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Math Help - Simple Plane and Point problem.

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    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.
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    Re: Simple Plane and Point problem.

    Quote Originally Posted by aqualiary View Post
    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*}.
    Last edited by Prove It; September 11th 2012 at 07:50 AM.
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    Lightbulb Re: Simple Plane and Point problem.

    Quote Originally Posted by aqualiary View Post
    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}
    Last edited by MaxJasper; September 11th 2012 at 11:28 AM.
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    Re: Simple Plane and Point problem.

    Quote Originally Posted by aqualiary View Post
    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.)
    Last edited by johnsomeone; September 12th 2012 at 05:45 PM.
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