Find All Points on a Paraboloid where Tangent Plane is Parallel to a Given Plane Find all points on the paraboloid $z=x^2+y^2$ where tangent plane is parallel to the plane $x+y+z=1$  and find equations of the corresponding tangent planes. Sketch the graph of these functions. 

I have its answer. I don't really understand such type of questions. And I am really willing to learn. Also I added its answer as a picture. Please teach me how to solve.

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 A: To get a normal vector to the paraboloid at a point (x,y,z), we can take the gradient $\nabla f(x,y,z)=-2xi-2yj+k$.  Since we want the tangent plane at the point to be parallel to the plane $x+y+z=1$, the normal vector $\nabla f(x,y,z)=-2xi-2yj+k$ has to be parallel to the vector $i+j+k$  (since this is a normal vector to $x+y+z=1$).  This means that $-2xi-2yj+k$ must be a constant multiple of $i+j+k$, so $-2xi-2yj+k=c(i+j+k)$ for some constant c. Then
$-2x=c$, $-2y=c$, and $1=c$, so $x=-1/2$ and $y=-1/2$.  Therefore $z=x^2+y^2=1/4+1/4=1/2$ at the point of tangency, and the tangent plane has equation $x+y+z=-1/2$
at this point.
A: You can obtain a parametrisation of your submanifold by viewing it as a graph of a function, and the parametrisation's partial derivates give a basis of the tangent space at each point.
Let the paraboloid, as a submanifold, be denoted by $S$. We have $f(x,y) = (x, y, x^{2} + y^{2})$ as our parametrisation, hence $T_{f(x,y)}S = <(1,0,2x)^{t}, (0,1,2y)^{t}>$. 
You now get exactly two unit-length vectors that span the orthogonal complement of $T_{f(x,y)}S$. Chose such a vector, let's call it $\nu$. Now all you have to do is find the plane you want your $T_{f(x,y)}S$ to be parallel to, and check that the line defined by $c(t) = f(x,y) + t\nu$ meets the plane orthogonally.
Edit: 
For the last part, we use that in euclidian space, a hyperplane is parallel to another hyperplane iff there exists a line that meets both orthogonally.
