[Blindmath] Vector CrossMultiplication Question

[Dzhovani]

Hi Rick,

   I, J, K in the second example are just common multipliers. They make 
it fancy to simplify the potential computation, but that's all. The 
cross product is just that - product across of the components.

HTH,

Dzhovani


On 12.12.2016 г. 18:02, Rick Thomas via Blindmath wrote:
> Hi David:
>
> So they are just placeholder subscripts and I don’t have to do any decomposition before performing vector cross multiplication, not just scaler multiplication across 2 3-dimensional vectors?
>
> Rick USA
>
>   
>
> From: David Tseng [mailto:davidct1209 at gmail.com]
> Sent: Monday, December 12, 2016 7:52 AM
> To: Blind Math list for those interested in mathematics <blindmath at nfbnet.org>
> Cc: Rick Thomas <ofbgmail at mi.rr.com>
> Subject: Re: [Blindmath] Vector CrossMultiplication Question
>
>   
>
> Hey Rick, the notation is a little weird. i j k are usually used as subscripts as are the numbers 1, 2, 3 in your examples.
>
> The unit vectors in R^3 are:
> (1, 0, 0) (0, 1, 0) (0, 0, 1).
>
> Multiplying a vector with a scaler is just multiplying each component with that scaler. In other words, the result is another vector.
>
> Putting that back into the second formula, you can see you get back the first form by distributing the scaler into each of the unit vectors above.
>
> On Mon, Dec 12, 2016 at 4:08 AM Rick Thomas via Blindmath <blindmath at nfbnet.org <mailto:blindmath at nfbnet.org> > wrote:
>
> Hi:
> I am reviewing Vector Cross Multiplication.
> One article claims that step 1 is to decompose any vector into Unit Form
> using (i,j,k)
> Others seem to use (i,j,k) only as component place holders in their
> equations unless they are assuming readers will know to do the unit
> decomposition before using their formulas.
> One article did not use (i,j,k) in their formula at all.
> Below are 2 solution formulas:
> Can you put into words whether decomposition needs to be performed to
> perform cross multiplication on 2 vectors prior to using them in the formula
> or show an example to explain a solution?
> Note: I use * to denote multiplication.
> First without (i,j,k)
> If the components for vectors A and B are known, then we can express the
> components of their cross product, C = A*B:
> Cx = (Ay*Bz - Az*By)
> Cy = (Az*Bx - Ax*Bz)
> Cz = (Ax*By - Ay*Bx)
> Second Article using (i,j,k)
> To take the cross product of two general vectors, we first decompose the
> vectors using the unit vectors i,j,k.
> Then proceed to distribute the cross product across the sums, using the
> rules to do the cross products between unit vectors.
> We can do this for arbitrary vectors
> u = u1, u2, u3)
>   and
> v = (v1, v2, v3)
>   to get a general formula:
> u = u1i + u2j + u3k
> v = v1i + v2j + v3k
> =
> (
> u1*v2 - u2*v1)k
> +
> (u3*v1 - u1*v3)j
> +
> (u2*v3 - u3*v2)i
> OK, so the above 2 methods look pretty similar but for the use of (i,j,k)
> Can you clear up this confusion for me either in words or by an example of 2
> vectors with component numbers not in unit form via a stepwise solution?
> I have not been able to figure this out in several days of googling.
> Are examples they give just specifying (i,j,k) just using them as place
> holders or do they actually calculate (i,j,k) and multiply the calculations
> by their values in the second example form?
> Rick USA
>
>
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