Definition of the Cross Product?

[2^Oscar]

Hi Guys,

I realize that this may seem like a really simple question but it's really driving me crazy. In my A-level maths we've just started looking at the cross product and I understand how to calculate it, it's application in calculating areas ect but I find myself not understanding what the product itself actually represents.

Could someone please help clarify this for me? I do not understand why the product represents a vector perpendicular to the vectors being crossed.


Thanks,
Oscar

 

[Studiot]

Vectors are either

1) Parallel

2) Skew

3) Crossed

In cases 1 and 2 they never intersect. (Skew lines are lines or vectors in 3 D that are not parallel and do not intersect.)

Crossed lines or vectors intersect at a point.

Now to the cross product.
As you have probably been told there are two ways we can multiply vectors. So we need 2 names.

The dot product of two vectors yields a scalar.

The cross (or sometimes called the wedge) product yields another vector.
It does more than this as it provides a mathematical 'in' to the idea of rotation since we have a natural rotation direction provided by the right hand rule.

So you see the two products appear for completeness, not some deeper mysticism.

You have probably been told that

the dot product of A.B is mod(a) x mod(B) cos\theta
and
the cross product is mod(a) x mod(B) sin\theta

 

[Peeter]

I liked Eli's write up here:

http://behindtheguesses.blogspot.com/2009/04/dot-and-cross-products.html

where he introduces the cross product as an oriented area.

 

[Landau]

So you see the two products appear for completeness, not some deeper mysticism.

You have probably been told that

the dot product of A.B is mod(a) x mod(B) cos\theta
and
the cross product is mod(a) x mod(B) sin\theta

That's a pretty bad explanation. Now it seems that, like the dot product of two vectors, the cross product of two vectors is a scalar! The cross product of two vectors is another vector, whose norm equals |a|.|b|.sin t (with t the angle between a and b).

 

[Studiot]

That's a pretty bad explanation. Now it seems that, like the dot product of two vectors

In my A-level maths we've just started looking at the cross product and I understand how to calculate it

There is no intention to mislead, just a desire not to open a beyond A level can of worms.

I debated with myself whether to introduce the area aspect, but left this to Oscar's teachers. But the area aspect is at least is in the plane of the generating vectors.

Skew lines and areas appear in A level in 'Vectors and Matrices 1'
The necessary apparatus to set vectors in R^{3} does not appear until later when vector spaces are introduced in 'Abstract Algebra'

Oscar if you have covered the latter there is more worth saying.

 

[HallsofIvy]

Not wanting to "open a beyond A level can of worms" is not a reason for saying something that is not true!

"mod(A) mod(B) sin(\theta)" is NOT the cross product, it is only the length of the cross product which is a vector.

 

[Studiot]

Not wanting to "open a beyond A level can of worms" is not a reason for saying something that is not true!

I agree what I said was not rigorous and that Landau's presentation is better.

But it was the literal truth since you want to play pedanticism. Look again at the whole sentence I wrote, please do not state something taken out of context to be untrue when it is actually true within the context stated.

Now to open the can of worms.

1) The cross product does not ever lead to vector within the vector space spanned by the generating vectors. You need to enlarge the supporting vector space to introduce the product 'vector.' This is a very fundamental idea that needs and is worth a deal of discussion. It is the most amazing thing. This may even be the crux of the OPs question and I gently enquired to try to determine this.

2) In some cases the cross product does not even produce a vector it produces a pseudovector. These have been discussed in other threads here.

 

[2^Oscar]

Hi guys,

Thank you for the replies. I apologise if my question has caused all the 'beyond A-level fuss', but I guess in a sense that is really what I'm wanting to be explained (all though granted I understand if it cannot be explained without me having high-level knowledge to back-up the explanation then it is probably best if I just accept the result of crossing two vectors for the time being). The main fact that is confusing me is that when two vectors are crossed the vector produced is perpendicular to both; I always thought that two vectors which are perpendicular to one another, such as two perpendicular forces, have no effect on each other, so I was confused how two vectors when crossed could produce a perpendicular vector? I assume this is something to do with the actual process of crossing two vectors, so it is really this that I'm trying to understand.

From the link Peeter posted I've gotten the impression that the vector product comes from wanting to find a measure of how perpendicular two intersecting vectors are to each other. If I've understood it correctly does this mean that the perpendicular vector is basically like a point of reference, being perpendicular itself to both vectors, that allows us to do this?

I've dealt with some of the area material, only so far as parallelograms and triangles, and have begun dealing with the triple scalar product and its applications to volume.

This is part of the AQA exam board's FP4 module. A group of people in my sixth form, myself included, are studying this as an 'extra' module to supplement our A-levels in mathematics. We receive weekly talks on some of the topics during lunchtime and the others we must teach ourselves. The vector product topic is one which we have covered in these sessions, but we have currently broken up for a bank holiday and so I have no contact with my tutor to make this query for the next fortnight, and I was hoping to have completely understood the vector product section before returning to sixth form and the above query was the main issue I felt was preventing me from thoroughly understanding it. If it is an abstract concept well beyond the scope of A-level mathematics then I must accept it as a given and work towards practising questions instead.


