Find Unit Vector perpendicular to the Surface, x^{3}+zx=1 at the point P=(1,2,-1) I figures that the perpendicular vector would be, N(X)=grad(x^{3}+zx) = (3x^{2}+z, 0, x) N(P)= (3,0,1) Then the unit vector would be, n=N(P)/||N(P)|| n=(3/5^{1/2},0,1/5^{1/2}) The answer i get is not the right answer and i dont see where im going wrong. Thanks,
Thanks for pointing that out Dick, minor error. (2,0,1) Its still not the right answer which is 1/27^{1/2}(5,1,1)
Then there's probably a typo in the problem. The surface x^3+zx=1 equation doesn't have a 'y' in it. That means the y direction is tangent to the surface. The normal vector can't possibly have a nonzero y component.