MHB Reflection Matrices: Find Components of R

Ciaran
Messages
71
Reaction score
0
Hi there,

I've got a unit vector u^, arbitrary vector v, and a vector w which is the reflection of v in a line in the direction of u. I have already proved that w= 2 (u^.v)u^ - v. However, the next part of my question asks me to write w= Rv and find the components of the matrix R, taking the components of u^ as (u_1, u_2) and likewise with v. I've done questions like these before but I'm not really sure how to do this one. Any help would be much appreciated!
 
Physics news on Phys.org
Ciaran said:
Hi there,

I've got a unit vector u^, arbitrary vector v, and a vector w which is the reflection of v in a line in the direction of u. I have already proved that w= 2 (u^.v)u^ - v. However, the next part of my question asks me to write w= Rv and find the components of the matrix R, taking the components of u^ as (u_1, u_2) and likewise with v. I've done questions like these before but I'm not really sure how to do this one. Any help would be much appreciated!

Hi Ciaran! Welcome to MHB! :)

I suspect that your formula should be:
$$\mathbf w = \mathbf v - 2(\mathbf{\hat u} \cdot \mathbf v)\mathbf{\hat u}$$

Suppose we simply substitute $\mathbf{\hat u}=(u_1,u_2)$ and $\mathbf v=(v_1,v_2)$ and simplify a bit.
What is then the result? (Wondering)
 
Thanks for your reply- the formula is indeed as I stated in the question; I rechecked my question paper. I think the reason it is different to what you thought it may be is due to the way the vectors are pointing?
When substituting, I got w= 2(u1v1+u2v2) (u1,u2)- (v1,v2). I apologise for the way it is typed; I've still to learn how to use the software!
 
Ciaran said:
When substituting, I got w= 2(u1v1+u2v2) (u1,u2)- (v1,v2).
Now represent this as
\[
\begin{pmatrix}w_1\\w_2\end{pmatrix}=
\begin{pmatrix}
2(u_1v_1+u_2v_2)u_1-v_1\\
2(u_1v_1+u_2v_2)u_2-v_2
\end{pmatrix}=
\begin{pmatrix}
2u_1^2-1&2u_1u_2\\
2u_1u_2 & 2u_2^2-1
\end{pmatrix}
\begin{pmatrix}v_1\\v_2\end{pmatrix}.
\]
 
Ciaran said:
Thanks for your reply- the formula is indeed as I stated in the question; I rechecked my question paper. I think the reason it is different to what you thought it may be is due to the way the vectors are pointing?

Ah, I see it now.
The formula I mentioned is for reflection in a (hyper)space that has $\mathbf{\hat u}$ as its normal vector.
Your formula is for reflection in the line with the direction $\mathbf{\hat u}$.
When substituting, I got w= 2(u1v1+u2v2) (u1,u2)- (v1,v2). I apologise for the way it is typed; I've still to learn how to use the software!

No sweat.
If you're interested, click Reply With Quote on any of our posts to see how it is done.
Basically, it's just a matter of putting your formula between [math]...[/math] or \$...\$ tags.
 
Back
Top