Question

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)?

Answer 1

Visualization of my question

Answer 2

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?

Answer 3

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

Answer 4

also kudos to your superb equation typing skills :)

Answer 5

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.

Answer 6

As far as where the far RHS of the equation is derived from... Let me see what I can figure out.

Answer 7

cos θ = sqrt2/2 = sin θ ... but i'm still confused about the LHS!

Answer 8

Alright. I don't understand how the middle part of the equation was derived. It seems to be missing a factor of r.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.