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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The shortest distance of a point (x1, y1, z1) from a plane ax + by + cz + d = 0 is given by |a*x1 + b*y1 + c*z1 + d|/sqrt ( a^2 + b^2 + c^2)

Here the point is (1,0,-2) and the line is x + 2y + z - 4 = 0

The shortest distance between the two is

D = |1*1 + 2*0 + 1*(-2) - 4|/sqrt ( 1 + 4 + 1)

=> D = |1 - 2 - 4|/sqrt 6

=> D = 5/ sqrt 6

The required distance is 5 / sqrt 6

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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