Find a unit vector normal to the surface cos(xy) = ez − 2 at (1. π. 0).

Find a unit normal to the surface cos(xy)=e^z-2 at (1,pi,0)

In this problem i have used the fact that < a b c> is the
normal vector to the plane ax + by + cz = d.
cos(xy) = e^z – 2
–> z = ln(2+cos(xy))
now, equation of tangent plane to z = f(x,y) at point (xo,yo,zo)
is given by:
z-zo = fx(xo,yo)(x-xo) + fy(xo,yo)(y-yo)
here, fx = -y sin(xy) / (2+cos(xy), fy = -x sin(xy) /
(2+cos(xy)
at (x,y) = (1,pi): fx = 0, fy = 0,
equation of tangent plane is:
z -0 = 0
–> z = 0 …(that is whole xy plane, and unit normal to xy
plane is k(cap)
other way:
z = 0
–> 0x + 0y + 1z = 0
that is whole xy plane, and unit normal to xy plane is
k(cap),
vector form is : < 0 0 1>

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