Proving Every Normal Line to a Sphere Passes Through Center

[gikiian]

Homework Statement

Prove that every normal line to a sphere passes through the centre of the sphere.

Homework Equations

Equation of the sphere with it's center at O(a,b,c):

(x-a)²+(y-b) ²+(z-c)²=r²

or

(x-a)²+(y-b) ²+(z-c)²-r² = 0 = F(x,y,z)Parametric equations of the normal line at any point P(x₀,y₀,z₀) on the sphere:

x=x₀+(Fx)t
y=y₀+(Fy)t
z=z₀+(Fz)t ; where Fx, Fy and Fz are the partial derivatives of F(x,y,z).

The Attempt at a Solution

Fx = 2(x-a)
Fy = 2(y-b)
Fz = 2(z-c)

At P(x₀,y₀,z₀) on the sphere, the partial derivatives will be as follows:

Fx = 2(x₀-a)
Fy = 2(y₀-b)
Fz = 2(z₀-c)

By putting these values in the parametric equations of the normal line, we get:

x=x₀+2(x₀-a)t
y=y₀+2(y₀-b)t
z=z₀+2(z₀-c)t

Now these equations tell that the line originates form the point P(x₀,y₀,z₀) and is along the vector <x₀-a,y₀-b,z₀-c> which passes from the center O(a,b,c) of the circle! does this prove the statement? Does the proof require something else?

Thanks.

 

[Char. Limit]

Looks good to me. You might need to prove that such a vector actually passes through the center of the circle, but I doubt that will be necessary.

 

[HallsofIvy]

Yes, rather than just saying "which passes from the center". Show that
$a= x_0+ 2(x_0-a)t$, $b= y_0+ 2(y_0- b)t$, and $c= z_0+ 2(z_0- c)t$ all for the same value of t.