Check that the point (1,-1,2) lies on the given surface. Then, viewing the surface as a level surface for a function f(x, y,z), find a vector normal to the surface and an equation for the tangent plane to the surface at (1,-1,2).

2x^2-3y^2+z^2=3.

Vector normal? and tangent plane?

(1,-1,2) lies on the surface.

Therefore a vector to the normal to the surface is

Therefore the equation of tangent plane is

2x+3y+2z=3

Step-by-step explanation:

Given equation of surface is

2x²-3y²+z²=3

Given point is (1,-1,2)

To check that whether the point lies on the surface or not .

We have to put x=1 , y= -1 and z= 2 in the given surface.

L.H.S

2(1)²-3(-1)²+2²

=2-3+4

=3 = R.H.S

Since the point (1,-1,2) point satisfies the equation.

Therefore (1,-1,2) lies on the surface.

Here f(x,y,z)= 2x²-3y²+z²

To find the a vector normal to the surface we have to find

   [ where only variable is x]

   [ where only variable y]

  [ where only variable z]

  = 4x

The gradient at (1,-1,2)

  [ putting x=1,y=-1 and z=2 in

                                      ]

Therefore a vector to the normal to the surface is

      [ remove the common part= 2]

The equation of tangent plane is

= normal vector

= the position vector of the given point

Here    and  

Therefore the equation of tangent plane is

⇒2.x+3.y+2.z=(1.2)+(-1)(3)+2.2

⇒2x+3y+2z=2-3+4

⇒2x+3y+2z=3

b. x- 48 =120

step-by-step explanation:

range is infinite on both x and y axis positive and negative

step-by-step explanation: