Question

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?

Answer 1

Answer:

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

Therefore a vector to the normal to the surface is

$\hat{n}=2\hat{i}+3\hat{j}+2\hat{k}$

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

$f_x=\frac{\partial f }{\partial x}$    [ where only variable is x]

$f_y=\frac{\partial f }{\partial y}$    [ where only variable y]

$f_z=\frac{\partial f }{\partial z}$   [ where only variable z]

$f_x=\frac{\partial f }{\partial x}=\frac{\partial }{\partial x}(2x^2-3y^2+z^2)$   = 4x

$f_y=\frac{\partial f }{\partial y}=\frac{\partial }{\partial y}(2x^2-3y^2+z^2)=-6y$

$f_z=\frac{\partial f }{\partial z}=\frac{\partial }{\partial z}(2x^2-3y^2+z^2)=2z$

The gradient at (1,-1,2)

$\bigtriangledown f(1,-1,2)\\= (4\times 1)\hat{i} +[-6\times (-1)]\hat{j}+(2\times 2)\hat{k}$   [ putting x=1,y=-1 and z=2 in

$=4\hat{i}+6\hat{j}+4\hat{k}$                                       $f_x,f_y \ and \ f_z$]

Therefore a vector to the normal to the surface is

$\hat{n}=4\hat{i}+6\hat{j}+4\hat{k}$

$or,\hat{n}=2\hat{i}+3\hat{j}+2\hat{k}$       [ remove the common part= 2]

The equation of tangent plane is

$\vec{r}.\hat{n}=\vec {a}.\hat{n}$

$\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}$

$\hat{n}$ = normal vector

$\vec{a}$ = the position vector of the given point

Here $\hat{n}=2\hat{i}+3\hat{j}+2\hat{k}$    and   $\vec a= \hat i-\hat j+2\hat k$

Therefore the equation of tangent plane is

$( x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+3\hat{j}+2\hat{k})=( \hat i-\hat j+2\hat k). (2\hat{i}+3\hat{j}+2\hat{k})$

⇒2.x+3.y+2.z=(1.2)+(-1)(3)+2.2

⇒2x+3y+2z=2-3+4

⇒2x+3y+2z=3