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Math Help - Reflection of Point onto Plane

  1. #1
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    Reflection of Point onto Plane

    If P(1,-2,4) is reflected in the plane with equation 2x-3y-4z+66=0, determine the coordinates of its image point, P'. (Note that the plane 2x-3y-4z+66=0 is the right bisector of the line joining P(1,-2,4) with its image).
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  2. #2
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    Hi skeske1234

    Let : A is the point on the plane that is perpendicular to P, then \vec {AP} = \vec {P'A}

    So, you need to find A first.
    That's what I can come up with. Maybe there are other shorter methods.

    Thanks
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  3. #3
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    how do i find A? could you show me plz?
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  4. #4
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    Hi skeske1234

    A is the intersection between the line L (that passes through A and P) and the plane so you need to find the eqution of line L.

    I'm pretty sure you've learnt about equation of line. Can you fiind it? Then, have you learnt to find the intersection between line and plane ?
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  5. #5
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    so am i supposed to find the intersection between the line and the plane or am i supposed to find the equation of the line.. sorry confused here

    also,
    i'm actually not sure how to find the equation of the line if I am only given a point

    But this is my attempt or guess so far:

    l is the eqtn of the line i am looking for right now (right?)
    l: (x,y,z)= (a,b,c) + t(2,-3,-4).
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  6. #6
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    Hi skeske1234

    You need to find the equation of the line first, then find the intersection between the line and plane.

    l is the eqtn of the line i am looking for right now (right?)
    l: (x,y,z)= (a,b,c) + t(2,-3,-4).
    Yes, you're on the right track. (a,b,c) is the point that lies on the line L. Do you know what the point is ?
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  7. #7
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    no, i don't know where to go from here.. do i have the direction vector correct?
    can you show me the next step? please
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  8. #8
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    Hi skeske1234

    Yes, the direction vector is correct (because the line is parallel to the normal of the plane). The line L passes through point P so (a,b,c) = (1,-2,4)

    Next step : find A
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  9. #9
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    l1: (x,y,z)=(1,-2,4)+t(2,-3,-4)
    l2: (x,y,z)=(a,b,c)+t(2,-3,-4)

    so when i made them equal to each other with their parametric equations, i got A(1,-2,4).
    This is the same as l1.. this doesn't make sense. What did I do wrong? can you show me how you get the correct answer?
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  10. #10
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    No

    There is only one equation of line, which is L : (x,y,z)=(1,-2,4)+t(2,-3,-4). This is the line that passes through A and P and is perpendicular to plane 2x - 3y -4z + 66 = 0.

    Now you need to find the point of intersection between the line and plane. Do you get what I mean ?
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  11. #11
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    so I got
    x=-3
    y=4
    z=12
    as my POI, this answer doesn't match the back of the book. Here's my work:

    x=1+2t
    y=-2-3t
    z=4-4t

    2(1+2t)-3(-2-3t)-4(4-4t)+66=0
    t=-2

    x=1+2(-2)
    y=-2-3(-2)
    z=4-4(-2)

    (-3,4,12) << does not match the answer they have in the textbook, where did i go wrong here?
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  12. #12
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    Songoku, their answer is (-7,10,20) if that helps..
    compared to my (-3,4,12) this is wrong.


    hmm... is your method wrong or is the back of the book wrong?
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  13. #13
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    Hi skeske1234

    What you found is point A, the point on the plane that is perpendicular to P, not the reflection. Your answer for A is right.

    Now, final step : read post #2
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  14. #14
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    Quote Originally Posted by songoku View Post
    Hi skeske1234

    A is the intersection between the line L (that passes through A and P) and the plane so you need to find the eqution of line L.

    I'm pretty sure you've learnt about equation of line. Can you fiind it? Then, have you learnt to find the intersection between line and plane ?


    What is the next step? I found the eqtn of the line and the intersection already. Can you be more specific?


    EDIT:

    Nevermind, I got the answer songoku, thanks for your help
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