find a tangent plane on a sphere

**szpengchao** · April 26th 2008, 10:13 AM

a sphere has equation: x^2+y^2+z^2=2
find the equation of the tangent plane at (0,1,1) on the sphere

**flyingsquirrel** · April 26th 2008, 10:51 AM

Hi

The tangent plane at a point $M$ of a sphere is perpendicular to $\vec{CM}$ where $C$ is the centre of the sphere. This gives you a vector which is normal to the plane. As we also have that $M$ is a point of the plane, you can determine the equation you're looking for...

**earboth** · April 26th 2008, 10:56 AM

Originally Posted by szpengchao

a sphere has equation: x^2+y^2+z^2=2
find the equation of the tangent plane at (0,1,1) on the sphere

If

$f(x,y,z=x^2+y^2+z^2-2$

then the normal vector of the tangent plane can be calculated by:

$\frac{\partial f}{\partial x}i +\frac{\partial f}{\partial y}j +\frac{\partial f}{\partial z}k = 0$

With your question you get:

$2xi+2yj+2zk=0$

Plug in the coordinates of the tangent point and the normal vector is:

$\vec n = 2j+2k$

Since the point T(0, 1, 1) is located on the plane the equation of the plane becomes:

$2y+2z-4=0$

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