Find locus of points for which $\nabla f$ is parallel to the $y$-axis

Question

If $f(x, y, z)=x^2 + y^2$, what is the locus of points for which $\nabla f$ is parallel to the $y$-axis?

I know that $\nabla f=(2x, 2y, 0)$ and I know that the $y$-axis is the set of all points such that $x=z=0$ and $y$ is any real number. To find out when $\nabla f=(2x, 2y, 0)$ is parallel to the $y$-axis, I need to find out when the cross product of $\nabla f$ and the $y$-axis is equal to zero. I just don't know what to take the cross product of $(2x, 2y, 0)$ with. Is the $y$-axis represented by $(0, y, 0)$?? I need help! Thank you!!

Answer 1

Two vectors are parallel if they are scalar multiples of each other. The direction of the $y$-axis is $(0,1,0)$. At any point in the set $$\{ (0,s,t)|s,t\in\mathbb{R}\}$$

the gradient vector satisfies

$$(0,2s,0)=2s(0,1,0)$$

and so is parallel to the $y$ axis. (Note that the locus of points where the vector $\nabla f$ is just the $yz$-plane.)

Answer 2

One do not need to do cross product because let $x = 0$, $y$ arbitrary number then the gradient is parallel to the y-axis.

To show how to do cross product with $(x,y,0)$ and $(0,w,0)$

$$\begin{vmatrix} i & j & k \\ x & y & 0 \\ 0 & w & 0 \\ \end{vmatrix} = xw \times k$$ set $x=0$ then the condition satisfies.