What would be a unit normal vector to the surface 2x^2 + 3y^2 + 2z^2 = 22 at the point (1, 2, 2) ?

Question

possible solutions.....


a. (14^-1/2 )(i + 2 j + 3 k)

b. (53^-1/2 )(4 i + j + 6 k)

c. (14^-1/2 )(i + 3 j + 2 k)

d. (11^-1/2 )(3 i + j + k)

e. (41^-1/2 )(4 i + 4 j + 3 k)

f. (34^-1/2 )(3 i + 4 j + 3 k)

or is it none of these?

Answer 1

If you have a surface given by f(x, y, z) = c then any direction v tangent to this surface at a given point must satisfy v . ∇f = 0, since otherwise by moving in the direction v we would be staying on the surface but changing the value of f. It follows that ∇f is a normal vector to the surface.

Here ∇f = (4x, 6y, 4z) = (4, 12, 8) = 4 (1, 3, 2). So (1, 3, 2) is a normal vector to the surface, with magnitude √(1 + 9 + 4). So the answer is 14^(-1/2) . (i + 3j + 2k), i.e. answer (c).

Answer 2

We know from theory that for a Level Surface:

φ(x, y, z) = k, (constant), [= 2x^2 + 3y^2 + 2z^2 = 22, in our case],

i∂F/∂x + j∂F/∂y + k∂F/∂y.................(1)

is a vector in the direction of normal to the surface at any point P(x1, y1, z1), [in our case A(1, 2, 2)] on it.

Now:

∂F/∂x = 4x = 4 at A,
∂F/∂y = 6y = 12 at A and
∂F/∂z = 4z = 8 at A.

Putting all this info in (1), a vector normal to the surface φ(x, y, z) = 2x^2 + 3y^2 + 2z^2 = 22 at A(1, 2, 2) will be 4i + 12j + 8k. Its magnitude = sqroot (4² + 12² + 8²) = 4*sqroot(14).

Hence the unit vector in this direction will be:

[4i + 12j + 8k]/4*sqroot(14) or (i + 3j + 2k)/sqroot(14) i.e. "c" is the correct answer to the given problem.