Problem on finding a tangent vector to a surface

Consider the surface given by $z = x^2 + y^2$. What is the (3d) tangent vector at $(1,1)$ that has an $\mathbf{i}$ component of 0 and a $\mathbf{j}$ component of 2?

Solution using Directional Derivatives
Let $\mathbf{v} = 0 \mathbf{i} + 2 \mathbf{j} + c \mathbf{j}$ be the vector we seek. We would like to find the value of $c$ such that $\mathbf{v}$ is tangent to $z(x,y) = x^2 + y^2$ at $(1,1)$.
Level curves and surfaces

To get a graphical sense of the problem, we draw the tangent vector in relation to the level curves of $z$ (left) and the $x=1$ cross section (right):
Recall that:
Using directional derivatives to find a tangent vector to a surface

Directional derivative definition

A tangent vector to a surface has a slope (rise in $z$ over run in $xy$) equal to the directional derivative of the surface height $z(x,y).$
The vector $\mathbf{v}$ has a rise of $c$ and a run in the $xy$ plane of $| 0\mathbf{i} + 2 \mathbf{j}| = 2$. Hence, the 'slope' of the vector $\mathbf{v}$ is $\frac{c}{2}$.
(see details)

We can compute that the directional derivative of $z(x,y)$ at $(1,1)$ in the direction of $0 \mathbf{j} + 2 \mathbf{j}$ is 2.
Equating, we get $$\frac{c}{2} = 2.$$
Thus, $c= 4$, and the tangent vector is $$\mathbf{v} = 2 \mathbf{j} + 4 \mathbf{k}.$$

Problem on finding a tangent vector to a surface

Solution using Directional Derivatives

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