Say a sphere equation like this: $x^2+y^2+z^2=5$. I want to find a point on the sphere whose tangent vector is perpendicular to the vector $\begin{bmatrix}
2\\
3\\
4
\end{bmatrix}$.
I go through the partial derivatives to get the tangent vector as $\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
2x\\
2y\\
2z
\end{bmatrix}$.
Now, I put the equation together this way $$\begin{bmatrix}
2x\\
2y\\
2z
\end{bmatrix}\cdot
\begin{bmatrix}
2\\
3\\
4
\end{bmatrix}=0$$
But I can't solve for the point because there are 3 unknowns and there will be many possible solution to $x,y,z$. Is my tangent vector correct in the first place? What should I do to to solve for the exact point?
Edit
I happen to find the same equation of the sphere in a book. Somehow, it says that the tangent vector of a point on the sphere is $\begin{bmatrix}
0\\
1\\
\frac{y}{\sqrt{5^2-y^2}}
\end{bmatrix}$. But how come it has this tangent vector different from mine. Which is correct? And how was this derived?
Best Answer
There is not one tangent vector at a point on a sphere. A sphere, being a two-dimensional surface, has tangent planes, and any vector lying in the tangent plane at a point is tangent to the sphere at that point.
What you calculated is not a tangent vector but a normal vector, the gradient of the function $x^2+y^2+z^2-5$ whose locus of zeros the sphere is.