How do you find the steepest direction on a plane P wrt to the x-y plane? Isn't this direction unique? The question says that the steepest direction makes the largest acute angle with the x-y plane?
So if you have a plane P that contains the vectors u = (3, −1, −2) and v = (−1, 0, 5). What is the steepest direction on P wrt x-y plane? Is this an angle or another vector? I am confused.
Without proof, rather based on intuition: if your two vectors are linearly independent and so define the plane, you can calculate the plane normal n = u cross v. From the xy-plane (eg. u_0 = (1, 0, 0), v_0 = (0, 1, 0)) you get the plane normal n_0 = (0, 0, 1). Now, you can build the cross product of the 2 normals, n_0 cross n, which yields an vector w orthogonal to them, which depends on their angle. To turn this vector w into the plane in the direction of the steepest decent you build the cross product with the plane normal and receive the vector of steepest decent g = w cross n. Your data yield:
n = (-5, -13, -1)
w = (13, -5, 0)
g = (5, 13, -194) or norm(g) = (0.026, 0.067, -0.997)
normalized vectors u(green), v(orange), g(black)