Question

Find the shortest distance, d, from the point (5, 0, −6) to the plane x + y + z = 6. d

Answer 1

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

x1 = 10, y1 = 0, z1 = -6 

d = sqrt[(x-10)^2 + y^2 + (z+6)^2] 

The point in plane have the coordinate z = 6 - x - y 

We'll re-write d: 

d = sqrt[(x-10)^2 + y^2 + (6 - x - y+6)^2] 

d = sqrt[(x-10)^2 + y^2 + (12 - x - y)^2] 

The distance d becomes the shortest if minimize the expression: 

d^2 = f(x,y) = [(x-10)^2 + y^2 + (12 - x - y)^2] 

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

fx = 2(x-10)-2(12 - x - y) 

fx = 0 

2x - 20 - 24 + 2x + 2y = 0 

4x + 2y = 44 

2x + y = 22 ...(1) 

fy = 2y -2(12 - x - y) 

fy = 0 

2y - 24 + 2x + 2y = 0 

2x + 4y = 24 

x + 2y = 12 .....(2) 

From (2) x=12-2y substitute this in (1) 

2(12-2y) + y = 22 

24 -4y+y=22 

-3y=-2 

y=2/3 

x=12-2(2/3)=32/3 

There is only one critical point (32/3 ; 2/3). 

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

d = sqrt[(x-10)^2 + y^2 + (6 - x - y)^2] 

d = sqrt[(2/3)^2 + (2/3)^2 + (-18/3)^2] 

d = sqrt(332)/3=2/3 sqrt(83)