I am reading "history of vector analysis":(adsbygoogle = window.adsbygoogle || []).push({});

https://www.amazon.com/History-Vect...=sr_1_3?ie=UTF8&s=books&qid=1256515819&sr=8-3

We have two quaternions [tex]\alpha , \alpha' [/tex] with "scalar" component equal to zero, these are their expressions:

[tex]\alpha' = x' \vec{i} +y' \vec{j} + z' \vec{k}[/tex]

[tex]\alpha = x \vec{i} +y \vec{j} + z \vec{k}[/tex]

with i,j and k fulfilling these known relations :

ij= k , jk=i , ik= -j ... and so on.

Then Hamilton wrote:

S(scalar, which is our -dot product) of [tex]\alpha*\alpha' = -(xx' +yy'+zz') [/tex] and

V(vector, which is our cross product) of [tex]\alpha*\alpha'[/tex] = i(yz'-zy') -j(zx' -xz') + k(xy'-yx')

Well, let's now compare this equations with the equations of our modern vector calculus, for example, if we have the vector position R of a particle, expressed in the base of the vectors i,j and k (orthogonal unit basis vectors) :

R= xi +yj +zk

R' =x'i + y'j +z'k

RR'=xx'+yy'+zz'

and R x R' =i(yz'-zy') -j(zx'-xz') + k(xy'-yx')

But now it is clear that we are using a basis that is conmutative , because ij=ji .

I don't understand this, if our vector calculus started with the calculations of Hamilton cited above ¿why can we use basis vectors that are conmutative? .

Perhaps you think that is a stupid question, but I would like to understand exactly what is the relation between the vectors that I use in my physics classes and these numbers discovered by Hamilton.

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# Precise relation between quaternion and vectors

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