Finding the shortest distance from a point to the plane

[Specter]

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 $\sqrt \frac {896}{49}$

 

[Charles Link]

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.

 

[Specter]

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.

 

[Charles Link]

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: $d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.

 

[ehild]

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 $\vec n = (n_1, n_2, n_3)$
n_i = 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
$t=\frac{Ax_1+Bx_2+Cx_3+D}{\sqrt{A^2+B^2+C^2}}$

 

[Specter]

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!

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

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

$d=\frac {-4} {\sqrt 14}$

 

[Charles Link]

$B=-3$ in this problem. You will get $|-16|=16$ for the numerator.

 

[Specter]

$B=-3$ in this problem. You will get $|-16|=16$ for the numerator.

Ah yes whoops... Thank you.