Consider the surface f (x comma y comma z )f(x,y,z)equals=negative 2 x squared plus 2 y squared minus 3 z squared plus 3 equals 0−2x2+2y2−3z2+3=0, which may be regarded as a level surface of the function wequals=f (x comma y comma z )f(x,y,z). The point P(negative 2 comma 2 comma 1 )(−2,2,1) is on the surface. a. Find the (three-dimensional) gradient of f and evaluate it at P. b. The heads of all the vectors orthogonal to the gradient with their tails at P form a plane. Find an equation of that plane.

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rodolforodriguezr rodolforodriguezr · Answer 1 · 2019-10-14T19:06:16+02:00

Answer:

a) (8,8,-6)

b) 4x+4y+3z = -3

Step-by-step explanation:

a)

The surface is given by the equation

f(x,y,z) = 0 where

[tex]f(x,y,z)=-2x^2+2y^2-3z^2+3[/tex]

The gradient of this function is the vector

[tex](\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})=(-4x,4y,-6z)[/tex]

If we evaluate it in the point P = (-2,2,1) we obtain the point

(8,8,-6)

b)

The vectors with their tails at P are of the form

(-2,2,1)-(x,y,z) = (-2-x, 2-y, 1-z)

as they must be orthogonal to the gradient, they must be orthogonal to the vector (8,8,6) so their inner product is 0

[tex](-2-x,2-y,1-z)\cdot(8,8,6)=0\Rightarrow -16-8x+16-8y+6-6z=0\Rightarrow 4x+4y+3z=-3[/tex]

and the equation of the desired plane is

4x+4y+3z = -3