[Blindmath] Vector CrossMultiplication Question

[Niels Luithardt]

Hi Rick,

write the cross product as e determinant and use the Laplace formula.

Laplace's formula expresses the determinant of a matrix in terms of
its minors. The minor Mi,j is defined to be the determinant of the
(n−1) × (n−1)-matrix that results from A by removing the ith row and
the jth column.


\vec{a}\times \vec{b}

\vec{a}:=(a_x;a_y;a_z)
\vec{b}:=(b_x;b_y;b_z)
\vec{i}:=(1;0;0)
\vec{j}:=(0;1;0)
\vec{k}:=(0;0;1)

Now, write the cross Product as a determinant and use laplace formula:
Det[\vec{i};a_x;b_x $\vec{j};a_y;b_y$\vec{k};a_z;b_z]
=\vec{i}*det[a_y;b_y$a_z;b_z]-\vec{j}*det[a_x;b_x$
a_z;b_z]+\vec{k}det[a_x;b_x $ a_y;b_y]
=\vec{i}*(a_y*b_z-a_z*b_y)+\vec{j}*(a_z*b_x-a_x*b_z)+\vec{k}*(a_x*b_y -a_y*b_x)
=\vec{i}*c_x+\vec{j}*c_y+\vec{k}*c_z
=\vec{c

I hope it helps

Niels}

2016-12-12 17:22 GMT+01:00, Dzhovani via Blindmath <blindmath at nfbnet.org>:
> 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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