What is the cross product of two vectors?

[erocored]

I understand that dot product gives us a number and cross product gives a vector. Why is this vector orthogonal to the others two, and why it has magnitude |a|*|b|*sinΘ? How to use cross product? What does it give to us?

 

[PeroK]

You could research this yourself, using the resources on the Internet. You could start here:

https://en.wikipedia.org/wiki/Cross_product

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/2-4-products-of-vectors/

 

[jedishrfu]

This is an interesting question.

If you recall the definition of a dot product:

$\vec A \cdot \vec B = |A||B|cos(\theta)$ where $\theta$ is the angle between vectors A and B.

The interpretation is that the dot product gives you the projection of $\vec A$ on $\vec B$ or vice-versa.

The projection of A on B is: $|A|cos(\theta)$ and similarly the projection of B on A is $|B|cos(\theta)$

Notice, the symmetry of the dot product.

Notice that we might also want to know $|A|sin(\theta)$ or $|B|sin(theta)$ which would be the components of A or B that are NOT projected that are perpendicular and now you have the basis for the cross product:

$\vec A \times \vec B = |A||B|sin(\theta)$

If you consider A and B as the edges of a parallelogram then the cross product defines the area of the parallelogram.

And so noting that the sin() function is positive or negative depending on whether you measure it clockwise or counterclockwise, you can begin to see that the definition of the cross product is:

$\vec A \times \vec B = \vec {|A||B|sin(\theta)}$ where the $\vec A \times \vec B$ vector is perpendicular to both $\vec A$ and $\vec B$.

which provides a very useful vector whose magnitude is the area and whose sign provides an oriented area depending on how you traverse it.

 

[etotheipi]

$\vec A \times \vec B = |A||B|sin(\theta)$

Either the left hand side ought to be $| \vec{A} \times \vec{B}|$, or the right hand side ought to be multiplied by a $\hat{n}$

 

[jedishrfu]

But you missed the vector arrow over the |A||B|sin() expression.

Yes, you're right I should have used the normal vector but I was trying to limit what I said in case the OP didn't know what a normal unit vector was.

And I was trying to derive it to its final definition as being perpendicular to both A and B.

@etotheipi Glad to have you back but don't you still have some studying to do...

 

[PeroK]

But you missed the vector arrow over the |A||B|sin() expression.

I've never seen that before!

 

[jedishrfu]

I did it via \vec { |A||B|sin(\theta) }

It didn't come out very well though, I expected it to extend across the expression. The $\vec n$ is a better approach but hey I'm only a robot.

 

[PeroK]

but hey I'm only a robot.

But you got to hang out with Anne Francis!

 

[etotheipi]

Okay well IMO that is a truly awful notation that gives me a lot of anxiety, but I know that you know what you're talking about and were just trying to simplify things for the OP. Fine.

@etotheipi Glad to have you back but don't you still have some studying to do...

haha, yes. ain't no national lockdown going to stop me from having to do a 3hr physics exam over zoom

 

[Buzz Bloom]

I was surprised looking through the posts that no one mentioned that the context is limited to 3-space, and that the forum is linear and abstract algebra, which of course can deal with higher dimensions.

 

[jedishrfu]

Remember the Feynman story of the physicist asking a mathematician for help with a problem?

Well, that's like this question.

Feynman discusses the applicability of mathematics, contrasted by the interests of most physicists:

If you say "I have a three-dimensional space" [...] and you ask mathematicians about theorems then they say "now look, if you had a space of n dimensions" then here are the theorems". "Yeah, well I only want the case of three dimensions..." "Well, then substitute n = 3!". It turns out that very many of the complicated theorems they have are much simpler because they happen to be special cases. The physicist is always interested in the special case. He's never interested in the general case. He's talking about SOMETHING. He's not talking abstractly about anything. He knows what he's talking about, he wants to discuss the new gravity law, he doesn't want the arbitrary force case, he wants the gravity law! And so, there's a certain amount of reduction because the mathematicians have prepared these things for a wide range of problems which is very useful and later on it always turns out that the poor physicists have to come back and say "excuse me, you wanted to tell me about these four dimensions.."

 

[jedishrfu]

But you got to hang out with Anne Francis!

Not only that some movie posters show me holding her in a very suggestive manner. I was a perfect gentleman and never stooped to such shenanigans.

 

[robphy]

"The Geometry of the Dot and Cross Products"
https://www.maa.org/sites/default/files/images/upload_library/4/vol6/Dray2/Dray.xml

UPDATE:
follow the link at the top of that page
to get to https://www.maa.org/sites/default/files/images/upload_library/4/vol6/Dray2/Dray.pdf
which has symbols that may be missing.

 

[jedishrfu]

A great resource for the OP @robphy . Thanks for sharing.

 

[Mark44]

$\vec A \times \vec B = |A||B|sin(\theta)$

$\vec A \times \vec B = \vec {|A||B|sin(\theta)}$ where the $\vec A \times \vec B$ vector is perpendicular to both $\vec A$ and $\vec B$.

