Show that the tangent plane of the of the cone intersects the cone in a line - help in communicating idea correctly.

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.