Find point on sphere with directional tangent vector

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?

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Um... every point on a sphere has at least one tangent vector that is perpendicular to $(2,3,4)$. If you want a point such that all tangent vectors are perpendicular to $(2,3,4)$, then just scale $(2,3,4)$ such that its length becomes $5$. –  Henning Makholm Nov 18 '11 at 17:33
Also, what you get from "going through the partical derivatives" is not a tangent vector, but a normal vector. –  Henning Makholm Nov 18 '11 at 17:34

1 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.

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ohh...I must have have remember it wrongly. What is the correct way to find the tangent vector? Also, I am thinking why isn't there a point on the sphere that has a normal vector of $[2, 3, 4]? Visually, there seems to have only one point. And that's why I thought I could use the tangent vector to find it. – xenon Nov 18 '11 at 17:45 @xEnOn: Please read my answer more carefully. Talking about "the tangent vector" makes no sense. – joriki Nov 18 '11 at 17:46 Sorry, I missed the tangent plane thing. I think I was so into that tangent vector thing that I missed it. Sorry. So the tangent plane is$2x(x-x_0)+2y(y-y_0)+2z(z-z_0)=0$. I updated the question earlier with an equation that I have found that claims to be the "tangent vector" of a point on the sphere. How was that derived? – xenon Nov 18 '11 at 17:58 @xEnOn: You can cancel the$2$s in that equation. That's a slightly unorthodox way of writing the equation; you're denoting the point on the sphere by$(x,y,z)$and the point on the plane by$(x_0,y_0,z_0)\$; it would be more usual to do it the other way around. Regarding the "tangent vector" you found in a book: I have no idea where that comes from. You might want to add more of the context that you found it in, or perhaps first reexamine the context yourself: Is it perhaps a particular tangent vector satisfying a particular condition? –  joriki Nov 18 '11 at 18:08