \ (x_2y_3 \ - \ x_3y_2)w_2 \ = \ 0 [/latex]

Obviously
[latex] w_1 \ = \ (x_2y_3 \ - \ x_3y_2) [/latex]
[latex] w_2 \ = \ -(x_1y_3 \ - \ x_3y_1) \ = \ (x_3y_1 \ - \ x_1y_3)[/latex]

will give us zero here!

If we now take

[latex] x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0 [/latex]

from

[latex] ^{ x_1w_1 \ + \ x_2w_2 \ + \ x_3w_3 \ = \ 0}_{ y_1w_1 \ + \ y_2w_2 \ + \ y_3w_3 \ = \ 0} [/latex]

and solve for [latex] x_3w_3[/latex] we get;

[latex] 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 [/latex]

[latex] x_3w_3 \ = \ x_1x_3y_2 \ - \ x_2x_3y_1 [/latex]

[latex] w_3 \ = \ x_1y_2 \ - \ x_2y_1 [/latex]

We now have

[latex] \overline{w} \ = \ (x_2y_3 \ - \ x_3y_2) \hat{i} \ + \ (x_1y_3 \ - \ x_3y_1) \hat{j} \ + \ w_3 \hat{k} [/latex]

[latex] \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} [/latex]

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

ray giraffe

sponsoredwalk wrote: »

[latex] (x_2y_1 \ - \y_2x_1)w_2 \ [/latex][latex]+[/latex] [latex]\ (x_3y_1 \ - \ x_1y_3)w_3 \ = \ 0[/latex]

[latex]w_2 \ = \ (x_3y_1 \ - \ x_1y_3) [/latex]
[latex]w_3 \ = \ - (x_2y_1 \ - \y_2x_1) \ = \ (y_2x_1 \ - \ x_2y_1) [/latex]

ab + cd = 0 does not imply b=c and d=-a.

Although the vector 'a x b' is perpendicular to a and b, it is not "the" vector perpendicular to a and b - there are many (all scalar multiples of 'a x b'). So you can't "find" it using that condition.

sponsoredwalk

If you choose [latex] w_2 [/latex] and [latex] w_3 [/latex]to be the said
values then you've got equality.

[latex] (x_1y_2 \ - \ y_1x_2)w_1 \ + \ (x_3y_2 \ - \ y_3x_2)w_3 \ = \ 0[/latex]

[latex] w_1 \ = \ (x_3y_2 \ - \ y_3x_2) [/latex]

[latex] w_3 \ = \ - \ (x_1y_2 \ - \ y_1x_2) \ = \ (y_1x_2 \ - \ x_1y_2) [/latex]

[latex] (x_1y_2 \ - \ y_1x_2) (x_3y_2 \ - \ y_3x_2) \ - \ (x_3y_2 \ - \ y_3x_2)(x_1y_2 \ - \ y_1x_2) \ = \ 0 \Rightarrow \ (x_1y_2 \ - \ y_1x_2) (x_3y_2 \ - \ y_3x_2) \ = \ (x_1y_2 \ - \ y_1x_2) (x_3y_2 \ - \ y_3x_2)[/latex]

sponsoredwalk

In this pdf the guy who wrote that book does pretty much the exact same thing as well...

ray giraffe

e.g. If I tell you x-2y = 0, you cannot conclude that (x,y)=(2,1) even though that is a solution

ray giraffe

sponsoredwalk wrote: »

In this pdf the guy who wrote that book does pretty much the exact same thing as well...

Not quite. Notice the fractions on the top of page 2. He chooses a particular solution in choosing a value for z_3.

sponsoredwalk

Well I went back and just reversed my choices for [latex] w_1 [/latex] & [latex] w_3 [/latex] when trying to find the solution to [latex] w_2 [/latex] 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.

sponsoredwalk

This is a terrible trend I've seen, making derivable things definitions as if they
come out of thin air. It's so unfortunate

If you see some problem with this please let me know, I don't want to walk
off, again, with false hope

FoxT

I came across this years ago when I was studying magnetism.

k, The cross product of 2 vectors v and w, is a vector orthogonal to v and w such that

1 -> |k| = |v||w| sin (theta). where theta is the angle between v and w.

( ie magnitude of k = product of the magnitudes of the other 2 vectors and the sin of the angle between them)

and

2 -> the direction of k is such that v,w, k form a right handed system.

By 'Right-handed system' it is meant that if you take a flat head screw & align the head so that it is parallel to v and then turn it so it is parallel to w, then the screw will travel in the direction of k.

from the above

ixi = jxj = kxk = 0 ( because sin (theta) is zero)

and

ixj = k, ( you are turning the screw clockwise )

but jxi= -k ( you are turning the screw counterclockwise )

and in general, the X operation is not commutative.


Hope this helps to give a non-mathematical but more intuitive understanding of it.

- FoxT