The question I have is directly related to the following question that has been answered:
Show that the tangent plane of the cone $z^2=x^2+y^2$ at (a,b,c)$\ne$0 intersects the cone in a line
The original expression is $z^{2} = x^{2} + y^{2}$
and $(a,b,c) \neq (0,0,0)$ is the main condition.
I went about solving this question in the same exact way as the author of the original to the extent that setting the surface equal to the tangent plane I obtained
$$(x-a)^{2} + (y-b)^{2} – (z-c)^{2} = 0$$
So what I envisioned happening based on a previous exercise I had done was that I would be able to solve the above expression for a specific $(x,y,z)$. I could then put this into $f(x,y,z) = x^{2} + y^{2} – z^{2}$ and would have the equation of a line come out of this. But this won't be the case in this situation. Assuming I understand things correctly the equation of a line in $\mathbb{R}^{3}$ should be a vector of the form $t(x,y,z)$ for some scalar $t \in \mathbb{R}$. I know the mechanics is correct, but I'm missing how to express the idea of the intersecting line properly. How would I do this in this situation?
Best Answer
The equation of the (upside down) cone is: $$ f(x,y,z) = x^2+y^2-z^2 = 0 $$ Note that the top of the cone is at the origin $(0,0,0)$. Take another point $(a,b,c)\ne(0,0,0)$ at the surface of the cone: $$ a^2+b^2-c^2 = 0 $$ The equation of the OP's line in $\mathbb{R}^{3}$ is required to be of the form: $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} t $$ So the vector $(a,b,c)$ should preferrably be normed: $$ a^2+b^2+c^2 = 1 $$ Two equations with three unknowns; we have one degree of freedom. A suitable solution is: $$ \begin{cases} a = \cos(\phi)/\sqrt{2} \\ b = \sin(\phi)/\sqrt{2} \\ c = 1/\sqrt{2} \end{cases} $$ To be interpreted as a vector to any point (angle $\phi$) on the circle $\,x^2+y^2=1/2\,$ at height $\,z=1/\sqrt{2}\,$. The end-result is: $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} \cos(\phi) \\ \sin(\phi) \\ 1 \end{bmatrix} t $$ EDIT. Before I forget, this line is, of course, in the tangent plane. $$ \vec{\nabla} f(x=a,y=b,z=c) = \begin{bmatrix} 2x \\ 2y \\ -2z \end{bmatrix} = 2\begin{bmatrix} a \\ b \\ -c \end{bmatrix} $$ So the equation of the tangent plane is (norming factors omitted): $$ ax+by-cz=0 \quad \mbox{with} \quad \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} t \quad \Longrightarrow \quad (a^2+b^2-c^2)t^2=0 $$ which is the same as saying that $(a,b,c)$ is at the surface of the cone.