I have read example in a math text book with the following problem:
The surface $z=f(x,y)=\sqrt{9-2x^2-y^2}$ and the plane $y=1$ intersect. Find parametric equation for a tangent line at $(\sqrt{2},1,2$)!
I know that the function $z$ is differentiated partially with respect to $x$ as $y$ is remained constant implying the slope of the tangent line at $(\sqrt{2},1,2)$, however what I am confused is that why the solution shows that the direction vector of the tangent line is $(1, 0, -\sqrt{2})$.