Again, thank you for your replies and the patience in explaining this to me!
Oscar

 

[Peeter]

The question of why the cross product is perpendicular to both, can also be looked at in different ways. One such way is by noting that one does not have to define a vector product as a vector, but can introduce a different sort of geometric object that directly encodes the plane spanned by the pair of vectors:

http://en.wikipedia.org/wiki/Bivector

(The wedge product used in this article has, in three dimensions, exactly the same magnitude as the cross product, and has some other similar properties).

It just happens to be that in three dimensions one can also uniquely define a normal to this plane, and that is convenient for many purposes. It is also unnatural in some instances, for quantities such as torque that ought to be perfectly expressible in 2 dimensions (or greater than 3) but using the duality notion (perpendicular to the plane) only allows for a 3D encoding.

 

[Studiot]

Hi Oscar we haven't abandoned you.

I will try to explain further what I was banging on about.

Firstly you might like to know that mathematicians and physicists have a slightly different definition of vector. Unfortunately biologists and computer scientists have yet other ones.

Anyway mathematicians say that vectors are the inhabitants of a vector space. That is they are objects that obey certain well specified rules or axioms. Mathematical vectors include the objects you and physicists call vectors as a special case and call them cartesian vectors, or sometimes line vectors, because they are referred to the normal x,y and possibly z axes. These are the ones we are dealing with here.

You might be suprised at some of the other objects that mathematicians call vectors. Not all of these have a process of multiplication so do not support a vector product.

However there are some useful properties, fundamental to all vectors.

A vector space is a collection of all vectors of a particular type.
Let us look at my attached diagram of a plane. This is a vector space. I have drawn 3 vectors, A, B and C.

Now first consider C. As I rotate it about the origin I can align it with A or B and by stretching it I can make it as long or short as I like.
In other words I can make it into any vector in the plane.

Now look at A and B.
The dashed lines look a bit like resolution into components, but the resolved parts are shorter than either A or B.
No matter I can fix this by multiplying A and B by a factor or number as I have shown in the equation underneath.

However I have turned it round and not shown it as a decomposition, rather an assembly so that

\alpha A + \beta B = C

In other words if I add the alpha times A to beta times B I will get another vector, C.

Since, as noted earlier, C can represent any vector in the plane I can represent any vector in the plane in this way.

The point of all this (which I won't prove) is that every vector that can be obtained from adding some portion of A to some portion of B is in the plane and that all vectors in the plane can be obtained this way.

Mathematically we have our vectors in our vector space (the plane) and an operation (addition). We say that the space is 'closed under addition'. Which means that adding any two members of the space will always yield a third member of the space, and that all members of the space can be accessed this way.

I can state, without proof, that A and B do not have to be at right angles for the above to be true. So long as they are not parallel we can pick any two vectors and the above will still be true.
This is known as generalisation, something mathematicians do all the time.

OK so we have a space and an operation that can get us any of the vectors in the space, but planes are 2 dimensional and exist in 3D space.

We can't use our addition operation to access any vectors that extend into the 3rd dimension, or to link these with those in our plane. This is clearly unsatisfactory.

Enter the cross product, designed specially for this purpose. You will find that the triple products you have commenced studing further extend the relationships between line vectors in 3D.

This is why they are so useful in Physics. I would recommend accepting that they are designed for this purpose (they would be pretty useless if they weren't) and just use them for the time being.

 

[2^Oscar]

Hi Guys,

Having read the two latest posts I find that my understanding of the concept is a lot clearer now. I see why it is that we use them and the reasons for actually having the vector product itself which was one of my main concerns in the topic.

I'll take your recommendation, Studiot, and accept for the time being that this is the purpose for which they are intended and concentrate on using them in solutions to problems.


Thank you very much for everybody's time on this thread; it has really helped me grasp the concept of the cross product.


Many thanks again,
Oscar

 

[Rasalhague]

OK so we have a [2 dimensional] space and an operation [addition of scalar multiples of basis vectors] that can get us any of the vectors in the space, but planes are 2 dimensional and exist in 3D space.

We can't use our addition [and scalar multiplication] operation to access any vectors that extend into the 3rd dimension, or to link these with those in our plane. This is clearly unsatisfactory.

Enter the cross product, designed specially for this purpose.

A cross product isn't necessary for this though, is it? You could designate any vector in the 3 dimensional vector space--provided that it isn't a linear combination of the other two (i.e. can't be obtained by adding and scaling them)--as your third basis vector, \textbf{G}. If there's a concept of angles in this space, \textbf{G} doesn't have to be at a right angle to \textbf{A} or \textbf{B}, any more than they have to be at right angles to each other.

Then you can represent any vector in the 3 dimensional space as a sum of scalar multiples:

\textbf{C} = \alpha \textbf{A} + \beta \textbf{B} + \gamma \textbf{G}