# Thread: Vector Calculus equation for planes

1. ## Vector Calculus equation for planes

I am having real trouble understanding this concept and I am hoping that maybe someone would be willing to help me out.

I need to find the equation for a plane that just touches the sphere (x - 2)^2 + (y - 3)^2 = (z - 3)^2 = 16 and is parallel to 1) xy plane 2) yz plane and 3) the xz plane. If someone can walk me through one of these I should be able to apply the info to the other two. Thanks very much!!!

Also, if the function (x,y,z) = x^2 + y^2 + z^2 - 2x + 3y + z I need to find a point (a,b,c) such that the function can be expressed in terms of distance from (a,b,c)

I know I need to complete the square but I am lost........

Can someone recommend a good online 3D graphing program? I need to really work on understanding what different equations look like in 3D. Thanks again, this site is WONDERFUL!

3. Originally Posted by Frostking
I need to find the equation for a plane that just touches the sphere (x - 2)^2 + (y - 3)^2 + (?) (z - 3)^2 = 16 and is parallel to 1) xy plane 2) yz plane and 3) the xz plane.
A plane parallel to the x-y plane has equation of the form $z=k$ for some constant k.

The sphere $(x - 2)^2 + (y - 3)^2 + (z - 3)^2 = 16$ has centre (2,3,3) and radius 4. So at its highest point z will be 3+4, and at its lowest point z will be 3–4. There are therefore two planes parallel to the x-y plane that touch the sphere, namely $z=7$ and $z=-1$.

Originally Posted by Frostking
Also, if the function (x,y,z) = x^2 + y^2 + z^2 - 2x + 3y + z I need to find a point (a,b,c) such that the function can be expressed in terms of distance from (a,b,c)

I know I need to complete the square
That is correct. Write the function as f(x,y,z) = x^2 – 2x + y^2 + 3y + z^2 + z. Complete the squares in the x, y and z terms, and you will get f(x,y,z) = (x–a)^2 + (y–b)^2 + (z–c)^2 – k (for some constants a, b, c, k). That tells you that f(x,y,z) is the square of the distance from (x,y,z) to (a,b,c), less k.