find the distance from the sphere to the plane

[Smily]

find the distance from the sphere x^2+ y^2+ z^2+ 2x+ 6y- 8z= 0 to the plane 3x + 4y + z= 15
:roll:
thanks for any help you can give me

 

[galactus]

sphere x^2+ y^2+ z^2+ 2x+ 6y- 8z= 0

Here's a start.

Group and complete the squares:

$\displaystyle (x+1)^{2}+(y+3)^{2}+(z-4)^{2}=26$

Center at (-1, -3, 4), radius $\displaystyle \sqrt{26}$

 

[pka]

If $\displaystyle \Pi$ is a plane with normal $\displaystyle N$ then if $\displaystyle A \notin \Pi$ then the distance from $\displaystyle A$ to $\displaystyle \Pi$ is given by $\displaystyle D(\Pi ;A) = \frac{{\left| {\vec{PA} \cdot N} \right|}}{{\left\| N \right\|}}$ where $\displaystyle P \in \Pi .$