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Math Help - Tangent Planes

  1. #1
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    Tangent Planes

    The question:

    Show that the curve r(t) = t^2\underline{i} + t\underline{j} + (5t - 4)\underline{k} is normal to the surface 2x^2 + y^2 + 5z^2 = 8 at the point (1, 1, 1).

    My attempt:
    I found the tangent plane for the second surface, which is:
    2x + y + 5z = 8

    I then noticed that the curve is at point (1, 1, 1) when t = 1. I differentiated the curve, substituted t = 1, and got:

    2\underline{i} + \underline{j} + 5\underline{k}

    Which is the same as the tangent plane for the second surface. I'm confused, how can it be normal to the surface if it has the same tangent? Or have I calculated it wrong?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Glitch View Post
    Which is the same as the tangent plane for the second surface. I'm confused, how can it be normal to the surface if it has the same tangent? Or have I calculated it wrong?

    Take into account that (2,1,5) is a perpendicular vector to 2x+y+5z=8 .
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  3. #3
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    Huh, now that I think about it, it is isn't it? >_<

    Thanks.
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