[Blindmath] Vector CrossMultiplication Question

David Tseng davidct1209 at gmail.com
Mon Dec 12 16:27:31 UTC 2016


Hi Rick,

It's all equivalent. The second form decomposes by using vector notation
while the first does it explicitly.

The second form is more in line with something you would see in linear
algebra.

Just so we're clear, i, j, k are not subscripts at all but vectors
themselves.

If you have a vector
a = [x, y, z]

you can separate the vector into its components using unit vectors (i, j, k)

a = x * i + y * j + z * k
 = x * [1 0 0] + y * [0 1 0] + z * [0 0 1]
 = [x 0 0] + [0 y 0] + [0 0 z]
 = [x y z]

Doing this linear separation gets formalized in linear algebra. For
example, the above values of x, y, z can be used to express any vector in
R^3.

As for the cross product, setting x, y, and z to the formulas given before
gets you back to the first formula.


On Mon, Dec 12, 2016 at 8:02 AM, Rick Thomas <ofbgmail at mi.rr.com> 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> 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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