Understanding Reflection in 3D Space: Finding Point Q | Homework Equations

  • Thread starter Thread starter canadian_beef
  • Start date Start date
  • Tags Tags
    Reflection
Click For Summary

Homework Help Overview

The discussion revolves around finding the coordinates of point Q, which is the reflection of point P(-1,3,4) across the plane defined by the equation 2x-y+z=1. Participants are exploring the concept of reflection in three-dimensional space.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the geometric interpretation of reflection, noting that Q is on the opposite side of the plane at an equal distance from it as P. There are attempts to derive the parametric equations of the line perpendicular to the plane and to find the intersection point with the plane.

Discussion Status

Some participants have provided guidance on how to find the reflection point, including the method of using parametric equations and identifying the intersection with the plane. There is an acknowledgment of confusion regarding the next steps after finding the intersection point.

Contextual Notes

There is a discrepancy in the plane equation mentioned by the original poster (2x-y+z=1) and the one referenced later (2x-y+z=11), which may affect the discussion. Participants are also navigating the challenge of understanding the reflection concept without seeking direct solutions.

canadian_beef
Messages
12
Reaction score
0

Homework Statement



The point Q is the reflection of P(-1,3,4), in the plane with equation 2x-y+z=1.

Determine the coordinates of Q

Homework Equations


The Attempt at a Solution



Well I am not really looking for a solution. I just want to know what I am trying to find. What is a reflection in three space. Where is point Q
 
Last edited:
Physics news on Phys.org
Q is on the opposite side of the plane, at the same distance from the plane as P.

There's only one straight line from P to the plane that is perpendicular to the plane. Follow the continuation of that line (through the plane) until you are as far away from the plane as P. That point is Q.
 
What Fredrik said. And it is called a reflection because that is how your reflection in a mirror appears: your reflection appears to be on the other side of the mirror in such a way that the mirror is a perpendicular bisector of any straight line from a point on your body to the reflection of that point.

Find the parametric equations of the line through P perpendicular to the plane (that's the easy part and it is very easy to choose the parameter so that P corresponds to parameter equal to 0. Find the parameter where the line intersects the plane (solve a single linear equation). If P corresponds to parameter 0, then Q will have parameter equal to twice that of the point of intersection.
 
Last edited by a moderator:
Thanks for the help so far

This is what I have done so far

I know that the plane is 2x-y+z=11
Therefore all line perpindicular to that plane have a direction of (2,-1,1)

The parametric equation of the line which passes through point P and hits the plane are
x=-1+2t
y=3-t
z=4+t

I substituted those values into the equation of the plane (2x-y+z=11)
and got t=2

I plugged t=2 back into the parametric equations and got the intersection point between between the line and the plane to be (3,1,6)

Now I am a little confused on what to do from here.

What i guessed to do is subtract the point (-1,3,4) from (3,1,6), and got (4,-2,2,) and i added that to (3,1,6) to get (7,-1,8)
if this is right why, and if it is wrong, how do i approach this..thanks
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K