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erocored
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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?
jedishrfu said:##\vec A \times \vec B = |A||B|sin(\theta)##
I've never seen that before!jedishrfu said:But you missed the vector arrow over the |A||B|sin() expression.
But you got to hang out with Anne Francis!jedishrfu said:but hey I'm only a robot.
jedishrfu said:@etotheipi Glad to have you back but don't you still have some studying to do...
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.."
PeroK said:But you got to hang out with Anne Francis!
Both of the above are incorrect.jedishrfu said:##\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##.
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?etotheipi said:But maybe it's more natural to view the vector product as an element of the space ##\Lambda^2 V##.
If you've been working with differential forms (as I think you have been) you might know more than you realize!etotheipi said: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 said: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?
erocored said: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?
Mark44 said:@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.
Not in a thread marked at "B" level.ChinleShale said:Yes but at the same time bilinearity and skew symmetry are key features of the cross product
Again, this is a "B" level thread.ChinleShale said:If you want to examine the cross product a little further I suggests looking how the cross product can be defined using quaternions.
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.wrobel said:I just recall that the cross product is not a vector it is a pseudo vector or the 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}:etotheipi said:@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.
Cross products are typically introduced in intro physics courses and in precalculus mathematics courses -- and there is not usually any mention of tensors.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.
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.suremarc said:I mean, I prefer to think of the cross product as a pseudovector, but I guess it’s probably not helpful for OP.
The cross product of two vectors is a mathematical operation that results in a vector that is perpendicular to both of the original vectors. It is also known as the vector product or outer product.
The cross product is calculated by taking the determinant of a 3x3 matrix formed by the two vectors and the unit vectors in the x, y, and z directions. The resulting vector is the cross product of the original two vectors.
The cross product can be interpreted geometrically as the area of a parallelogram formed by the two vectors, with the direction of the resulting vector determined by the right-hand rule.
The cross product and the dot product are both mathematical operations involving vectors, but they result in different types of values. The dot product results in a scalar (a single number), while the cross product results in a vector.
The cross product has various applications in physics, engineering, and computer graphics. It is commonly used to calculate torque, magnetic fields, and 3D rotations. It is also used in vector calculus to solve problems in areas such as electromagnetism and fluid dynamics.