Determine the shortest distance from the point (1,0,-2) to the plane x+2y+z=4?

We'll write the distance formula form a point to a point
from plane:

d = sqrt[(x-x1)^2 + (y-y1)^2 +
(z-z1)^2]

x1 = 1, y1 = 0, z1 =
-2

d = sqrt[(x-1)^2 + y^2 +
(z+2)^2]

The point in plane have the coordinate z = 4 - x -
2y

We'll re-write d:

d =
sqrt[(x-1)^2 + y^2 + (4  - x - 2y+2)^2]

d = sqrt[(x-1)^2 +
y^2 + (6  - x - 2y)^2]

The distance d becomes the shortest
if minimize the expression:

d^2 = f(x,y) =  [(x-1)^2 + y^2
+ (6  - x - 2y)^2]

To minimize the function f, we'll have
to determine the critical points. For this rason, we'll determine the partial
derivatives:

fx = 2(x-1)-2(6  - x -
2y)

fx = 0

2x - 2 - 12 + 2x +
4y = 0

4x + 4y = 14

2x + 2y =
7 (1)

fy = 2y -4(6  - x -
2y)

fy = 0

2y - 24 + 4x + 8y =
0

4x + 10y = 24

2x + 5y = 12
(2)

(2) - (1) => 2x + 5y - 2x - 2y = 12 -
7

3y = 5

y =
5/3

2x + 10/3 = 7

2x = 7 -
10/3

2x = 11/3

x =
11/6

There is only one critical point (11/6 ;
5/3).

We'll calculate the shortest distance from the given
point to the plane:

d = sqrt[(x-1)^2 + y^2 + (6  - x -
2y)^2]

d = sqrt[(5/6)^2 + (5/3)^2 +
(5/6)^2]

d =
5sqrt6/6

The shortest distance form the point
(1,0,-2) to the plane x+2y+z=4 is d = 5sqrt6/6.