Derive cross product from dot product

[persia7]

can you show me derive cross product from dot product?

 

[arildno]

No.
They are differently defined operations.

 

[mathman]

Cross product is defined only for 3 dimensional vectors. Dot product is defined for vectors in any dimension, including infinite (Hilbert space).

 

[persia7]

im sorry for u
look at this
http://heaveninthebackyard.blogspot.co.uk/2011/12/derivation-of-cross-product-formula.html

 

[voko]

So what is your question, really?

 

[persia7]

why did u say what don't know

 

[HallsofIvy]

That site derives part of the formula for the calculation of the cross product from the condition that the cross product of vectors u and v is perpendicular to both. The rest requires the condition that the length of the cross product be [itex]|u||v| sin(\theta)[/itex].

It looks to me like a fairly detailed (and, so, tedious!) derivation. Do you have a specific question about it? Where do you have difficulty?

 

[arildno]

why did u say what don't know

It seems you are not quite familiar with the freedom we have in what we choose to DEFINE, and what is then to be DERIVED.

For example:
From the definition of the cross product, we then DERIVE the orthogonality property.

But, if we, as the the blogger does, CHOOSES as a condition how to derive a vector orthogonal to two others, you will find a vector PROPORTIONAL to the standardly defined cross product.

If we place a further condition on the vector we seek, that its magnitude should be the area spanned by the two others, then you no longer have mere proportionality of the vector you'll find.
---
However, and here's your flaw:
You seem to think the procedure outlined in the blog post is in some sense TRUER than definitions. They're not.

 

[persia7]

u define cross product for what?

 

[arildno]

Well, then it is much easier to derive additional properties, such as distributivity, anticommutativity and non-associacivity of the cross product.

Quite simply:
It is very practical to define it in the usual way, and several of its more non-intuitive properties are more readily seen then.

 

[arildno]

I would like to emphasize, though, that being able start out at an Other End of maths, and get over to a more familiar place has its own value.

It is a bit about developing a flexibility in your mind, and the blogger had a nice little post on how to start out geometrically and fiddle out the correct formula for the dot product.

 

[persia7]

blogger has shown how u can achieve a formula first before know the definition

 

[arildno]

blogger has shown how u can achieve a formula first before know the definition

And? So what?

We tend to use clever, time-honed definitions in maths, which are poised to develop more interesting relationships than already known ones.

 

[D H]

t is a bit about developing a flexibility in your mind, and the blogger had a nice little post on how to start out geometrically and fiddle out the correct formula for the dot product.

I found that site rather ugly. The analysis was overly long and inelegant, and left hand rule? Please!

blogger has shown how u can achieve a formula first before know the definition

Read your private messages. Stop using text speech.

The blogger did not do that. The blogger started with the definition of the cross product as ##a \times b = |a|\,|b|\sin\theta\,\hat n## and showed that this was equivalent to the alternate definition. He did so in a very roundabout manner, and (YECH!) he used a left handed rule and left handed coordinate system in doing so.

 

[ZeroPivot]

cross product only works in 3d space its used for mechanics. dot product works in all dimensions.

HOWEVER you can have a 3d dot product which is the link you showed.

 

[D H]

persia7, read your private messages. Now.