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Math Help - reflection of curve

  1. #1
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    Apr 2011
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    reflection of curve

    Hi
    I have the following question:
    there is a plane in R^3 given by:
    2x+z+1 = 0

    Let \psi:R^3 -> R^3 be the reflection along the plane, and let C be a curve in R^3 given by:
    x = t
    y = t^2
    z = t^3

    Find the equation of \psi(C)

    I know the relection equation and I have got down to the following:

    (t, t^2, t^3) - (2/5)(2t + t^2 + 1)(2, 0, 1)

    How do I simplify this down to the x, x and z components?

    Thanks
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by srose34 View Post
    Hi
    I have the following question:
    there is a plane in R^3 given by:
    2x+z+1 = 0

    Let \psi:R^3 -> R^3 be the reflection along the plane, and let C be a curve in R^3 given by:
    x = t
    y = t^2
    z = t^3

    Find the equation of \psi(C)

    I know the relection equation and I have got down to the following:

    (t, t^2, t^3) - (2/5)(2t + t^2 + 1)(2, 0, 1)

    How do I simplify this down to the x, x and z components?

    Thanks
    1. I didn't check if your calculations are OK.

    2. I changed the notation so you can see what is a scalar factor and what is a vector.

    \langle t, t^2, t^3 \rangle - (2/5)(2t + t^2 + 1)\langle2, 0, 1\rangle

    3. You'll get:

    \begin{array}{l}x = t - \frac25 (2t+t^2+1) \cdot 2 \\ y = t^2 - \frac25 (2t+t^2+1) \cdot 0 \\ z = t^3 - \frac25 (2t+t^2+1) \cdot 1\end{array}

    4. Simplify the RHSs.
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  3. #3
    Newbie
    Joined
    Apr 2011
    Posts
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    ok, thanks that helps.
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