Unit Vectors for Polarization and Wave Vector Directions

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
roam
Messages
1,265
Reaction score
12

Homework Statement


I am having difficulty understanding the very first step of the following solved problem (I understand the rest of the solution).

How did they obtain the expressions for ##\hat{n}## (the direction of polarization), and ##\hat{k}## (the unit vector pointing in the direction of the wave vector)? :confused:

problem1.jpg

Homework Equations



##k=\frac{\omega}{\lambda f} = \frac{\omega}{c}=\frac{2 \pi}{\lambda}##

##E(r, t) = E_0 \ cos (k.r - \omega t) \hat{n}##

##B(r,t) = \frac{1}{c} E_0 \ cos (k.r - \omega t) (\hat{k} \times \hat{n})##

The Attempt at a Solution



What technique did they use to find the expression ##\frac{1}{\sqrt{6}} (\hat{x}+2\hat{y}+\hat{z})## for the unit vector perpendicular to ##x+y+z=0## plane?

Likewise, how did they get the expression ##\frac{1}{\sqrt{5}} (\hat{y}-2 \hat{z})## for the unit vector parallel to the y-z plane?

I could not find any explanations in my Linear Algebra textbook. So any explanation would be greatly appreciated.
 

Attachments

  • problem1.jpg
    problem1.jpg
    44.5 KB · Views: 419
on Phys.org
blue_leaf77 said:
Where do you get that solution from?

This was the solution provided by my teacher. I don't understand, where did he get he get the expressions for ##\hat{n}## and ##\hat{k}## from?
 
Well that looks strange to me. If the wavevector should be perpendicular to ##x+y+z=0## plane then this plane must be parallel to the planes of constant phase ##\mathbf{k} \cdot \mathbf{r}=C## with ##C## a constant, in fact this plane is one of them. Which means any plane with equation ##x+y+z=C## is traversed by the beam perpendicularly, and we see the possible unit vector of ##k## that that can form such equation by the dot product with ##\mathbf{r}## must subtend the same angle with all three axes.
But I would like to hear the other's opinion.