# Laser reflecting off a mirror to a point on a plane

• Nov 17th 2009, 08:33 AM
Laser reflecting off a mirror to a point on a plane
The problem is;
Tim is pointing a lazer beam from the point A in the direction u towards a mirror whose shape is given by G(x,y,z)=0. After reflecting from the mirror, where will the beam show up on the plane X=10
A=(2,8,3) u=<1,-3,0> and G(x,y,z) = X^2 + Y^2 - Z^2 -11

I have a few ideas on how to do it but i keep running into walls with them, any help would be appreciated
• Nov 17th 2009, 06:10 PM
shawsend
Hey. I'm not sure. I went through it quick but this is a start:

First find where the light beam hits the surface. If we have a point $\displaystyle (p_x,p_y,p_z)$ which is the green point in the picture and where the light beam starts, and a vector direction $\displaystyle <a_x,a_y,a_z>$, then a parametric representation of the line is $\displaystyle \{p_x+a_x t, p_y+a_y t,p_z+a_z t\}$. That line is the blue line in the figure. It's easy then to find where it hits the surface $\displaystyle x^2+y^2-z^2-11=0$ as we just substitute the parametric line into that equation and I get $\displaystyle t=2$. Then it hits at the point $\displaystyle (4,2,3)$. Now we need to calculate the normal at that point via $\displaystyle \nabla f\large]_p=<8,4,-6>$. That's the red line. Then I'd calculate the angle between the blue and red which I calculated as $\displaystyle \pi-\arccos\left(\frac{-4}{\sqrt{10} \sqrt{64+16+36}}\right)$ which is almost 90 degrees.

Then the reflected line will have the same angle and follow the plane of the incident and normal (shown as the hatched surface). Then we'd need to parameterize the reflected line and then compute a t at which x=10 to find where it hits the plane x=10.
• Nov 17th 2009, 08:08 PM
I was able to get as far as that after going through it with a group a classmates, the part i am now stuck at is finding the equation of the reflected line, i know that if i can find it i can find where it intersects the X=10 plane. thanks for you help so far
• Nov 18th 2009, 02:59 AM
shawsend
I corrected my work from an earlier attempt. Still a little unsure and thinking about a 500 million space walk but I digress. Here goes:

This is what I have so far:

I can calculate the vector angle between the incident light ray and the normal of the surface (the blue and red lines) at the impact point of $\displaystyle (4,2,3)$:

$\displaystyle \cos(t)=\frac{<-1,3,0>.<8,4,-6>}{|-1,3,0| |8,4,-6|}=\sqrt{2/145}$

Now, let the reflection ray be at the point $\displaystyle (10,y,z)$. Then it will satisfy the following two equations:

$\displaystyle -18(10-4)-6(y-2)-28(z-3)=0$

$\displaystyle \frac{<10,y,z>.<8,4,-6>}{|10,y,z| |8,4,-6|}=\sqrt{2/145}$

and the (numerical) solution is $\displaystyle (10,-11.9,2.12)$

Do you know what the correct answer is?
• Nov 18th 2009, 07:18 AM
Quote:

Originally Posted by shawsend
I corrected my work from an earlier attempt. Still a little unsure and thinking about a 500 million space walk but I digress. Here goes:

This is what I have so far:

I can calculate the vector angle between the incident light ray and the normal of the surface (the blue and red lines) at the impact point of $\displaystyle (4,2,3)$:

$\displaystyle \cos(t)=\frac{<-1,3,0>.<8,4,-6>}{|-1,3,0| |8,4,6|}=\sqrt{2/145}$

Now, let the reflection ray be at the point $\displaystyle (10,y,z)$. Then it will satisfy the following two equations:

$\displaystyle -18(10-4)-6(y-2)-28(z-3)=0$

$\displaystyle \frac{<10,y,z>.<8,4,6>}{|10,y,z| |8,4,-6|}=\sqrt{2/145}$

and the (numerical) solution is $\displaystyle (10,-11.9,2.12)$

Do you know what the correct answer is?

I understand what you did, but could you show me the method you used to find the Y or Z, i know if you have one you can find the other but im unsure as to how to find one of them
• Nov 18th 2009, 09:27 AM
shawsend
You can do that right? I used Mathematica but we could do it manually. First write out the vector angle formula:

$\displaystyle \frac{80+4y-6z}{\sqrt{100+y^+z^2}\sqrt{64+16+36}}=\sqrt{2/145}$

Simplify that to obtain the equation:

$\displaystyle 145(40+2y-3z)^2=58(100+y^2+z^2)$

Expand that and substitute from the other one $\displaystyle y=\frac{-14z-6}{3}$.

You'll obtain a quadratic in z giving two values but only one results in a vector angle that is equal to the vector angle between the surface normal and incident light beam which is the one I chose.

I still have some doubts about this answer however although in the plot below (click on it to get a larger picture), it does appear to be reflecting about right. Would be nice if we could verify this answer.
• Nov 18th 2009, 06:35 PM