Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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)?
 one year ago
 one year ago
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)?
 one year ago
 one year ago

This Question is Closed

littlecatBest ResponseYou've already chosen the best response.0
Visualization of my question
 one year ago

XishemBest 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?
 one year ago

littlecatBest 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
 one year ago

littlecatBest ResponseYou've already chosen the best response.0
also kudos to your superb equation typing skills :)
 one year ago

XishemBest 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.
 one year ago

XishemBest 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.
 one year ago

littlecatBest ResponseYou've already chosen the best response.0
cos θ = sqrt2/2 = sin θ ... but i'm still confused about the LHS!
 one year ago

XishemBest 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.
 one year ago

littlecatBest 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...?
 one year ago

XishemBest 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.
 one year ago

littlecatBest 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
 one year ago

littlecatBest 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
 one year ago

XishemBest 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.
 one year ago

XishemBest 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.\[\hat r \neq a\hat r_x+a\hat r_y\]
 one year ago

XishemBest 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.
 one year ago

littlecatBest 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!
 one year ago

XishemBest 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.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.