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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.

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 :)

fatty cakez
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1 Answers1

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The matrix $$ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ is clearly a reflection in the $x$-axis. Therefore what we want to do is map $y=2x$ to the $x$-axis with a rotation, reflect, and then undo the rotation. Suppose our line has angle $\tan{\theta}$ to the $x$-axis (here, $\tan{\theta}=2$, from which we can find cosine and sine using $\cos^2+\sin^2=1$ and $\cos^2 = 1/(1+\tan^2)$, as $\cos{\theta}=1/\sqrt{5}$, $\sin{\theta}=2/\sqrt{5}$ ). Then a rotation that sends $y\cos{\theta}=x\sin{\theta}$ to $x=0$ is $$ \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{pmatrix}. $$ Then the whole reflection is given by $$ \begin{pmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{pmatrix}, $$ which when you multiply out and simplify gives $$ \begin{pmatrix} \cos{2\theta} & \sin{2\theta} \\ \sin{2\theta} & -\cos{2\theta} \end{pmatrix} $$ Putting $\theta=\arctan{2}$ and re-expanding should give $$ \begin{pmatrix} -3/5 & 4/5 \\ 4/5 & 3/5 \end{pmatrix}.$$

Chappers
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  • thanks man, is there any particular way you figured out that third matrix? – fatty cakez May 05 '15 at 01:03
  • The third matrix? – Chappers May 05 '15 at 01:03
  • "Then a rotation that sends ycosθ=xsinθ to x=0 is:" sorry I haven't learned how to format properly on this site yet – fatty cakez May 05 '15 at 01:06
  • Ah, well, you can look at what it does to the line and its normal: one is sent to $(1,0)$, the other to $(0,1)$. It is then easy to solve the equations to find the correct matrix. (And you have listed the rotation matrix in your question, so I assumed you knew how that worked.) – Chappers May 05 '15 at 01:09
  • @fattycakez That second matrix is a common matrix that I guess you should know – Skeleton Bow Dec 20 '16 at 08:49