- #1

- 277

- 4

**Mod note**: Moved from a technical forum section, so missing the homework template.

@fab13 -- please post homework problems in the appropriate section under Homework & Coursework.

I have the following exercise to solve : I have to find all the points on the surface ##x^2+y^2+z^2=36## (so a sphere of radius = 6) where tangent planes are parallel to plane ##3x+4y+5z=0##

For the moment, I can get this reasoning :

if ##M_{0}=(x_0,y_0,z_0)## is a point that staisfies the wanted property above and ##(x,y,z)## a point of its tangent plane, I have (with ##C## the center of the sphere) :

[tex]\vec{CM_{0}}\cdot \vec{MM_{0}}=0[/tex]

So with the sphere of equation : ##(x-a)^2+(y-b)^2+(z-c)^2=R^2##, I can deduce :

##(x_0-a)(x-x_0)+(y_0-b)(y-y_0)+(z_0-c)(z-z_0)=0## and especially :

##(x_0-a)(x-a)+(y_0-b)(y-b)+(z_0-c)(z-c)-R^2=0##

In my case, ##R^2=36## and ##a=b=c=0##, so I have :

##x_0 x +y_0 y +z z_0 -36=0##

From this point, how can I deduce ##(x_0,y_0,z_0)## ( I recall, the tangent plane at this point has to be parallel to plane ##3x+4y+5z=0##) ?

Thanks for your help

Last edited by a moderator: