– 1,4, – 9, 16, – 25, ..?
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Let . The tangent plane to the surface at (0, 0, 8) is
The gradient is
so the tangent plane's equation is
The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation
or ,
, and
.
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
