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littlecat
Group Title
Charge q1 is at origin and charge q3 is at (a,a).
Why is unit vector r1,3 written as r1,3 = cosθi + sinθ j = sqrt2/2 (i + j)?
 2 years ago
 2 years ago
littlecat Group Title
Charge q1 is at origin and charge q3 is at (a,a). Why is unit vector r1,3 written as r1,3 = cosθi + sinθ j = sqrt2/2 (i + j)?
 2 years ago
 2 years ago

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littlecat Group TitleBest ResponseYou've already chosen the best response.0
Visualization of my question
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Writing...\[\vec{r}_{13} =\cos \theta \hat i + \sin \theta \hat j\]is the same thing as writing...\[\vec r_{13}=r_x\hat i +r_y\hat j\]Can you see where this relationship is derived now?
 2 years ago

littlecat Group TitleBest ResponseYou've already chosen the best response.0
hmm.. if it is the same, why was it not written as \[r⃗ 13=iˆ+jˆ\] ? I attempted to use polar coordinates which gave me the answer r^ = a (i^ + j^), this eventually leads to a wrong answer in this superposition question
 2 years ago

littlecat Group TitleBest ResponseYou've already chosen the best response.0
also kudos to your superb equation typing skills :)
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
ihat and jhat only give an indication of the direction of the vector. By writing...\[\vec r_{13}=\hat i+\hat j\]You are implying that \[\vec r_{13}=1\hat i+1\hat j\]Which is not the case. The ihat is just a unit vector, only implying direction. If you project the vector r_13 onto the x and y axes and break it down into its component vectors you can see where the expression it gives is derived from.
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
As far as where the far RHS of the equation is derived from... Let me see what I can figure out.
 2 years ago

littlecat Group TitleBest ResponseYou've already chosen the best response.0
cos θ = sqrt2/2 = sin θ ... but i'm still confused about the LHS!
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Alright. I don't understand how the middle part of the equation was derived. It seems to be missing a factor of r.dw:1345972919614:dwI would think that it would be, then...\[\vec r_{13}=\vec r_{13}\cos \theta \hat i+\vec r_{13}\sin \theta \hat j\]I'm not sure where that r_13 factor is lost.
 2 years ago

littlecat Group TitleBest ResponseYou've already chosen the best response.0
the figure shows an isosceles triangle.. why is the projection of the hypothenuse r1,3 not equal to the length a...?
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
The projection would be equal to a.\[r_x=a \hat i; r_y=a\hat j\]I believe.
 2 years ago

littlecat Group TitleBest ResponseYou've already chosen the best response.0
yes! then r_13 = a(i^ + j^) ? but the answer shows r_13 = sqrt2/2 (i^ + j^) :E
 2 years ago

littlecat Group TitleBest ResponseYou've already chosen the best response.0
the factor of r may have been lost due to the fact that it is a unit vector... i'm not sure myself
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Oh! It is a unit vector, so that would explain why the factor is lost... And for the other part, hmm.
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
I think the issue is that you can't express a unit vector in terms of a constant. Since r is a unit vector and not a standard vector, the following aren't true.\[\hat r \neq a\hat r_x+a\hat r_y\]
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
If r were a standard vector, it would be true, but since it's a unit vector, things work out differently. My understanding of unit vectors is shaky, so, that's about as much as I can tell from it.
 2 years ago

littlecat Group TitleBest ResponseYou've already chosen the best response.0
OH that actually helps :) I'll think of it as basic polar coordinates x = rcosθ = r sqrt2/2 y = rsinθ = r sqrt 2/2 without the factor r but keeping sqrt2/2 since it is referenced to θ THANKS!
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Yep. I think that the constant is worked into the unit vector somehow. Unit vectors are really something I should review :P.
 2 years ago
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