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Math Help - geometry, parallel vectors...

  1. #1
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    geometry, parallel vectors...

    QUESTION:

    Show that the tangent planes of the unit sphere at the points (x,y,0) are parallel to the z axis.

    MY WORK:

    I get the tangent planes as x^2 + y^2 = 1 ...

    in R3 that's the vector (cos(t), sin(t), z) right?

    And the z axis is the vector (0,0,z)

    so the dot product is z^2

    And the product of the magnitudes is sqrt(1 + z^2)*z

    cos(theta) = z/sqrt(1+z^2) ??

    But I need cos(theta) to be 1 for those vectors to be parallel...

    HELP PLEASE! What have I done?
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  2. #2
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    Quote Originally Posted by RanDom View Post
    QUESTION:

    Show that the tangent planes of the unit sphere at the points (x,y,0) are parallel to the z axis.

    MY WORK:

    I get the tangent planes as x^2 + y^2 = 1 ...

    in R3 that's the vector (cos(t), sin(t), z) right?

    And the z axis is the vector (0,0,z)

    so the dot product is z^2

    And the product of the magnitudes is sqrt(1 + z^2)*z

    cos(theta) = z/sqrt(1+z^2) ??

    But I need cos(theta) to be 1 for those vectors to be parallel...

    HELP PLEASE! What have I done?
    x^2 + y^2 = 1 is definitely not the equation of a plane. I've barely glanced at this question but from my brief glance I think you'd be well advised to consider the coordinates of the points as (x_1, \, y_ 1, \, 0) rather than (x,y,0) .......
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  3. #3
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    OK, so...

    the unit sphere is x^2 + y^2 + z^2 = 1

    f(x,y,z) = x^2 + y^2 + z^2 - 1

    At a point P = (p_1,p_2,p_3) the tangent plane is

    2p_1(x-p_1) + 2p_2(y-p_2) + 2p_3(z-p_3) = 0

    Cancel a factor of 2 and use p_1^2 + p_2^2 + p_3^2 = 1

    p_1x + p_2y + p_3 = 1

    for the point (x_1,y_1,0) that's x_1x + y_1y = 1

    ... and then?

    I'm really lost
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  4. #4
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    Quote Originally Posted by RanDom View Post


    OK, so...

    the unit sphere is x^2 + y^2 + z^2 = 1

    f(x,y,z) = x^2 + y^2 + z^2 - 1

    At a point P = (p_1,p_2,p_3) the tangent plane is

    2p_1(x-p_1) + 2p_2(y-p_2) + 2p_3(z-p_3) = 0

    Cancel a factor of 2 and use p_1^2 + p_2^2 + p_3^2 = 1

    p_1x + p_2y + p_3 {\color{red}z} = 1 Mr F adds a small correction in red.

    for the point (x_1,y_1,0) that's x_1x + y_1y = 1

    ... and then?

    [snip]
    And then you're nearly done. A normal to the plane x_1x + y_1y = 1 is x_1 \, i + y_1 \, j ....
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  5. #5
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    ah, which is perpendicular to (0,0,1), making the plane parallel... thanks
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