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Math Help - Point on paraboloid at which the tangent plane is parallel to plane

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    Point on paraboloid at which the tangent plane is parallel to plane

    Find pt on paraboloid x = 5y^2 + 7z^2, if it exists, at which the tangent plane is parallel to plane -x + y + z = 3.

    Not completely sure how to approach this problem. I'm pretty sure it involves the gradient, so I set f(x,y,z) = x - 5y^2 - 7z^2 and found that gradient which was \nabla f = i - 10yj - 14zk.

    Do I need the equation for the tangent plane of the paraboid? And are -1, 1, 1 the directional numbers for the plane -x + y + z = 3?
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    Re: Point on paraboloid at which the tangent plane is parallel to plane

    Quote Originally Posted by deezy View Post
    Find pt on paraboloid x = 5y^2 + 7z^2, if it exists, at which the tangent plane is parallel to plane -x + y + z = 3.

    Not completely sure how to approach this problem. I'm pretty sure it involves the gradient, so I set f(x,y,z) = x - 5y^2 - 7z^2 and found that gradient which was \nabla f = i - 10yj - 14zk.

    Do I need the equation for the tangent plane of the paraboid? And are -1, 1, 1 the directional numbers for the plane -x + y + z = 3?
    Two planes are parallel if their normal vectors are parallel.

    The normal vector of the plane is - \mathbf{i}+ \mathbf{j}+ \mathbf{k}

    or  \mathbf{i}- \mathbf{j}- \mathbf{k}

    Now just set this equal to the normal vector of the surface that you have found and find what the variables have to be.

     \nabla f = \mathbf{i} - 10y\mathbf{j} - 14z \mathbf{k}
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    Re: Point on paraboloid at which the tangent plane is parallel to plane

    Ok, here's what I've done:

    -10y = 1
    -14z = 1
    y = -1/10
    z = -1/14

    I got confused at this point. I wasn't sure what I do with the x. \nabla f = i -(10y)j-(14z)k. There wasn't an x in the equation, and 1 \neq -1, so does the point exist?

    And is there a way to check, if I found a point, that it was a correct point?
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    Re: Point on paraboloid at which the tangent plane is parallel to plane

    Quote Originally Posted by deezy View Post
    Ok, here's what I've done:

    -10y = 1
    -14z = 1
    y = -1/10
    z = -1/14

    I got confused at this point. I wasn't sure what I do with the x. \nabla f = i -(10y)j-(14z)k. There wasn't an x in the equation, and 1 \neq -1, so does the point exist?

    And is there a way to check, if I found a point, that it was a correct point?
    You have forgotten that every plane has two normal vectors, as I pointed out in my original post. If you muliply the normal by -1 you get get another normal vector but with opposite direction.
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