Shortest Distance from point to a plane

Question

Q: Find the shortest distance from the point $A(1,1,1)$ to the plane $2x+3y+4z=5$.

I understand that we need to pick a point P on the plane such that we can form a vector PA.

Can someone explain where to go from here, including a conceptual explanation of deriving the appropriate distance formula?

Thanks

Answer 1

We need a line which is orthogonal to $$2x+3y+4z=5$$

The line $\vec{n}=<a,b,c>$ is normal to the plane $ax+by+cz=d.$ So that's pretty convenient... And then we get the rest of the way there from Bernard's hint!

$A+t\vec{n}$ certainly goes through A when $t =0$ we just need to figure out when it goes through our plane. Then we will have our two points and we will exploit distance formula. Using Bernard's hint we will use some substitutions to achieve $$2x+3y+4z=5$$ $$2(1+2t)+3(1+3t)+4(1+4t)=5$$ $$(2+3+4)+(4+9+16)t=5$$ $$9+29t=5$$ $$t=-\frac{4}{29}$$

Plugging this into $A+t\vec{n}$ finds the point we need and now we can just compare our points using distance formula.

Let's call this new point $B$ $$B=$$ $$A+t\vec{n}$$ $$A(1,1,1)-\frac{4}{29}\vec{n}(2,3,4)$$ $$(1,1,1)-\frac{4}{29}(2,3,4)$$ $$(1-\frac{8}{29},1-\frac{12}{29},1-\frac{16}{29})$$

$$(\frac{21}{29},\frac{17}{29},\frac{13}{29})$$

Then the distance between $A$ and $B$ is $$d(A,B)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$ where $$A=(x_1,y_1,z_1)=(1,1,1) \text{ and }B=(x_2,y_2,z_2)=(\frac{21}{29},\frac{17}{29},\frac{13}{29})$$

Answer 2

So we are looking for some analogy to this problem in the 2d. Let's start with a problem in the 2d. How about the distance from $A(2,2)$ to $2x+3y=5$?

I originally chose $(1,1)$ for the example... but that won't work... whoops.

The first theorem we need to invoke: If we draw a line segment S connecting a point $P$ and a line $L$ we will draw a line which is perpendicular to $P$. What would you do from there?

So we would find the normal line right? There is some line which is perpendicular to $L$ and also that travels through $A$. That would mean that the slope of this line would be the opposite reciprocal of slope of $L$. The slope of $L$ is $-\frac{2}{3}$ which means the slope of the line normal line would be $\frac{3}{2}$ and now we need a line $y=\frac{3}{2}x+ b$ that travels through $A(2,2)$ but that would mean $b=-1$. Now what is the point on $L$ that this line goes through? Well we need to satisfy $y=\frac{3}{2}x-\frac{1}{2}$ and $2x+3y=5$ at the same time! But this can only be one point. And once you have $(\frac{16}{13},\frac{11}{13})$ you apply the distance formula and we're done.

Your puzzle will be to consider what this whole mess looks like in $3d$! But the steps are the same: We need a line orthogonal to the plane $2x+3y+4z=5$ that travels through $A(1,1,1)$. I'll race you to it! See who gets it first!

Answer 3

Hint:

You can parametrise the line through $A$ perpendicular to the plane with normal vector $\vec n$, writing its vector equation as $$M=A+t\mkern2mu\vec n$$ You have to determine the intersection point $P$ of this line with the plane. So the coordinates of $P$ must satisfy the equations: $$\begin{cases} x=1+2t,\quad y=1+3t,\quad z=1+4t ,\\ 2x+3y+4z=5. \end{cases}$$ The shortest distance is just the distance $MP$, i.e. $\,|t|\lVert\vec n\rVert$. You should rediscover the classic formula for the distance $d(M)$ from a point $M(x,y,z)$ to the plane with equation $ax+by+cz+d=0$: $$d(M)=\frac{|ax+by+cz+d|}{\sqrt{a^2+b^2+c^2}}.$$

Answer 4

For any n dimensional point $\in \mathbb{R}^{n}$ we can say the equation of a plane will be $a_1x_1+a_2x_2 + \dots a_nx_n + b = 0$ where $a_i, b \in \mathbb{R}$. Now to find distance from a n-dimensional point $(y_1, y_2, \dots , y_n)$ we will take the the from the point to the plane which is parallel to the gradient and take its magnitude.

The gradient of the plane is $<a_1, a_2, a_3, \dots, a_n>$. Consider a vector from the n dimensional point to a point on the plane : $<y_1-x_1, y_2-x_2, \dots, y_n-x_n>$ which will be parallel to the gradient of the plane and hence will be normal to the plane.

Now we have to solve the equations $$a_1x_1+a_2x_2 + \dots a_nx_n + b = 0$$ and the equation $$\frac{y_1-x_1}{a_1} = \frac{y_2-x_2}{a_2} = \dots \frac{y_n - x_n}{a_n} = \theta$$. Using these 2 equation we can solve for the point $(x_1, x_2, x_3 \dots x_n)$ as follows.

$$y_1-x_1 = a_1\theta$$ $$\vdots$$ $$y_n-x_n = a_n\theta$$

The distance we want is $$d = \sqrt{(y_1-x_1)^2 + (y_2-x_2)^2 + \dots (y_n-x_n)^2}$$

Which is nothing but $$d = |\theta|\sqrt{a_1^2 + a_2^2 + a_3^2 + \dots a_n^2}$$ Subtituting for $\theta$ we get $$\frac{a_1y_1 + a_2y_2 + \dots a_ny_n + b }{(a_1^2+\dots + a_n^2)} =\theta$$

Hence finally the distance is

$$d = \frac{|a_1y_1 + \dots a_ny_n + b|}{\sqrt{a_1^2 + a_2^2 + a_3^2 + \dots a_n^2}}$$