vector and breaking out terms

[Mimmo]

u = (2x/Sqrt[xyz], 2y/Sqrt[xyz],2z/Sqrt[xyz])
v = (1, 1, 1)

to check if u || v then I simply take the cross product and check if it equals 0. But to make the calculation easier "2" can be broken out of "u". But can I also break out "Sqrt[xyz]" without changing anthing? i.e. is it generally OK to instead of taking
(1, 1, 1)x(2x/Sqrt[xyz], 2y/Sqrt[xyz],2z/Sqrt[xyz]) taking (1, 1, 1)x(2x, 2y,2z). Basically, can you break out terms containing variables and by taking the cross product expect it to give you the correct answer?

 

[soroban]

Hello, Mimmo!

\(\displaystyle \vec{u}\;=\;\left\langle \frac{2x}{\sqrt{xyz}}.\;\frac{2y}{\sqrt{xyz}},\;\frac{2z}{\sqrt{xyz}}\right\rangle\;\;\;\;\vec{v} \;=\;\langle1,\,1,\,1\rangle\)

to check if \(\displaystyle \vec{u}\,\parallel\,\vec{v}\), I simply take the cross product and check if it equals 0. . . . correct
But to make the calculation easier, "2" can be broken out of "u". . . . right
But can I also break out \(\displaystyle \sqrt{xyz}\) without changing anything? . . . Yes, you can!

i.e. is it generally OK to instead of taking: \(\displaystyle \,\langle1,\,1,\,1\rangle\,\times\,\left\langle\frac{2x}{\sqrt{xyz}},\,\frac{2y}{\sqrt{xyz}},\,\frac{2z}{\sqrt{xyz}\right\rangle\)
\(\displaystyle \;\;\)taking \(\displaystyle \,\langle1,\,1,\,1\rangle\,\times\,\langle2x,\,2y,\,2z\rangle\)

Basically, can you break out terms containing variables and by taking the cross product,
expect it to give you the correct answer? \(\displaystyle \;\) . . . Yes!

If the components of a vector have common factors, they can be factored out
\(\displaystyle \;\;\)whether they are constants or variables.

We can write: \(\displaystyle \,\vec{u}\;=\;\frac{2}{\sqrt{xyz}}\,\langle x,\,y,\,z\rangle\)
\(\displaystyle \;\;\)But I recommend that you don't discard the common factor.