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Math Help - problem involving vectors, planes and points

  1. #1
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    problem involving vectors, planes and points

    question:
    1. 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 31.8.09 by an explosion that will spontaneously occur at the so-called ' armageddon point' A = (1,1,1) in space. it so happens that melbourne (considered as a point) will be the first place on earth that is destroyed by this explosion.
    a) calculate the coordinates of melbourne
    b) what is the distance between A(the armageddon point) and melbourne. what is the distance between A and the earth?

    2. a) according to the flat mars society. mars is also a plane, given by the equation: 4x +3y - z = -3.
    find a parametric equation of the line in which earth (as in question 1) intersects mars.
    b) how far away is canberra, given by the point (-5,10,13), from this line?

    3. Two perfectly round pieces of rock are hurtling through space. The first, Supermanís
    holiday asteroid, has radius 0.3, and is traveling on the line given by the parametric
    equation
    x(t) = 2 + t, y(t) = −1 − t, z(t) = t.
    The second rock, made of kryptonite and set in motion by one of Supermanís enemies,
    has radius 0.1 and is traveling on a line given by the parametric equation
    x(s) = 3 − s, y(s) = 1, z(s) = 1 + s.
    Calculate the distance between the two lines (8 marks) and use this distance to prove
    that Superman has nothing to worry about. (2 marks)
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  2. #2
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    Quote Originally Posted by milkyway View Post
    question:
    1. 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 31.8.09 by an explosion that will spontaneously occur at the so-called ' armageddon point' A = (1,1,1) in space. it so happens that melbourne (considered as a point) will be the first place on earth that is destroyed by this explosion.
    a) calculate the coordinates of melbourne
    b) what is the distance between A(the armageddon point) and melbourne. what is the distance between A and the earth?

    2. a) according to the flat mars society. mars is also a plane, given by the equation: 4x +3y - z = -3.
    find a parametric equation of the line in which earth (as in question 1) intersects mars.
    b) how far away is canberra, given by the point (-5,10,13), from this line?

    3. Two perfectly round pieces of rock are hurtling through space. The first, