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littlecat
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)?
Visualization of my question
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?
hmm.. if it is the same, why was it not written as \[r⃗ 13=iˆ+jˆ\] ? I attempted to use polar co-ordinates which gave me the answer r^ = a (i^ + j^), this eventually leads to a wrong answer in this superposition question
also kudos to your superb equation typing skills :)
i-hat and j-hat 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 i-hat 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.
As far as where the far RHS of the equation is derived from... Let me see what I can figure out.
cos θ = sqrt2/2 = sin θ ... but i'm still confused about the LHS!
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:dw|I 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.
the figure shows an isosceles triangle.. why is the projection of the hypothenuse r1,3 not equal to the length a...?
The projection would be equal to a.\[r_x=a \hat i; r_y=a\hat j\]I believe.
yes! then r_13 = a(i^ + j^) ? but the answer shows r_13 = sqrt2/2 (i^ + j^) :E
the factor of r may have been lost due to the fact that it is a unit vector... i'm not sure myself
Oh! It is a unit vector, so that would explain why the factor is lost... And for the other part, hmm.
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.\[\hat r \neq a\hat r_x+a\hat r_y\]
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.
OH that actually helps :) I'll think of it as basic polar co-ordinates 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!
Yep. I think that the constant is worked into the unit vector somehow. Unit vectors are really something I should review :P.