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Thanks,

Rocky

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Thanks,

Rocky

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Doc Al

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They haveI thought vectors had only length and direction.

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Thx

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Doc Al

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If the vectors A and B have units of length, then C = A X B will literally be a vector representing an area (units of length*length). But in general, the units of the cross product will be the product of the units of A and B. For example: Torque = r X F. r is a position vector with units of length and F is a force with units of force, thus torque has units of length*force. It's still useful to think of the cross-product as an "area" in a more general sense.I quess I don't know why I'd ever want a vector that is proportional to area in the first place.

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HallsofIvy

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One reason we would want "a vector that is proportional to area" is to be able to define the "vector differential of surface area", [itex]d\vec{S}[/itex], normal to the surface and whose length is the differential of surface area.

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tiny-tim

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Hi Rocky! Welcome to PF!… how it can be stated that given C=AXB (all vectors), that C is equivalent to the area of the rectangle formed by A and B in the A,B plane. How can you equate length and area.

AxB is technically the dual of the 2-form (exterior product) A∧B …

we can also make 3-forms and so on (A∧B∧C∧…), and a k-form

the dual of a k-form is an (n-k)-form (where n is the dimension of the space)

and there's an isomorphism between all the k-forms and all their duals (the (n-k)-forms) …

in this case, n = 3, so the vector AxB is the dual of something that is two-dimensional.

(and just to make you even more confused … and I admit this has virtually nothing to do with your question …

AxB technically is a

for that reason, we can never add vectors and pseudovectors. )

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C=AXB (equation should be bold)

C is a pseudovector since you can't have a negative area

The units of C are the combination of the units of A and B as in dist and force in torque

So basically I am thinking that C is a kind of graph of whats going in the the plane that contains the two vectors A and B and creates the parallogram formed by them.

Just a crazy question: Since C is a sort of graph of the changes of the area of the parallelogram, is C a kind of integral of what is no longer an equation but a pair of vectors??

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And why would the vector have to be normal to the plane that contains the two vectors? Is that arbitrary?

Can I now create a new vector calculus rule called the ROCKY PRODUCT that defines a paralellogram fromed by A and B in a plane, but the pseudovector is at 45 degrees to the plane holding the vectors. Or an infinite amount of new rules at differing angles?

Anyone see what I'm driving at?

Thanks

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tiny-tim

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Hi Rocky!

(are you the rocky product of your parents? )

You certainly can …Can I now create a new vector calculus rule called the ROCKY PRODUCT that defines a paralellogram fromed by A and B in a plane, but the pseudovector is at 45 degrees to the plane holding the vectors. Or an infinite amount of new rules at differing angles?

but it wouldn't satisfy certain very useful rules, such as:

(A+B)xC = AxC + BxC

and (AxB).C wouldn't be the volume of the paralellepiped (the wedgy thingy)

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I think there is a world of as of yet untapped inbetween rules that will in the future all of a sudden become markedly useful. Call me the Nostrodamus of Vector Calculus but I believe that there is significance in currently insignificant correlations.

Here's a quick question. If AXB is the perp pseudovector that implies the area of the two vectors A and B in a certain plane for torque let's say. What is the significance of the normal vector C to torque? Cannot the torque magnitude plus the combined units of A and B, convey the same exact information about torque not matter what angle the pseudovector has in relation to the plane of A and B? for example. A=1 meter, B=1 Newton, so C=1 Newton/meter.

I would say that C's relative angle to A and B is insignificant.

Comments?

:)

Rocky

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tiny-tim

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Hi Nostrodamus of Vector Calculus!Call me the Nostrodamus of Vector Calculus …

(btw, i wouldn't go aroundIf AXB is the perp pseudovector that implies the area of the two vectors A and B in a certain plane for torque let's say. What is the significance of the normal vector C to torque? Cannot the torque magnitude plus the combined units of A and B, convey the same exact information about torque not matter what angle the pseudovector has in relation to the plane of A and B? for example. A=1 meter, B=1 Newton, so C=1 Newton/meter.

I would say that C's relative angle to A and B is insignificant.

The significance is that torque

torque isn't just a number, you

the basic torque equation is torque equals rate of change of angular momentum, and both torque and angular momentum are vectors (pseudovectors).

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The Karnac of Vector Calculus (downgrade lol)

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tiny-tim

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The Karnac of Vector Calculus (downgrade lol)

Nope, you've lost me …I will buy that but the cross product seems to imply that torque would neccessarily be perpendicular to the plane of A and B by definition. That's highly unlikely that all torque is applied in that direction so what about torque that is applied at an angle NOT perp to the plane of A and B?

the A and B of torque are the force and the distance to the axis …

with that A and B, torque

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HallsofIvy

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"Have done with it" and just ignore the fact that it gives theWhy is the Cross Product needed in the first place? As a simplification, I would think that if one is presented with two vectors and asked to find the parallelogram formed, it's much simpler to just multiply the two magnitudes and have done with it. Just like in non-vector mathematics.

