# Plane tangent to a sphere

• Dec 30th 2009, 03:53 PM
empireruler
Plane tangent to a sphere
If anyone could help me with this problem, it would be most appreciated.

Find an equation for the plane tangent to the sphere x^2+y^2+z^2=81 at (1,4,8).

I have tried finding the vector perpendicular to [1,4,8] but I end up with a+4b+8c=0 and can't figure out a way to find a, b, and c.

Thanks.
• Dec 30th 2009, 04:17 PM
Plato
Quote:

Originally Posted by empireruler
Find an equation for the plane tangent to the sphere x^2+y^2+z^2=81 at (1,4,8).
I have tried finding the vector perpendicular to [1,4,8] but I end up with a+4b+8c=0 and can't figure out a way to find a, b, and c.

Don't stop! You are right there. Finish it off.
The vector $<1,4,8>$ is the normal.
Use that normal and the point $(1,4,8)$ to write the equation.
• Dec 30th 2009, 04:48 PM
empireruler
Wait do you mean like x+4y+8z= 81... wouldn't that just give me the perpendicular plane... the one that shares the normal vector. Sorry if I missed the point you were getting at.
• Dec 30th 2009, 05:01 PM
Plato
Quote:

Originally Posted by empireruler
Wait do you mean like x+4y+8z= 81... wouldn't that just give me the perpendicular plane... the one that shares the normal vector. Sorry if I missed the point you were getting at.

What do you think a tangent plane means?
Think about a tangent to a circle.
It is perpendicular to a radial segment.
• Dec 30th 2009, 05:12 PM
empireruler
wouldnt x+4y+8z = 81 be the plane that is parallel to the plane that is tangent to the sphere? Sorry I am not understanding
• Dec 31st 2009, 04:59 AM
HallsofIvy
Quote:

Originally Posted by empireruler
wouldnt x+4y+8z = 81 be the plane that is parallel to the plane that is tangent to the sphere? Sorry I am not understanding

If it is parallel to the tangent plane and contains the point (1, 4, 8), which it does because 1+ 4(4)+ 8(8)= 1+ 16+ 64= 81, it is the tangent plane!