Question

the equation of the plane that is tangent to the surface xyz =8 at the point (1,2,4) is?

Answer 1

Normal vector n = < yz, zx, xy >
At the point (1, 2, 4), n = < 8, 4, 2 >
Thus Tangent plane at ( 1, 2, 4 ):
8 ( x -1) + 4 ( y -2 ) + 2 ( z -4) = 0
8x + 4y + 2z - 24 = 0