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Thread: Tangent Planes

  1. #1
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    Tangent Planes

    I am trying to find all planes that are tangent to both x^2 + y^2 + z^2 = 1 and x^2 + y^2 + 2z = 0.

    I tried using gradient to get the general equation of the planes for each but after that I got stuck.

    Any suggestions would be greatly appreciated. Thanks!
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  2. #2
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    Here's one way.

    The general tangent plane to the sphere at the point $\displaystyle (x_1,y_1,z_1)$ is $\displaystyle xx_1+yy_1+zz_1-1=0$, and the general tangent plane to the paraboloid at $\displaystyle (x_2,y_2,z_2)$ is $\displaystyle xx_2+yy_2+z+z_2=0$.

    If these planes are identical, we must have $\displaystyle \frac{x_1}{x_2}=\frac{y_1}{y_2}=\frac{z_1}1=\frac{-1}{z_2}$.

    So $\displaystyle x_1=-\frac{x_2}{z_2}$, $\displaystyle y_1=-\frac{y_2}{z_2}$ and $\displaystyle z_1=-\frac1{z_2}$. Using $\displaystyle x_1^2+y_1^2+z_1^2=1$ we deduce that $\displaystyle x_2^2+y_2^2+1=z_2^2$.

    But $\displaystyle x_2^2+y_2^2=-2z_2$. Therefore $\displaystyle z_2^2+2z_2-1=0$ and $\displaystyle z_2<0$ implying $\displaystyle z_2=-1-\surd2$.

    Hence $\displaystyle z_1=-\frac1{z_2}=-1+\surd2$, $\displaystyle x_1^2+y_1^2=1-z_1^2=2(-1+\surd 2)$ and $\displaystyle x_2^2+y_2^2=-2z_2=2(1+\surd 2)$.

    To summarise: with $\displaystyle k=1+\surd 2$ the common tangent planes are $\displaystyle xx_2+yy_2+z=k$ where $\displaystyle x_2^2+y_2^2=2k$, touching the sphere at $\displaystyle (x_2/k,y_2/k,1/k)$ and the paraboloid at $\displaystyle (x_2,y_2,-k)$.
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  3. #3
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    $\displaystyle x^2 + y^2 + z^2 = 1~~ $ (1)
    $\displaystyle x^2 + y^2 + 2x = 0~~ $ (2)

    Use a parametric equation to replace (2) ,

    $\displaystyle x = r \cos{ \theta} , y = r \sin{\theta} , z = - \frac{r^2}{2}$

    And the normal of the paraboloid ( in vector form , but not unit vector) is :

    $\displaystyle \nabla (x^2 + y^2 + 2z) = 2x i + 2y j + 2k = 2r\cos{\theta} i + 2r\sin{\theta}j + 2k $

    And the tangent plane of it is :

    $\displaystyle 2(x-r \cos{ \theta})(r \cos{ \theta}) + 2(y-r \sin{\theta})(r \sin{\theta}) + 2z + r^2 = 0$

    OR

    $\displaystyle r \cos{ \theta} x + r \sin{ \theta} y + z - \frac{r^2}{2} = 0$

    Since the required planes touch the unit sphere , the distance of the plane from the oringin is 1 .

    $\displaystyle | \frac{r \cos{ \theta} (0) + r \sin{ \theta} (0) + (0) - \frac{r^2}{2}}{ \sqrt{ r^2 + 1} } | = 1$

    $\displaystyle r^4 - 4r^2 - 4 = 0$

    Solving the equation , gives $\displaystyle r = \sqrt{2 + 2\sqrt{2}}$

    Finally , substitute back to the equation of the plane ,

    the required plane is obtained :

    $\displaystyle r \cos{ \theta} x + r \sin{ \theta} y + z - \frac{r^2}{2} = 0$



    In fact , there are infinite planes which satisify your requirement
    Last edited by simplependulum; Jul 4th 2009 at 08:26 PM.
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