# Simple Plane and Point problem.

• Sep 11th 2012, 08:38 AM
aqualiary
Simple Plane and Point problem.
According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.
• Sep 11th 2012, 08:47 AM
Prove It
Re: Simple Plane and Point problem.
Quote:

Originally Posted by aqualiary
According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.

I assume that your explosion will take on the form of a sphere with centre at (1,1,1) and increasing radius. So you will be looking for the first point that

\displaystyle \begin{align*} (x - 1)^2 + (y - 1)^2 + (z - 1)^2 = r^2 \end{align*} intersects with \displaystyle \begin{align*} x + y + z = 18 \end{align*}.

From the first equation we have

\displaystyle \begin{align*} x^2 - 2x + 1 + y^2 - 2y + 1 + z^2 - 2z + 1 &= r^2 \\ x^2 + y^2 + z^2 - 2\left(x + y + z\right) + 3 &= r^2 \\ x^2 + y^2 + z^2 - 2(18) + 3 &= r^2 \\ x^2 + y^2 + z^2 - 36 + 3 &= r^2 \\ x^2 + y^2 + z^2 &= r^2 + 33 \end{align*}

So the intersection with the plane will happen on the sphere centred at (0,0,0) of radius \displaystyle \begin{align*} \sqrt{r^2 + 33} \end{align*}.
• Sep 11th 2012, 10:11 AM
MaxJasper
Re: Simple Plane and Point problem.
Quote:

Originally Posted by aqualiary
According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.

Plane: $A x+B y+C z+D=0$ with A=B=C=1, D=-18

Perpendicular line to the plane:

$\frac{a}{A}=\frac{b}{B}=\frac{c}{C}$ with a=b=c =perp line direction numbers.

Parametric equation of perp line passing through point (1,1,1):

$X\text{=}1+a t$
$Y\text{=}1+b t$
$Z\text{=}1+c t$

The perp line passes through (1,1,1) eliminating parameter t resulting in line: X=Y=Z...by sustituting in the plane's equation: 3X=3Y=3Z=18 :: X=Y=Z=6...: intersection point on plane (x,y,z)= (6,6,6).

Distance of point (X1,Y1,Z1)=(1,1,1) to the plane is:

$|\frac{A X_1+B Y_1+C Z_1+D}{\sqrt{A^2+B^2+C^2}}| = 5 \sqrt{3}$
• Sep 12th 2012, 05:52 PM
johnsomeone
Re: Simple Plane and Point problem.
Quote:

Originally Posted by aqualiary
According to the "Flat Earth Club", Earth is a plane described by the equation
x + y + z = 18:
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment
is due by an explosion that will spontaneously occur at the so-called \armageddon point"
A = (1; 1; 1). Point B will be the first point destroyed. Calculate Point B.

The problem amounts to finding the point B in the plane that's closest to A in $\mathbb{R}^3$. I'll do it using vectors.

Although this setup might seem long and involved, it's actually, I think, by far the easiest way to do the problem. It just takes doing several such problems to get used to all the vector-stuff, but once you do, it makes everything seem almost trivial. I consider it a vastly more simple and direct than any method that calculates the distance from the point to the plane.

A plane P in $\mathbb{R}^3$ (or any hyper plane in any $\mathbb{R}^n$) is uniquely determined by a point on it, and a non-zero vector normal to it, via

$Plane = \{ \vec{v} \in \mathbb{R}^n \ | \ (\vec{v} - \vec{v_0}) \cdot \vec{n} = 0 \}$, where $\vec{n}$ is a non-zero normal vector, and $\vec{v_0}$ is any vector in that plane.

For this problem, $P = \{ \vec{v} \in \mathbb{R}^3 \ | \ (\vec{v} - \vec{v_0}) \cdot \vec{n} = 0 \}$, where $\vec{n} = \hat{i} + \hat{j} + \hat{k}$, and, say, $\vec{v_0} = 18 \hat{i}$.

(There, I chose $\vec{n}$ using the coefficients (1, 1, 1) of the plane's equation x + y + z = 18, and $\vec{v_0}$ by picking any point in that plane - I choose (18,0,0).)

