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
Hi
The tangent plane at a point $\displaystyle M$ of a sphere is perpendicular to $\displaystyle \vec{CM}$ where $\displaystyle C$ is the centre of the sphere. This gives you a vector which is normal to the plane. As we also have that $\displaystyle M$ is a point of the plane, you can determine the equation you're looking for...
If
$\displaystyle f(x,y,z=x^2+y^2+z^2-2$
then the normal vector of the tangent plane can be calculated by:
$\displaystyle \frac{\partial f}{\partial x}i +\frac{\partial f}{\partial y}j +\frac{\partial f}{\partial z}k = 0$
With your question you get:
$\displaystyle 2xi+2yj+2zk=0$
Plug in the coordinates of the tangent point and the normal vector is:
$\displaystyle \vec n = 2j+2k$
Since the point T(0, 1, 1) is located on the plane the equation of the plane becomes:
$\displaystyle 2y+2z-4=0$