Both of the above are incorrect.
In the first, the expression on the right side is the magnitude of the cross product. Corrected, it would be
$|\vec A \times \vec B| = |A||B|sin(\theta)$
The second is incorrect, even with the vector arrow symbol, since it is "vectorizing" a scalar, sort of like writing $\overrightarrow {42}$.

 

[Infrared]

I taught linear algebra recently, and the way I introduced cross products was as follows (with no claim to originality):

Given $u,v\in\mathbb{R}^3$, the function $w\mapsto\det(u,v,w)$ is linear, and hence there is a unique vector in $\mathbb{R}^3$, which we denote by $u\times v$ such that $(u\times v)\cdot w=\det(u,v,w)$ for all $w\in\mathbb{R}^3.$ This definition has some nice features.

First of all, it is obvious that $u\times v$ is perpendicular to both $u$ and $v$ and also that $u,v,u\times v$ is a positively oriented basis when $u$ and $v$ are independent ("right hand rule").

It clarifies the mysterious mnemonic with unit vectors inside of a determinant that is often used by students to compute the cross product (just taking $w$ to be those unit vectors one at a time gives exactly that formula).

With a bit of thought, this definition gives you $||u\times v||^2=||u||^2 ||v||^2-(u\cdot v)^2$ and hence you get back the usual formula for length of a cross product.

Finally, it naturally generalizes to a "product" that given $n-1$ vectors in $\mathbb{R}^n$, returns a vector orthogonal to all of them.

 

[etotheipi]

One small thing I thought was cool was that somewhere in Landau/Liftshitz they mentioned that you can also view a vector product as a rank 2 antisymmetric tensor, e.g. for instance the $e_i \wedge e_j$ component of a torque is $G^{ij} = x^i F^j - x^j F^i$. And because of the antisymmetry, one of the indices is sort of redundant (we only have 3 degrees of freedom, e.g. imagine the matrix representation), and you can then define a single index object, or vector, via $\tau^i = \frac{1}{2} \varepsilon_{ijk} G^{jk}$. But maybe it's more natural to view the vector product as an element of the space $\Lambda^2 V$. I know basically nothing about exterior algebra yet but I have the feeling that it explains and generalises a lot of fun things from introductory vector algebra & calculus ☺

 

[Infrared]

But maybe it's more natural to view the vector product as an element of the space $\Lambda^2 V$.

A skew-symmetric map $V\times V\to k$ is an element of $\Lambda^2(V^*)$, so a skew-symmetric map $V\times V\to V$ (like the cross product) should be an element of $\Lambda^2(V^*)\otimes V$, right?

I know basically nothing about exterior algebra yet but I have the feeling that it explains and generalises a lot of fun things from introductory vector algebra & calculus ☺

If you've been working with differential forms (as I think you have been) you might know more than you realize!

 

[robphy]

In Tevian's paper, they emphasize the area interpretation of the cross-product.
When I teach, I emphasize this concept... the oriented parallelogram formed from the factors is more fundamental. For more advanced students, I'll use the word "bivector".
(They can be added and scalar multiplied... such bivectors form a [mathematical] vector space.)

In 3-D Euclidean space, which is equipped with a metric, we can then
form the Hodge dual and associate a vector perpendicular to this bivector.

 

[etotheipi]

A skew-symmetric map $V\times V\to k$ is an element of $\Lambda^2(V^*)$, so a skew-symmetric map $V\times V\to V$ (like the cross product) should be an element of $\Lambda^2(V^*)\otimes V$, right?

Ah yeah, sorry, I think I screwed up the notation a little bit. The map from $(u,v) \in V \times V \rightarrow k$, where $k$ is some field, will be of the form $A_{ij} e^i \wedge e^j$ where the $e^i \in V^*$ are some basis set of $V^*$, in which case this map is an element of $\Lambda^2 V^*$. And then like you said another map $V \times V \rightarrow V$ will be an element of the tensor product of $\Lambda^2 V^*$ and $V$. Meanwhile the object $u \wedge v = \frac{1}{2} (u^i v^j - u^j v^i) e_i \wedge e_j \in \Lambda^2 V$ (and, if we wanted, this I guess we could view as a map $V^* \times V^* \rightarrow k$, since $V$ and $V^{**}$ are canonically isomorphic). Hope that's right now

 

[Mark44]

@etotheipi and @Infrared, please keep in mind that this thread was posted at the "B" level. Notions of tensors and skew-symmetric maps are likely well above the OP's current level.

 

[Vanadium 50]

I agree with Mark. I don't think the problem the OP has is that things just aren't complicated enough.

I do like the idea of defining the cross product as Infrared describes, although simpler notation would be my strong preference in this case. If you define a × b ⋅ c as the determinant of the components, and the student has already seen the determinant formula for a parallelepiped then a lot of mysterious-looking formulas become less mysterious.

