Showing that normal line passes through a point.

Question

I need to show that a line passes through a point. How should I go about doing this? The question is:

Let $L$ be the normal line at $(1,1,1)$ to the level surface of $f(x,y,z) = x^2 - z$ that passes through $(1,1,1)$. Show that $L$ passes through the point $(3,1,0)$.

Thanks in advance!

Edit: I tried taking the gradient, where the gradient is (2,0,-1). So L would be 2(x-1)-(z-1) = 0. Is this right? If it is, how do I show that it passes through the point?

Edit 2: How do I show that it passes through (3,1,0)? (1+2(3),1,1-(0))?

Edit 3: So you're saying that t_0 = 1? I'm confused as to how does this show that L passes through the point.

Edit 4: Can you please show an example where the point is NOT in L? Thanks!

Edit 5: So conflicting values of t? Thank you very much for your help!

Answer 1

You should know that $\nabla f(1,1,1)$ is a normal vector to the surface at said point. Hence, the normal line is: $$L: \quad {\bf X}(t) = (1,1,1) + t \nabla f (1,1,1), \quad t \in \Bbb R.$$

To show the claim, you must find $t_0 \in \Bbb R$ such that ${\bf X}(t_0) = (3,1,0)$. Ok?

Answer 2

If the line passes through the two given points, its direction vector is $(3,1,0)-(1,1,1)=(2,0,-1)$, indeed parallel to the gradient.