Points, Lines, and Planes in 3D Space Given 3 points find a plane

[yjoung]

Homework Statement

A flashlight located at the origin, (0, 0, 0), shines a beam of light towards a flat mirror. The beam reflects off of the mirror at (8, 4, 1) and then passes through (10, 8, 5). What is the equation of the plane that contains the mirror? Express answer in linear form.

Homework Equations

Basically those pertaining to lines and planes in 3-D space.

The Attempt at a Solution

I've been struggling with this for quite a long time...my attempts won't help much at all since they really have been limited to fiddling with the vectors between points and getting tangled in a mess of dot products and cross products. All in all, I assume that if you know how to solve this problem then it might come to you in an instant..

I simply am stuck.

 

[Dick]

You have three points in the plane, right? Can you use those to find a normal vector to the plane? Showing your work for this part of the problem would be a good idea.

 

[yjoung]

I'm afraid that this isn't really trying to ask for a plane with all those points included.

The only point given, that would be in the mirror plane, should be (8, 4, 1)...at least to my knowledge.

If it contained all those points then I would've been able to solve this.

 

[Dick]

I'm afraid that this isn't really trying to ask for a plane with all those points included.

The only point given, that would be in the mirror plane, should be (8, 4, 1)...at least to my knowledge.

If it contained all those points then I would've been able to solve this.

Good point. Try this, the mirror plane is normal to the bisector of the angle between the incoming and outgoing rays, right?

 

[yjoung]

Well I don't think so...

because the plane of the light beam would just be a light beam stretched out like a plane...and it obviously wouldn't be hitting the mirror at a right angle, since it would just (it = light beam) sent in the opposite direction (<-, this becomes, ->, that).

And there are two light beams, one coming from the origin, and one bouncing off the mirror, so I am just not following you.

If you could insert the few beginning steps you used to solve this problem, that would probably help me understand much more faster than more conceptualization.

 

[yjoung]

Yes, I've gotten that far (from your updated/edited reply)

but I'm horribly stuck on how to get any sensible vector out of the angle (which is 48.19 degrees via Cross Product manipulation).

 

[Dick]

Maybe I'm not being clear. Look at http://www.physicsclassroom.com/class/refln/u13l1c.cfm If you find the average of the incoming and outgoing beam directions you will have a normal to the mirror plane.

 

[Dick]

Yes, I've gotten that far (from your updated/edited reply)

but I'm horribly stuck on how to get any sensible vector out of the angle (which is 48.19 degrees via Cross Product manipulation).

Find two unit vectors describing the incoming and outgoing beam directions. Point them in the same direction and average them (add one to the other and divide by two) to get the angle bisector. Does the picture help? This is hard to describe in words.

 

[yjoung]

ok so, this what I have done,

Vector I (incoming) -> the vector going from (0,0,0) to (8,4,1) -> (8-0, 4-0, 1-0) =

<8,4,1>

Vector O (outgoing) -> the vector going from (8,4,1) to (10,8,5) -> (10 -8, 8-4, 5-1) =

<2,4,4>

So what you meant by "two unit vectors describing the incoming and outgoing beam directions. Point them in the same direction and average them ", was to do the following?

Unit Vector I = <8,4,1> / 9
Unit Vector O = <2,4,4> / 6

(<8,4,1> / 9) + (<2,4,4> / 6) = <11/18, 10/18, 7/18>

and if I solve for it, assuming that what I got is the normal vector to the mirror plane,

<11/18, 10/18, 7/18> *(DP dot) <x-8, y-4, z-1> = 0

it solves for,

(11/18)*(x-8) + (10/18)*(y-4) + (7/18)*(z-1) = 0

11x + 10y + 7z = d-constant,

though, the answer given by my instructor is,

5x - 2y - 5z = 27.

Could you explain why that is?

 

[Dick]

ok so, this what I have done,

Vector I (incoming) -> the vector going from (0,0,0) to (8,4,1) -> (8-0, 4-0, 1-0) =

<8,4,1>

Vector O (outgoing) -> the vector going from (8,4,1) to (10,8,5) -> (10 -8, 8-4, 5-1) =

<2,4,4>

So what you meant by "two unit vectors describing the incoming and outgoing beam directions. Point them in the same direction and average them ", was to do the following?

Unit Vector I = <8,4,1> / 9
Unit Vector O = <2,4,4> / 6

(<8,4,1> / 9) + (<2,4,4> / 6) = <11/18, 10/18, 7/18>

and if I solve for it, assuming that what I got is the normal vector to the mirror plane,

<11/18, 10/18, 7/18> *(DP dot) <x-8, y-4, z-1> = 0

it solves for,

(11/18)*(x-8) + (10/18)*(y-4) + (7/18)*(z-1) = 0

11x + 10y + 7z = d-constant,

though, the answer given by my instructor is,

5x - 2y - 5z = 27.

Could you explain why that is?

Yes. What I meant by "point in the same direction" is either reverse the incoming beam direction so it's pointing outward, and then average or reverse the outgoing. They both have to be pointing in the same direction from the mirror direction before you can average them to get the bisector. Look at the picture again. Do you see why? Try changing <8,4,1>/9 to <-8,-4,-1>/9. You will get a normal direction of <-5,2,5>. I tried it, it's true. Nice work BTW.

 

[yjoung]

wow, ok, thank you so much Dick.

Finally got it.