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

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08-03-2018
Mathematics

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

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dominic24 dominic24 · Answer 1 · 2018-03-09T18:15:43+01:00

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)

xero099 xero099 · Answer 2 · 2021-09-30T16:22:43+02:00

The shortest distance from the point [tex](x,y,z) = (5,0,-6)[/tex] to the plane [tex]x+y+z = 6[/tex] is [tex]\frac{5\sqrt{3}}{3}[/tex] units.

In this question we shall use the minimum distance formula for a point and a plane, which is defined below:

[tex]d = \frac{|a\cdot x + b\cdot y + c\cdot z + d|}{\sqrt{a^{2}+b^{2}+c^{2}}}[/tex] (1)

Where:

[tex]d[/tex] - Distance.
[tex]a, b, c, d[/tex] - Coefficients of the plane.
[tex](x,y,z)[/tex] - Coordinates of the point.

If we know that [tex]a = b = c = 1[/tex], [tex]d = 6[/tex] and [tex](x,y, z) = (5, 0, -6)[/tex], then the shortest distance is:

[tex]d = \frac{|5 + 0 -6 + 6|}{\sqrt{3}}[/tex]

[tex]d = \frac{5}{\sqrt{3}}[/tex]

[tex]d = \frac{5\sqrt{3}}{3}[/tex]

The shortest distance from the point [tex](x,y,z) = (5,0,-6)[/tex] to the plane [tex]x+y+z = 6[/tex] is [tex]\frac{5\sqrt{3}}{3}[/tex] units.

We kindly invite to see this question on the minimum distance formula: https://brainly.com/question/8622663