Given a sphere with equation $$x^2+y^2+z^2=9$$ and the parametric form of a line, $$(x,y,z)=(1+2t, 1+2t, 1+2t),$$ find the equation of a plane tangent to the sphere and perpendicular to the line.
I know that there are two planes, but I only need to find one of them. However, I am having difficulty figuring out how to do this.
On
Since the line is perpendicular to the plane,
and the line is parametrized as $(x,y,z)=(1+2t,1+2t,1+2t)$,
the equation of the plane is $2x+2y+2z=C$.
The points on the line satisfy $x=y=z$.
The line intersects the sphere when $x^2+y^2+z^2=9$;
i.e., $x^2+x^2+x^2=9$; i.e., $x^2=3$; i.e., $x=\pm\sqrt3.$
Therefore, the equations of the planes tangent to the sphere are $2x+2y+2z=\pm6\sqrt3$.
As said by Jean Marie, the equation of the plane will be as
$$x+y+z-k=0$$
the distance from the sphere center to this plane should be the radius
$$\frac{|0-k|}{\sqrt{1+1+1}}=3$$
which gives $$k=\pm 3\sqrt{3}$$