Question

Given the surface X^2+3y^2+2Z^2=6. Find the unit vector normal to the surface at Q(2,0,1)

Answer 1

Answer:

Explanation:

Calling

f

(

x

,

y

,

z

)

=

x

3

+

y

3

+

3

x

y

z

−

3

=

0

The gradient of

f

(

x

,

y

,

z

)

at point

x

,

y

,

z

is a vector normal to the surface at this point.

The gradient is obtained as follows

∇

f

(

x

,

y

,

z

)

=

(

f

x

,

f

y

,

f

z

)

=

3

(

x

2

+

y

z

,

y

2

+

x

z

,

x

y

)

at point

(

1

,

2

,

−

1

)

has the value

3

(

−

1

,

3

,

2

)

and the unit vector is

{

−

1

,

3

,

2

}

√

1

+

3

2

+

2

2

=

{

−

1

√

14

,

3

√

14

,

√

2

7

}