I Understanding the cross product and rotations

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Hi

I have used cross products thousands of time without really knowing what it actually does; I know how to compute it, but I don't feel like I understand it. Also, when it shows up in physics/kinematics contexts, it's only because the magnitudes of the vectors involved have to be multiplied and with the sine of the angle between them (and have to point in a direction perpendicular to both vectors). I know this is equal to [itex]X\times Y[/itex], but I don't know quite why.

I am looking for one or more resources which can explain it both rigorously and intuitively (not necessarily simultaneously).
 

fresh_42

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The answer to your question depends on what you expect as an answer to "what is?". There is certainly a geometric and an algebraic answer, maybe even an analytic one regarding the Lie algebra structure it defines. So what is allowed to be a what is answer? An interesting read is the following link: https://arxiv.org/pdf/1205.5935.pdf
 
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Well, we can start off with a book that includes a proof of the equality of [itex] |X\times Y| = |X||Y|sin(\delta )[/itex]. Also, a proof that the new vector will be perpendicular to both X and Y and point according to the right hand rule.
 

fresh_42

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Before we can do any of these ...
Well, we can start off with a book that includes a proof of ...
... we can start off with a definition: What is the cross product? You can define it by what you want to prove. Hence a proof would be obsolete, hence my question: how is it defined? Without a definition you cannot prove anything!

Have a look on the pages 4,8 and 27 in the link I gave you! The author explains your questions step by step from the start. If you do not like to read it, then again: provide definition and expectation.
 

WWGD

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Hi

I have used cross products thousands of time without really knowing what it actually does; I know how to compute it, but I don't feel like I understand it. Also, when it shows up in physics/kinematics contexts, it's only because the magnitudes of the vectors involved have to be multiplied and with the sine of the angle between them (and have to point in a direction perpendicular to both vectors). I know this is equal to [itex]X\times Y[/itex], but I don't know quite why.

I am looking for one or more resources which can explain it both rigorously and intuitively (not necessarily simultaneously).
What definition would you like other than it is an operation that takes a pair of vectors v,w and spits out a vector z perpendicular to both v,w? EDIT: Not being sarcastic, just wondering what you are after.
 
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Before we can do any of these ...

... we can start off with a definition: What is the cross product? You can define it by what you want to prove. Hence a proof would be obsolete, hence my question: how is it defined? Without a definition you cannot prove anything!

Have a look on the pages 4,8 and 27 in the link I gave you! The author explains your questions step bystep from the start. If you do not like to read it, then again: provide definition and expectation.
Alright, I'll read through that document. Regarding definition:
[itex]
\begin{align}

{} &(a_2b_3 - a_3b_2)\mathbf{\color{ProcessBlue}{i}} + (a_3b_1 - a_1b_3)\mathbf{\color{red}{j}} + (a_1b_2 - a_2b_1)\mathbf{\color{Dandelion}{k}}\\
\end{align}

[/itex]
I always thought this was the definition. I guess what I wanna know is why this has all the properties I listed above... Or, rather, why a vector with magnitude equal to [itex]|x||y|sin(\delta )[/itex] that is perpendicular to x and y would look like this.
 

fresh_42

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I always thought this was the definition. I guess what I wanna know is why this has all the properties I listed above... Or, rather, why a vector with magnitude equal to |x||y|sin(δ)|x||y|sin(δ)|x||y|sin(\delta ) that is perpendicular to x and y would look like this.
Three vectors span a parallelepiped. Look up how to calculate the volume with exterior products, and then compute it with Pythagoras. The answer lies in these three objects and how they are related.
 

Cryo

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rather, why a vector with magnitude equal to [itex]|x||y|sin(\delta )[/itex] that is perpendicular to x and y would look like this.

Why not to proove it yourself? You have ##\mathbf{c}=\mathbf{a}\times\mathbf{b}##. Take the dot products ##\mathbf{a}.\mathbf{c}=a_x c_x + \dots = a_x (a_y c_z - a_z c_y) + \dots## and ##\mathbf{b}.\mathbf{c}##. You will quickly establish that ##\mathbf{c}## is perpendicular to both ##\mathbf{a,b}##.

Next choose some nice values for ##\mathbf{a,b}## that lie in the xy-plane (so ##\mathbf{c}## will be along z-axis), say ##\mathbf{a}=\left(a, 0, 0\right)## and ##\mathbf{b}=\left(q,p,0\right)##. You will find that there always is an angle, ##\theta## such that ##p=q\tan\theta##, which will make ##\mathbf{b}=\left(q, q\tan\theta,0\right)##. Now try to find the cross product and bear in mind that ##|\mathbf{b}|=|q|\sqrt{1+\tan^2\theta}=|q|/\cos\theta##
 
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I always thought this was the definition. I guess what I wanna know is why this has all the properties I listed above... Or, rather, why a vector with magnitude equal to ##|x||y|\sin(\delta)## that is perpendicular to x and y would look like this.
Perhaps you will be satisfied if I offer up the following explanation. Suppose we were to define the cross product in the latter way, i.e. ##\mathbf{x}\times \mathbf{y}## is perpendicular to ##\mathbf{x}## and ##\mathbf{y}##, and it has magnitude ##|\mathbf{x}||\mathbf{y}|\sin(\delta)##. (I believe that in order to guarantee uniqueness, we have to require that ##\mathbf{x}\times\mathbf{y}## follows the right-hand rule, but I'm not certain on that.) This is sufficient to prove that ##\mathbf{x}\times \mathbf{y}## is linear in each of ##\mathbf{x}## and ##\mathbf{y}##. Luckily for us, linear functions are relatively well-understood, so we can glean a lot more information about the cross product just from knowing that.