Doing that also lends itself to the idea of a cross-product as a sort of directed area. Why do you want such a thing? Well, it falls out of the definition, but now that it has, let's see how useful it can be.

Of course, if the student has never seen a determinant before, this is as useful as a Siamese-Swahili Dictionary. ("Once you know one, you can teach yourself the other!")

 

[atyy]

I understand that dot product gives us a number and cross product gives a vector. Why is this vector orthogonal to the others two, and why it has magnitude |a|*|b|*sinΘ? How to use cross product? What does it give to us?

The cross product is the area of a parallelogram.
https://mathinsight.org/cross_product

By convention the cross product is taken to be in the direction perpendicular to the other two. It turns out that this convention of giving surfaces a "direction" is useful, for example, when the laws of electromagnetism are stated in integral form.

 

[Ganesh Mahadevan]

The cross product of two vectors yields a third vector, the magnitude of which is equal to the area of the parallelogram enclosed between the vectors, and the direction is the same as where the primary face of the parallelogram points to. Since there are two faces on any parallelogram, a disambiguation convention is adopted. If the first of the two vectors is rotated about the vertex towards the second one, the direction in which a right hand screw would move if imparted a similar rotation, gives us the orientation of the primary face. If we flipped the order of the vectors, we will get the opposite direction. Cross product is, therefore, taken as a non commutative operation.

 

[ChinleShale]

@etotheipi and @Infrared, please keep in mind that this thread was posted at the "B" level. Notions of tensors and skew-symmetric maps are likely well above the OP's current level.

Yes but at the same time bilinearity and skew symmetry are key features of the cross product and should be illustrated perhaps with a simple explanation of what they mean. The definition of cross product by itself IMO leaves the question one partially answered.

 

[ChinleShale]

@erocored

If you want to examine the cross product a little further I suggests looking how the cross product can be defined using quaternions. This tells you something about quaternions as well.

 

[Mark44]

Yes but at the same time bilinearity and skew symmetry are key features of the cross product

Not in a thread marked at "B" level.

If you want to examine the cross product a little further I suggests looking how the cross product can be defined using quaternions.

Again, this is a "B" level thread.

 

[wrobel]

I just recall that the cross product is not a vector it is a pseudo vector or the tensor density. It should be taken into account when one chooses a coordinate frames. The standard formulas for the cross product are valid in the positively oriented frames.
For example, the angular velocity $\omega$ of a rigid body is also a pseudo vector. That is why in physically reasonable formulas $\omega$ is appeared just in the cross product. Cross product of $\omega$ and a vector gives a vector.

 

[Mark44]

I just recall that the cross product is not a vector it is a pseudo vector or the tensor density.

The cross product of two vectors in $\mathbb R^3$, as usually understood, is a vector, one that is orthogonal to the two given vectors.

The process that is used to calculate the cross product uses a pseudo-determinant, in which the unit vectors $\hat i, \hat j$, and $\hat k$ appear in the top row. Is that what you're thinking of?

 

[etotheipi]

@Mark44 I think what @wrobel is saying, is that the cross product of two vectors is not itself a vector, because it does not transform like a vector. Instead, it transforms with the additional multiplicative factor of the determinant of the transformation matrix, and it's just a tensor density.

 

[Mark44]

@Mark44 I think what @wrobel is saying, is that the cross product of two vectors is not itself a vector, because it does not transform like a vector. To get the right transformation properties (i.e. to turn it into a vector) you also need to multiply by the determinant of the transformation matrix. But before you do that it's still just a tensor density.

Here's what wikipedia has to say (https://en.wikipedia.org/wiki/Cross_product#:~:text=In mathematics, the cross product or vector product,is denoted by the symbol {displaystyle%20times%20}:

Given two linearly independent vectors a and b, the cross product, a × b (read "a cross b"), is a vector that is perpendicular to both a and b, and thus normal to the plane containing them.

Cross products are typically introduced in intro physics courses and in precalculus mathematics courses -- and there is not usually any mention of tensors.

 

[suremarc]

I mean, I prefer to think of the cross product as a pseudovector, but I guess it’s probably not helpful for OP.

I think the most succinct explanation of the cross product at this level probably comes from the volume formula, as others have said. Namely, that the “signed volume” $V(\mathbf{a,b,c})$ of a parallelepiped with sides $\mathbf{a,b,c}$ is linear in each of its arguments, and that $\mathbf{a\times b}$ is the unique (pseudo)vector such that $V(\mathbf{a,b,c})=(\mathbf{a\times b})\cdot\mathbf{c}.$ While most of us know from linear algebra that in any given basis such a vector exists, it can be verified in Cartesian coordinates.

 

[Mark44]

I mean, I prefer to think of the cross product as a pseudovector, but I guess it’s probably not helpful for OP.

I agree. Several posts that are well beyond B-level have been removed. Since the OP has not returned since his initial post, I'm closing this thread. If he/she has additional questions that haven't been addressed in this thread, I will reopen it.