Let S be the sphere of radius 1 centered at (2, 4, 6). Find the distance from S to the plane x + y + z = 0.

Answer:

5.928

Step-by-step explanation:

Given that:

The relation of the plane x+y+z= 0

Suppose (x,y,z) is any point on the plane.

Then the difference between (2,4,6) to (x,y,z) is:

∴

To estimate the maximum and minimum values of the function f(x,y,z) subject to the constraint g(x,y,z) = x+y+z =0

By applying Lagrane multipliers;

If we differentiate equation (1) with respect  to x; we have:

f(x,y,z) = 2x -4

If we differentiate equation (1) with respect  to y; we have:

f(x,y,z) = 2y - 8

If we differentiate equation (1) with respect  to z; we have:

f(x,y,z) = 2z - 12

Differentiating g(x,y,z) with respect to x, we have:

Differentiating g(x,y,z) with respect to y, we have:

Differentiating g(x,y,z) with respect to z, we have:

Calculating the equations

x+y+z = 0  - - - (5)

replacing x, y, z values in the given constraint

x + y + z = 0

Therefore, from equation (2)

x = 2 - 4

x = - 2

From equation (3)

x = 4 - 4

x =  0

From equation (3)

x = 6 -4

x = 2

i.e (x,y,z) = (-2, 0, 2)

∴

since we are taking only the positive integer because distance cannot be negative, then:

The distance from the center of the sphere to the plane is 6.928.

However, the distance from the surface S to the plane is:

6.928 - radius of the sphere.

where;

the radius of the sphere is given as 1

Then:

the distance from the surface S to the plane is:

6.928 - 1

= 5.928