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welle
Why does the cross product of two vectors produce a vector which is perpendicularto the plane in which the original two lie?(whenever i go to look it up it is already assumed that it is perpendicular)
Originally posted by welle
Why does the cross product of two vectors produce a vector which is perpendicularto the plane in which the original two lie?(whenever i go to look it up it is already assumed that it is perpendicular)
Originally posted by welle
I don't quite understand how that proves that the resulting vector is perpendicular and not at any other angle
Originally posted by welle
That's true, but you are talking about dot product while my question was about cross product(vector product). Ax B =C
If vectors A and B lie in a plane, why should the resulting vector C be perpendicular to that plane?
Just to muddy the waters even more, I'll add Spivak's general definition:What definition of cross-product are you using?
Originally posted by Welle
I tried taking the dot product of v x u with v as Ambitwistor and NateTG suggested and got it equal to v^(2) u sin(theta) cos(theta), and when i used the component method i still couldn't get it to zero although i probably did it wrong, can someone show me how to get it to zero( i apologize for my misunderstanding)
Originally posted by HallsofIvy
Once again, what is your definition of "cross product"??
One definition of cross product is:
(ai+ bj+ ck)x(ui+ vj+ wk)= (bw-cv)i-(aw-cu)j+(ab-bu)k which is what Ambitwistor was doing. Using that definition the proof is a tedious but direct calculation.
Another definition is the one I gave before:
The cross product of vector u and v is the vector with length equal to length of u times length of v time sine of the angle between u and v, perpendicular to both u and v and directed according to the "right hand rule".
Because you write that, for the dot product of u with uxv, you got
"v^(2) u sin(theta) cos(theta)" you appear to be using the second definition but that includes "perpendicular to both u and v" by definition.
Originally posted by welle
Why does the cross product of two vectors produce a vector which is perpendicularto the plane in which the original two lie?(whenever i go to look it up it is already assumed that it is perpendicular)
The resulting vector defined by the cross product is sometimes called a "pseudovector" because it is not a result of an agent such as force. For basically the same reason, the centripetal acceleration of a mass times the mass is a "pseudoforce" because it is a force that is not the result of a physical entity.
Originally posted by welle
I think that i am still a bit confused.How come a product of two vectors can be in one case scalar and in another a vector?Does the product depend on the nature of the vectors? By which i mean that the product of force and displacemnt vectors will produce a scalar while that of force and moment arm a vector.Then does it make sense to take a dot product or a cross product of any two vectors, and if it doesn't, can one really use both of those techniques to prove something as ambitwistor did?Mathematically ambitwistor's proof is perfect and leaves no doubt, yet i still don't understand whether the proof required a definition, an assumption, or nothing but pure algebra and geometry.If it was an assumption, and had purely physical grounds, is there no way of prooving this assumption?
Originally posted by J. D. Heine,
Is the direction of "preudoforce" dependent on other forces while the force itself is not?
Originally posted by matt grime
It is also called the inner product, and the cross or vector product is called the outer product too.
i think i saw selfAdjoint saying this in some thread somewhere too. that's two people who have called the cross product an outer product.
so perhaps i am wrong.
in my head, it works like this:
an inner product takes two vectors and makes a scalar
a vector product takes two vectors and makes a new vector
and outer product takes two vectors and makes a tensor
an inner product takes two vectors and makes a scalar
a vector product takes two vectors and makes a new vector
and outer product takes two vectors and makes a tensor
well, i would like to establish whether this terminology as i have it is correct (in context).Originally posted by saski
lethe writes:
I'm afraid the terminology depends on context.
nonsense. the inner product on a vector space exists between vectors and vectors (not dual vectors). there is no ambiguity here.The only ambiguity here is that sometimes only certain kinds of vector should have inner products formed with other kinds. Really, vectors should only have inner products formed with vectors from the dual space. When you have a metric defined, you can happily convert vectors to duals, so anything goes.
hmm... the dot product is a special case of an inner product, but they are not the same thing. contraction is also not the same thing. both of those are never applied to, for example, Hilbert spaces. Hilbert spaces have an inner product, not contraction (since there are no indices), and not a dot product (unless it is a finite dimensional Hilbert space)The inner product is also called a dot product or contraction.
nonsense. vector product has nothing to do with orthogonality or basis vectors. consider, for example, the Lie bracket in a Lie algebra. or the matrix multiplication in the algebra of matrices. what have these to do with orthogonal bases? nothing at all. remember, a vector product is a product which is a vector. just like it sounds.Yes, in 3-space with orthonormal basis vectors, i.e. rectangular coordinates.
ahh... now that answers my question. the vector product in a Clifford algebra is sometimes called an outer product.In other cases, we take two vectors and form a bivector. This is generally called a wedge product or exterior product. However, the Clifford algebra community, following Grassmann, call this an outer product.
you bet. i usually call it a tensor product too. that is a good name for a product that produces a tensor (just like vector product is a good name for a product that produces a vector)Mathematicians usually call that a tensor product. However, in computing usage, e.g. Wolfram Mathematica, it's called an outer product - or at least the operation on components is.
nonsense. the inner product on a vector space exists between vectors and vectors (not dual vectors). there is no ambiguity here.
