Question

Find the equation of the tangent plane and normal to the surface xyz =4 at the point (1,2,2)

Answer 1

Step-by-step explanation:

Hello student,

The equation of surface is F(x, y, z) = xyz – 4 = 0

Differentiating this equation partially w.r.t. x, y and z respectively

∂F/∂x = yz

∂F/∂y = xz

∂F/∂z = xy

Hence, at the point (1, 2, 2) we have

∂F/∂x = 4

∂F/∂y = 2

∂F/∂z = 2

Hence, the equation of tangent plane at (1, 2, 2) is

(x-1)4 + (y-2)2 + (z-2)2 = 0

So, 2x + y+z = 6

Equation of normal line at point (1, 2, 2) is

(x-1)/4 = (y-2)/2 = (z-2)/2

or (x-1)/2= (y-2)/1 = (z-2)/1