# Homework Help: Reflection in a plane

1. Sep 9, 2010

### mjbourquin

Okay so in a problem I have a vector reflecting off a mirror. I have the equation for the normal of the mirror at the coordinates where the vector of light hits the mirror. The normal of the mirror and the vector incident are in the same geometric plane. How do I find the refection of the vector incident in this same plane.

My vector incident is
5i + 10j + 10k
some coordinates in plane are (5,0,0)
angle between normal of mirror and vector incident is about 11.4 degrees
The normal of the mirror is
-i - j - k

equation for plane
-5y + 5z = 0
I think that's right. I have no idea how to find a new vector with this information.

2. Sep 9, 2010

### gabbagabbahey

Well, a 3D vector has 3 components, so to determine an unkown vector you will need 3 linearly independent equations (since there are 3 unkowns that you need to solve for). So, what 3 properties must the reflected vector have? What will be its magnitude? What will be the angle between it and the normal to the plane? What can you say about the cross product of the reflected vector and the normal to the plane?

3. Sep 9, 2010

### mjbourquin

The cross product of the normal to the plane and the reflected vector will be the the product of the two vectors magnitudes since sin(90) = 1. This seems like a lot of work for something you would think would not take so much. Having to use matrices to solve. Doing all of this you would still get the original vector though.

would this work
magnitude must equal original vectors magnitude so
15 = sqrt(x^2 + y^2 + z^2)
225 = X^2 + y^2 + z^2
arccos((v(r) dot product v(o))/(magnitude of v(o) * magnitude of v(r))) = 22.8
207.4 = 5x + 10y + 10z
arccos((v(r) dot product n/(magnitude of v(o) * magnitude of n)) = 11.4
25.46 = -x -y - z

so you have these 3 equations
-5y + 5z = 0 (dot product of normal to place and reflected vector equals 0)
207.4 = 5x + 10y + 10z
25.46 = -x - y - z

4. Sep 9, 2010

### gabbagabbahey

First, the angle between the reflected vector and the normal to the plane/mirror, will not be 90 degrees. Second, the cross product of two vectors results in another vector, not a scalar; yu need to worry about the direction of the resultant aswell ( Compare the cross product of the reflected vector with the normal to the cross product of the incident vector with the normal...are the magnitudes the same? How about the directions?)

Using decimals in your calculations can introduce rounding errors, so its better for you to use exact expression until you get your final result (i.e. use $\cos^{-1}(\frac{-5}{3\sqrt{3}})$ instead of 11.4 degrees). In fact, I don't think 11.4 degrees is correct.

Again, the angle between the normal and the reflected vector is not 90 degrees (Google "reflection" to see how the angle of incidence is related to the angle of reflection), so your first equation is incorrect.