...
sometimes the contraction of a (m,n) rank tensor (with n>0) with a vector is called an inner product in differential geometry, but this terminology is misleading; this is certainly not an inner product space.
hmm... the dot product is a special case of an inner product, but they are not the same thing.
contraction is also not the same thing. both of those are never applied to, for example, Hilbert spaces. Hilbert spaces have an inner product, not contraction (since there are no indices)[...]
nonsense. vector product has nothing to do with orthogonality or basis vectors.
the Lie bracket in a Lie algebra. or the matrix multiplication in the algebra of matrices. what have these to do with orthogonal bases?
i think the Clifford people must be following Clifford, not Grassman... eh?
do you mean to imply that Grassman also called his wedge product an outer product?
Where did you get the impression that orthogonality and metrics are linked, as in the phrase
'requires a metric, ie the definition of orthogonality'
It is perfectly possible to define exterior products without reference to a metric. Indeed, I cannot think of one off hand that uses a metric, but then I'm an algebraist.
Also, I would rather that quote about bra and ket said the physicists and some applied mathematicians used the terminology. The identification of the hilbert space and its dual implies reflexivity, which is not true for a general banach space (or, hilbert spaces I believe) where we also have linear fuctionals, though I don't believe you want to see the generalization of Rietz's Representation Theorem (actually, it isn't really a generalization, as much as the hilbert space version you're used to is a specialization).
contraction refers to what you do when you get rid of an index in tensor index notation by summing over one of them.Originally posted by saski
But I can quote MTW: "Contraction seals off two of the tensor's slots, reducing the rank by two." That has to include the contraction of the tensor product of a contravariant and a covariant vector. If we're not to call that a contraction, what should we call it?
well, i am not a mathematician, so perhaps i shouldn't speak for them, but as far as i can tell, mathematicians do not use bra ket notation at all, because it is extremely sloppy."Mathematicians term <B|A> the inner product of a bra and a ket." Bras are defined as linear functionals operating on kets, i.e. dual vectors. So <B|A> is a contraction by MTW's definition. Howvever, t there's a norm on the Hilbert space allowing one to convert between bras and kets, so <B|A> is also the inner product of |A> and |B>.
when you say "metric", do you mean Riemannian metric?Consider:
[itex] w_i = \epsilon_i_j_k u^j v^k [/itex]
It becomes a vector product only by raising the index on w, which requires a metric, i.e. definition of orthogonality. Or you can write the exterior product:
[itex] u^j v^k - v^j u^k [/itex]
but you need a Hodge star to make a vector out of it, and you need the metric for the Hodge star.
perhaps you should review the definition of Lie algebra.The Lie bracket expresses non-commutation of Lie derivative operators; it's not a simple matter of alternating tensor products, and it's certainly not the same thing as a vector product.
see above. please tell me your definition of vector product. if you define the vector product to be the cross product in R3, then of course anything else will not be.And matrix multiplication is entirely the multiplication of row with column vectors. Again, not a vector product.
I stand by what I said.
yeah, i know what a Clifford algebra is, but thanks.However, I've permitted confusion between that exterior product and the Clifford product.
What the Clifford people do is make a grand basis containing scalar unity, the unit vectors, unit bivectors, unit trivectors, etc; the whole graded sequence of exterior products. Then they define a super-product on the span of that, which they call the "associative" OR "geometric" OR "Clifford" product.
Originally posted by matt grime
- a gentleman only takes bases when he has to.
The cross product of two vectors is a vector that is perpendicular to both of the original vectors. It is also known as the vector product.
The cross product of two vectors can be calculated by taking the product of their magnitudes and the sine of the angle between them.
The perpendicularity of the cross product is significant because it allows us to determine the direction of the resulting vector. The direction is always perpendicular to the plane formed by the two original vectors.
The cross product has many applications in physics, engineering, and computer graphics. It is used to calculate torque, magnetic fields, and 3D rotations, among other things.
Yes, the cross product can be extended to any number of vectors. The resulting vector will still be perpendicular to all of the original vectors and its direction will depend on the order in which the vectors are multiplied.