Finding the shortest distance from a point to the plane

In summary, the formula for finding the shortest distance from a point to a plane is d = |ax<sub>0</sub> + by<sub>0</sub> + cz<sub>0</sub> + d| / √(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>), where (x<sub>0</sub>, y<sub>0</sub>, z<sub>0</sub>) is the coordinates of the point and ax + by + cz + d = 0 is the equation of the plane. The coordinates of the closest point on the plane to a given point can be determined by using the
  • #1
Specter
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Homework Statement


Find the shortest distance from P(-4,2,6) to the plane 2x-3y+z-8=0.

Homework Equations


##|proj_\vec n \vec PQ|=|(\frac {\vec PQ \cdot \vec n} {\vec n \cdot \vec n})| |\vec n|##

The Attempt at a Solution



I kind of had to guess some steps because it was done differently in my lesson and I couldn't figure it out that way. Am I on the right track?

The normal is ##\vec n =(2,-3,1)##

Find another point on the plane, let x=0 and y=0

2x-3y+z-8=0
2(0)-3(0)+z-8=0
z=0

Another point on the plane is Q(0,0,8).

I have to find the direction of ##\vec {PQ}##

##\vec {PQ}=\vec Q - \vec P##
=(0,0,8)-(-4,2,6)
=(4,-2,2)

To find the shortest distance, I think I would project ##\vec {PQ}## onto ##\vec n##.

##|proj_\vec n \vec PQ|=|(\frac {\vec PQ \cdot \vec n} {\vec n \cdot \vec n})| |\vec n|##

=##|\frac {(4,-2,2) \cdot (2,-3,1)} {(2,-3,1) \cdot (2,-3,1)}| |(2,-3,2)|##

=##|\frac {(4)(2)+(-2)(-3)+(2)(1)} {(2)(2)+(-3)(-3)+(1)(1)}| |(2,-3,1)|##

=##|\frac {16} {14}| |(2,-3,1)|##

=##|\sqrt (\frac {16} {7}), (-\frac {24} {7}), (\frac {8} {7})|##

=##\sqrt (\frac {16} {7})^2 + (-\frac {24} {7})^2 + (\frac {8} {7})^2##

=##\sqrt \frac {256} {49} + \frac {576} {49} + \frac {64} {49}##

=##\sqrt \frac {896}{49}##

The shortest distance is ##\sqrt \frac {896}{49}##
 
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  • #2
I learned the formula ## d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} ## and I get ## d=\frac{16}{\sqrt{14}} ##. With a little arithmetic, (your numerator is divisible by 7, and ## 896=64 \cdot 14 ##), your answer agrees with mine. And yes, your method works.
 
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  • #3
Charles Link said:
I learned the formula ## d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} ## and I get ## d=\frac{16}{\sqrt{14}} ##. With a little arithmetic, (your numerator is divisible by 7, and ## 896=64 \cdot 14 ##), your answer agrees with mine. And yes, your method works.
Thanks! Do you know what that formula is called? That seems like an easier/quicker way to solve these problems.
 
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  • #4
Specter said:
Thanks! Do you know what that formula is called? That seems like an easier/quicker way to solve these problems.
It's a fairly well-known result. It's simply the formula for the distance from a point to a plane in 3 dimensions. There is a 2-dimensional version of it for the distance from a point to a line in the x-y plane: ## d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} ##.
 
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  • #5
Specter said:
Thanks! Do you know what that formula is called? That seems like an easier/quicker way to solve these problems.
The formula is easy to derive.
The equation of the plane is Ax+By+Cz+D=0
The point P is (x1, y1, z1)
The unit normal vector of the plane is ##\vec n = (n_1, n_2, n_3) ##
ni = A/||n|| and ## ||n|| = \sqrt{A^2+B^2+C^2}##
Write the parametric equation of the line that is normal to the plane and goes through P.
x=x1-tn1
y=y1-tn2
z=z1-tn3
The magnitude of t is the distance along the line.
Q(x, y, z) is point of the plane. Substitute x, y, z into the equation of the plane and solve for t.
A(x1 - tn1)+B(y1 - tn2)+C(z1 - tn3)+D=0
##t=\frac{Ax_1+Bx_2+Cx_3+D}{An_1+Bn_2+Cn_3}##, but
##An_1+Bn_2+Cn_3=\frac{A^2+B^2+C^2}{||n||}=\sqrt{A^2+B^2+C^2}##
The shortest distance is the length of the normal line
##t=\frac{Ax_1+Bx_2+Cx_3+D}{\sqrt{A^2+B^2+C^2}}##

upload_2018-12-6_6-43-5.png
 

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  • #6
Charles Link said:
It's a fairly well-known result. It's simply the formula for the distance from a point to a plane in 3 dimensions. There is a 2-dimensional version of it for the distance from a point to a line in the x-y plane: ## d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} ##.

I tried solving it this way and got a bit of a different answer. Might be an obvious mistake but this is my first math course in a while!

##d= \frac {|A_x+B_y+C_z+D|} {\sqrt (A)^2+(B)^2+(C)^2}##

## d=\frac {|2(-4)+3(2)+1(6)-8|} {\sqrt (2)^2+(3)^2+(1)^2}##

##d=\frac {-4} {\sqrt 14}##
 
  • #7
## B=-3 ## in this problem. You will get ## |-16|=16 ## for the numerator.
 
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  • #8
Charles Link said:
## B=-3 ## in this problem. You will get ## |-16|=16 ## for the numerator.
Ah yes whoops... Thank you.
 
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Related to Finding the shortest distance from a point to the plane

1. What is the formula for finding the shortest distance from a point to a plane?

The formula for finding the shortest distance from a point to a plane is d = |ax0 + by0 + cz0 + d| / √(a2 + b2 + c2), where (x0, y0, z0) is the coordinates of the point and ax + by + cz + d = 0 is the equation of the plane.

2. How do you determine the coordinates of the closest point on the plane to a given point?

The coordinates of the closest point on the plane to a given point can be determined by using the formula (x0, y0, z0) = (-ad/a2 + b2 + c2, -bd/a2 + b2 + c2, -cd/a2 + b2 + c2), where (x0, y0, z0) are the coordinates of the closest point, and ax + by + cz + d = 0 is the equation of the plane.

3. Can the shortest distance from a point to a plane be negative?

No, the shortest distance from a point to a plane cannot be negative. Distance is a scalar quantity and is always positive. The shortest distance is the perpendicular distance from the point to the plane, which cannot be negative.

4. How does the position of the point relative to the plane affect the shortest distance?

The position of the point relative to the plane affects the shortest distance in that if the point lies on the plane, the shortest distance is 0. If the point is above or below the plane, the shortest distance is positive. The shortest distance will be negative if the point is below the plane, but this is not a valid distance as discussed in the previous question.

5. Can the shortest distance from a point to a plane be greater than the distance from the point to the origin?

Yes, the shortest distance from a point to a plane can be greater than the distance from the point to the origin. This can happen if the point is located on the opposite side of the plane from the origin. In this case, the shortest distance is the perpendicular distance from the point to the plane, while the distance from the point to the origin is the distance along the line connecting the two points, which can be longer.

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