Because the cross product is linear (to be more precise, it is bilinear), one can show that in any given coordinate basis, we can express ##\mathbf{x}\times \mathbf{y}## as a sum: ##(\mathbf{x}\times \mathbf{y})_k = \sum_{i,j} c_{ijk}\mathbf{x}_i \mathbf{y}_j## where ##c_{ijk}## is an array of constants. (More generally, in any coordinate basis we have that ##(\mathbf{x}\times \mathbf{y})_k = c_{ij}^{\ \ \ k}\mathbf{x}^i \mathbf{y}^j##, where ##c_{ij}^{\ \ \ k}## is now understood to be a tensor. But that's not important.... yet.)

From there, you can work out each component ##c_{ijk}## by computing the cross product on basis vectors: ##c_{ijk} = (\mathbf{e}_i\times \mathbf{e}_j)_k##. Then from there you can recover the general form of the cross product. In actuality this applies for any (multi)linear function, since any such function is determined by its values on basis vectors.
 

mathwonk

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Whenever we have n-1 vectors in n space, they will "generally" (i.e. if and only if they are independent) span a hyperspace, i.e. a subspace of codimension one (i.e. n-1). Thus there will generally be a complementary orthogonal space of dimension one, and often we would like to be able to find a basis of this complementary line. This is the same as finding a non trivial solution to a set of n-1 homogeneous linear equations in n variables. One way to express such a solution is by cramers rule, which gives a solution in terms of determinants. One can thus write this solution itself as a determinant of an n by n matrix whose first row is the basis vectors e1, e2, ..., en, and with the given n-1 vectors as the next n-1 rows. Then the determinant is a linear combination of the basis vectors, with coefficients which are determinants of the n choices of n-1 by n-1 dimensional cofactors formed from the last n-1 rows. In your case, when n=3, the determinant of the resulting 3by 3 matrix is the expression you wrote down as your definition of the cross product. By definition of dot product, notice also that the resulting determinant is the "dot product" of the vector of cofactors, with the "vector" whose entries are the n basis vectors. I.e. a linear combination has the same form as a dot product.

This means that if you replace the first row of your matrix with the entries of an actual vector, the resulting determinant will be the dot product of that vector with the cross product of your original n-1 vectors. I.e. the dot product between the cross product of some n-1 vectors, and another vector, is always the determinant of the nbyn matrix made up of those n vectors. That means it will be zero if the other vector is chosen to be any one of the given n-1 vectors in the cross product, because then you are taking the determinant of a matrix with a repeated row. So it follows immediately from your definition of a cross product as a determinant, that the cross product of a collection of vectors is always perpendicular to any one of those given vectors.

This works in any dimension, but we are used to talking only about the "product" of two things, so it is more usual to restrict attention to the cross product of two vectors in 3 space. In that case however the same discussion shows that the cross product of 2 vectors is always perpendicular to both of them. (Of course it is zero when those vectors are dependent by the same property of determinants.) When they are independent we don't really care which vector we get as a basis for the complementary space, just as long as it is not zero. We do want to know how long the one we choose is however and one can compute that this method gives one whose length is |x||y| sin(t), from which it is also clear that this is zero when the angle t is zero. Of course having chosen this one with this length we get nice relationships between the cross product and certain volumes. All this follows from the fact that determinants compute volumes. I.e. it seems natural to choose as cross product the vector whose length is determined by the given two vectors in some natural way, in fact its length equals the area of the parallelogram they span. This also shows why it equals zero when they are not independent. We also have to decide whether to choose either one vector of that length or its negative. It is natural in a right handed dominated world to choose one that has right hand orientation. That is also what comes out of this determinant definition.

So the cross product is just a succinct way to write down cramer's rule for a solution of a system of 2 homogeneous equations in 3 unknowns. It is useful for computing an orthogonal complement to a plane spanned by two vectors in 3 space, and also the plane area they span.

Another way to define the cross product of two vectors v,w in three space is to say it is the unique vector 1) perpendicular to both v and w, and 2) of length equal to the area of the parallelogram they span, and 3) when that length is non zero, it is oriented so that when appended to the vectors v,w in that order, i.e. in the order v, w, vxw, we get a right hand coordinate system.

then the problem is how to compute the coordinates of vxw in terms of the coordinates of v and w, which is solved by the formula you used as the definition of vxw. This approach is maybe more instructive, but then it is hard to explain how to make this computation, so most authors "cheat" and take the computation as the definition. That makes all the properties easy to prove, but the motivation is totally lost, hence your question. I.e. people who see your definition first, understandably have no idea what is going on. That is why math sometimes is taught badly in comparison to physics, where concepts are often respected. Well to be fair, in my personal opinion, physics books usually offer better intuition, and math books often provide more precision, so it helps to combine their lessons.
 
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robphy

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mathwonk

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the previously linked article reminded me that cross products are a particular case of hamilton's quaternions. i.e. those have form a + bi + cj + dk, and are also multiplied by the rule that ixj = k, jxk = i, .....

Hence in Maxwell's book on electricity and magnetism as i recall, the basic theorems of several variable calculus such as stokes etc are stated using quaternions rather than cross products. (Do people still read Maxwell? I myself have given up my copies of Spivak, baby Rudin, and Apostol, but I still have Maxwell. Well, I donated those other fine books to an undergraduate workspace library.)
 
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