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What is a reflection

  1. Jan 1, 2007 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    3. The attempt at a solution

    Well im 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: Jan 1, 2007
  2. jcsd
  3. Jan 1, 2007 #2


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    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.
  4. Jan 2, 2007 #3


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    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: Jan 2, 2007
  5. Jan 2, 2007 #4
    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

    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
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