Vector Reflection Across y=2x: Solving with Rotation and Change of Bases

  • Thread starter Thread starter fattycakez
  • Start date Start date
  • Tags Tags
    Reflection Vector
fattycakez
Messages
21
Reaction score
0

Homework Statement


Find the (exact) reflection of the vector v = (5, 1) across the line: y = 2x.
Hint: A sketch of v and the line may suggest an approach.

Homework Equations



The Attempt at a Solution


I found the matrix
-3/5 6/5
4/5 2/5
which seems like it gives the reflection across y=2x

But my question is: is there way to do this by rotating the axes and changing bases? (I'm pretty sure this is what the assignment is asking me to do)
I'm having a hard time visualizing it since no angle is given to put into the rotation equations for R2
i.e.
x'=xcosθ +ysinθ
y'=-xsinθ+ycosθ

Any help is greatly appreciated :)
 
Physics news on Phys.org
fattycakez said:

Homework Statement


Find the (exact) reflection of the vector v = (5, 1) across the line: y = 2x.
Hint: A sketch of v and the line may suggest an approach.

Homework Equations



The Attempt at a Solution


I found the matrix
-3/5 6/5
4/5 2/5
which seems like it gives the reflection across y=2x

But my question is: is there way to do this by rotating the axes and changing bases? (I'm pretty sure this is what the assignment is asking me to do)
I'm having a hard time visualizing it since no angle is given to put into the rotation equations for R2
i.e.
x'=xcosθ +ysinθ
y'=-xsinθ+ycosθ

Any help is greatly appreciated :)
What angle does the vector, <5, 1>, make with the line y = 2x ?
 
SammyS said:
What angle does the vector, <5, 1>, make with the line y = 2x ?
Man I'm slow, it makes an angle of 52.125! When I use that and the (5,1) in the rotation equations it looks like its reflecting in the wrong direction
(4th quadrant rather then second quadrant)
The new vector appears to be at a 90 degree angle with y=2x, do I need another rotation or something like that?
Thanks :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top