Tangent plane and normal line Find the equation of the tangent plane and the vectorial equation of the normal line of the surface
$$z = x^2 + y^2$$
at the point $(1,-2,5)$
For the vectorial equation of the normal line, should I consider the function $F(x,y,z) = x^2 + y^2 -z$ and $\vec{n} = \vec{\nabla}F(1,-2,5)$ as my normal vector?
I'm not sure of how should I do this.
Thanks!
 A: The surface
$z = x^2 + y^2 \tag{1}$
is in fact the level surface of the function
$F(x, y, z) = x^2 + y^2 -z \tag{2}$
on which $F(x, y, z)$ takes the value $0$.  Since the gradient of a differentiable function is normal to its level surfaces, we can indeed obtain a normal to (1) by computing the gradient of $F(x, y, z)$ as given by (2).  We have
$\nabla F(x, y, z) = (2x, 2y, -1) \tag{3}$
at any point $(x, y, z) \in \Bbb R^3$; thus, at the point $(1, -2, 5)$, which is easily seen to lie in the surface 
$F(x, y, z) = x^2 + y^2 - z = 0, \tag{4}$
($1^2 + (-2)^2 -5 = 0$) we have
$\nabla F(1, -2, 5) = (2, -4, -1). \tag{5}$
Now, I'm not exactly sure what it means to represent a line in vectorial form, but an expression for the line normal to the surface $z = x^2 + y^2$ at the point $(1, -2, 5)$ may now be written in terms of the normal vector $\vec n = \nabla F(1, -2, 5)$ and the position vector field $\vec r= (x, y, z)$ on $\Bbb R^3$; taking $\vec r_0 = (1, -2, 5)$, we have
$\vec r(t) = \vec r_0 + t\nabla F(1,2,5), \tag{6}$
or
$\vec r(t) = (1, -2, 5) + t(2, -4, -1). \tag{7}$
An expression for the tangent plane may be had in a roughly similar manner; $\vec r = (x, y, z)$ is a point in the tangent plane if and only if the vector $\vec r - \vec r_0$ lies in that plane and is hence perpendicular to $\nabla F(1, -2, 5)$; thus we may write
$(\vec r - \vec r_0) \cdot \nabla F(1, -2, 5) = (\vec r - \vec r_0) \cdot (2, -4, -1) =0 \tag{8}$
for the equation of the (tangent) plane passing through the point $\vec r_0 = (1, -2, 5)$.  A little algebra allows (8) to be re-written in another well-known form:
$2x -4y - z = \vec r_0 \cdot (2, -4, -1) = (1, -2, 5) \cdot (2, -4, -1) = 5. \tag{9}$
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: (1)You have a function $f(x,y,z)=x^2+y^2-z$, now to find the tangent plane of $f$ at $P=(1,-2,5)$ you want to first compute the $\nabla(f(p))$, which is the gradient of $f$ at $P$. Well the $$\nabla(f)=(2x,2y,-1) \implies \nabla(f(1,-2,5))=(2,-4,-1).$$
Now, you have your normal vector and you want to find a plane perpendicular to $N=\nabla(f(1,-2,5))=(2,-4,-1)$ that goes through $P$. 
Solve,
\begin{align*}
(X-P)\cdot N=0 &\implies X\cdot N = P\cdot N \\
&\implies (x,y,z)\cdot (2,-4,-1) = (1,-2,5)\cdot (2,-4,-1) \\
&\implies 2x-4y-z=2+8-5 \\
&\implies 2x-4y-z=5
\end{align*} and this is your tangent plane.
(2) Now for the vectorial equation I assume you want a parametric line? So, $$P+tA=(1,-2,5)+t(2,-4,-1).$$
