# The normal of a surface that passes through the origin

this is my first question here.

Suppose I have a surface as follow:

$$x^2+y^2+z^2=9$$

The gradient of the surface at a particular point $P = (x_0,y_0,z_0)$ is just $(2x_0,2y_0,2z_0)$ and the parametric equation of the line is $x(t) = x_0 + t(2x_0)$ and so on. Now how can we find the points that have their normal passing through the origin?

Normally I would set $P = (0,0,0)$ and the direction vector would remain. But here the direction vector is directly dependent on the point chosen, so if I were to set $P = (0,0,0)$, then there would be no line as well.

Could you guys help me with this?

Thanks

• The only point $(x,y,z)$ satisfying $x^2+y^2+z^2 = 0$ is $(0,0,0)$. – fuglede Jun 12 '14 at 9:43
• Sorry, let me change my equation so that the constant is equal to 9 . – vTx Jun 12 '14 at 9:46
• now $(0,0,0)$ is not a point of the surface. – mookid Jun 12 '14 at 10:08

Let $M(t)= (x_0 +2x_0 t,y_0 +2y_0 t,z0 +2z_0 t )$.
Check if the equation $M(t) = (0,0,0)$ has a solution. In you initial problem every point has their normal passing through the origin, because for every point the equation has a solution ($t=-1/2$).