Let $\vec{A} = \hat{i} + \hat{j} + \hat{k}$, corresponding to A = (1, 1, 1). The vector $\vec{B}$ that's closest to $\vec{A}$ in $P$ is given by:

$\vec{B} = P \cap L$, where $L$ is the perpenticular line from $\vec{A}$ to $P$. But $L \perp P \Rightarrow}$ "direction ("slope") of $L$" is parallel to $\vec{n}$, since $\vec{n} \perp P$.

Thus $L$ is given by $L = \{ \vec{L_0} + t\vec{n} \ | \ t \in \mathbb{R} \}$, where $\vec{L_0}$ is any vector in $L$. Choose $\vec{L_0} = \vec{A}$.

THUS THIS IS THE PROBLEM:
FIND
$\vec{B} \in P \cap L$, where $P = \{ \vec{v} \in \mathbb{R}^3 \ | \ (\vec{v} - \vec{v_0}) \cdot \vec{n} = 0 \}$, and $L = \{ \vec{A} + t\vec{n} \ | \ t \in \mathbb{R} \}$,

and where $\vec{A} = \hat{i} + \hat{j} + \hat{k}, \vec{n} = \hat{i} + \hat{j} + \hat{k},$ and $\vec{v_0} = 18 \hat{i}$.

SOLUTION:
Have $\vec{B} \in P \cap L$. Thus $\vec{B} \in P$, so $(\vec{B} - \vec{v_0}) \cdot \vec{n} = 0$. Also, $\vec{B} \in L$, so $\vec{B} = \vec{A} + t_0\vec{n}$ for some $t_0 \in \mathbb{R}$.

Finding $t_0$ will find $\vec{B}$. Substitute the second equation into the first to get and solve an equation for $t_0$.

Get: $( (\vec{A} + t_0\vec{n}) - \vec{v_0}) \cdot \vec{n} = 0 \Rightarrow ( (\vec{A} - \vec{v_0}) + t_0\vec{n}) \cdot \vec{n} = 0$

$\Rightarrow (\vec{A} - \vec{v_0}) \cdot \vec{n} + t_0 ( \vec{n} \cdot \vec{n} ) = 0 \Rightarrow (\vec{A} - \vec{v_0}) \cdot \vec{n} +|\vec{n}|^2t_0 = 0$

$\Rightarrow |\vec{n}|^2t_0 = (\vec{v_0} - \vec{A}) \cdot \vec{n} \Rightarrow t_0 = \frac{(\vec{v_0} - \vec{A}) \cdot \vec{n} }{|\vec{n}|^2}$.

Now plug in the values $\vec{A} = \hat{i} + \hat{j} + \hat{k}, \vec{n} = \hat{i} + \hat{j} + \hat{k}$, and $\vec{v_0} = 18 \hat{i}$ to find $t_0$:

Have $(\vec{v_0} - \vec{A}) \cdot \vec{n} = ((18 \hat{i}) -(\hat{i} + \hat{j} + \hat{k})) \cdot (\hat{i} + \hat{j} + \hat{k})$

$= (17\hat{i} - \hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = (17)(1) + (-1)(1) + (-1)(1) = 15$.

and $|\vec{n}|^2 = |\hat{i} + \hat{j} + \hat{k}|^2 = (1)^2 + (1)^2 + (1)^2 = 3$.

Thus $t_0 = \frac{(\vec{v_0} - \vec{A}) \cdot \vec{n} }{|\vec{n}|^2} = \frac{15}{3} = 5$.

Therefore $\vec{B} = \vec{A} + t_0\vec{n} = (\hat{i} + \hat{j} + \hat{k}) + 5(\hat{i} + \hat{j} + \hat{k}) = 6\hat{i} + 6\hat{j} + 6\hat{k}$.

The answer to the problem is that the closest point B in that plane to the point A is B = (6, 6, 6).

(There's a math-religion-geek joke there that I didn't catch at first. What's the "armageddon point"? It's 666.)