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electrokidBest ResponseYou've already chosen the best response.3
what does that "*" represent?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
* is equal to \(\times \)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
no.., it just times
 one year ago

niksvaBest ResponseYou've already chosen the best response.0
@gerryliyana solve for RHS first by using the values of dimensional vectors given in question
 one year ago

electrokidBest ResponseYou've already chosen the best response.3
\[\vec{a}\times\vec{b}=\left\begin{matrix} \hat{x}&\hat{y}&\hat{z}\\ u&v&0\\ x&y&0 \end{matrix}\right =\hat{z}(uyvx)\\ \vec{a}\cdot\vec{b}=ux+vy\\ ab = (u+iv)(x+iy)=ux+i(vx)+i(uy)+(i^2)(xy) \]
 one year ago

mykoBest ResponseYou've already chosen the best response.0
are you sure it's right? Because: a.b=ux+vy iz(axb)=i(uyxv) and a*b=uxvy+i(uy+vx) so it is not equal: uxvy+i(uy+vx) =/= ux+vy+i(uyxv)
 one year ago

electrokidBest ResponseYou've already chosen the best response.3
\[ab=(uyvx)+i(vx+uy)\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
its same with my answer @myko
 one year ago

electrokidBest ResponseYou've already chosen the best response.3
\[ab=\hat{z}\cdot(\vec{a}\times\vec{b})+i(\vec{a}\cdot\vec{b})\]
 one year ago

mykoBest ResponseYou've already chosen the best response.0
vx+uy =/= a.b @electrokid
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
so what the right answer??
 one year ago

electrokidBest ResponseYou've already chosen the best response.3
that "*" in the question must be a complex conjugate
 one year ago

electrokidBest ResponseYou've already chosen the best response.3
then it seems you might have it right.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
good idea!! oh my bad...,
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
yeah,"*" means conjugate
 one year ago

electrokidBest ResponseYou've already chosen the best response.3
always make sure that people are talking the same math language
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
thank you @electrokid
 one year ago

niksvaBest ResponseYou've already chosen the best response.0
@electrokid if "*" is complex conjugate then it will be rightly solved
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
i got it now.., thank you guys :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
by the way.., if all of you have a facebook., add me on facebook http://www.facebook.com/gerry.resmiliyana thank you so much!
 one year ago
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