I answered that before. Distinguishing the normal to a surface is an important ability in applications and that is an important use of cross product. If a surface is given in terms of two parameters, u and v, with "position vector", [itex]\vec{r}(u,v)[/itex], then the derivatives, [itex]\vec{r}_u[/itex] and [itex]\vec{r}_v[/itex], are two vectors in the tangent plane to the surface. The "fundamental vector product", [itex]\vec{r}_u\times\vec{r}_v[/itex] is a vector normal to the surface. It's magnitude, [itex]\left|\vec{r}_u\times\vec{r}_v\right|[/itex] times "dudv", [itex]\left|\vec{r}_u\times\vec{r}_v\right|dudv[/itex] is the "differential of surface area". To integrate a real-valued function over the surface, you write it as [itex]\int\int f(u,v)\left|\vec{r}_u\times\vec{r}_v\right| dudv[/itex]. Even simpler is integrating anAnd why would the vector have to be normal to the plane that contains the two vectors? Is that arbitrary?

Of course you can! Is itCan I now create a new vector calculus rule called the ROCKY PRODUCT that defines a paralellogram fromed by A and B in a plane, but the pseudovector is at 45 degrees to the plane holding the vectors. Or an infinite amount of new rules at differing angles?

Anyone see what I'm driving at?

Thanks

Apropos your post on "torque", you don't see to understand that torque is always measured relative to some

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As far as the torque analogy to the cross product, I'm still not grasping it properly. Could the entire thing be put into a non-mathematical statement for easier digestion? ie

The cross product vector (pseudovector), which is always normal to A and B, relates to the the axis of rotation around which the torque is applied and the magnitude of the cross product vector (pseudovector) is equal to the magnitude of the torque applied in whatever units the vector A and B are set to.

If that's pretty close to being accurate, I would still ask this really crazy question.

I would argue that the area of a parallogram is not A time B sin theta but simply A times B.

Here's my rational.

Vector A has a magnitude and B has a magnitude. Imagine a parallogram. Hold the bottom two points in place and rotate the top two points such that they line up with the bottom two points in the vertical axis. Now you have the same magnitudes just at different angles. And the area is not the actual magnitude's products. So you are saying that by shifting the top two points, you are actually changing the area inside the geometric object? It seems counter-intuitive. If you can explain how it does change the area, then I will believe it.

Thanks.

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Office_Shredder

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Calculating the area of a parallelogram is fairly easy and straightforward; why don't you try to do it?

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HallsofIvy

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As opposed to a Jewish way?:tongue:Railing against things in a gentile way

You can "argue" that 2+ 2= 5 if you like.is the best method for me to understand things I do not understand. It is rather like an exclamation point at the end of a sentence for executing emphasis. :)

As far as the torque analogy to the cross product, I'm still not grasping it properly. Could the entire thing be put into a non-mathematical statement for easier digestion? ie

The cross product vector (pseudovector), which is always normal to A and B, relates to the the axis of rotation around which the torque is applied and the magnitude of the cross product vector (pseudovector) is equal to the magnitude of the torque applied in whatever units the vector A and B are set to.

If that's pretty close to being accurate, I would still ask this really crazy question.

I would argue that the area of a parallogram is not A time B sin theta but simply A times B.

I don't know how to "explain" this to you since you don't seem to want to accept the evidence of your eyes. Instead of rotating so that the top two points "line up with the bottom two points in the vertical axis", rotate so that the lie along the same horizontal line. Do you not see that the area isHere's my rational.

Vector A has a magnitude and B has a magnitude. Imagine a parallogram. Hold the bottom two points in place and rotate the top two points such that they line up with the bottom two points in the vertical axis. Now you have the same magnitudes just at different angles. And the area is not the actual magnitude's products. So you are saying that by shifting the top two points, you are actually changing the area inside the geometric object? It seems counter-intuitive. If you can explain how it does change the area, then I will believe it.

Thanks.

I don't see any way you can "understand" things like the cross product if you do not know basic geometry.

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Tx

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So you would have maximum volume at 90 degree angles from what you say.

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No it is not. The argument is by contradiction, as pointed out in HallsofIvy's reply: let your vectors rotate so that the "top points" drop to coincide with the horizontal vector, giving an area of 0, clearly not the non-zero area as they had when a rectangle. The argument that all rotations of the upright vectors contain the same area is therefore not sound.

Tx

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Yes, I get that as I stated in the second from the last post. thx

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"The length of the cross product = the area of the parallogram with sides A nad B"

So he's say that for example

meters=meters^2

And someone replies and says "Well, he just means the number values are the same!"

I would have to pose the question

Are units not important in many math equations??

One of my profs always used to take off points when no units were used in a math equation.

Thx

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tiny-tim

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hmm … fair question …I just watched an MIT math professor …

Are units not important in many math equations??

One of my profs always used to take off points when no units were used in a math equation.

Thx

well, they're certainly important in

but why would a math prof be bothering with units?

In maths of vector spaces, you usually choose a basis (eg

But to return to physics …

angular momentum is position "cross" velocity,

and it's in units of metres*squared* per second.

angular momentum is position "cross" velocity,

and it's in units of metres

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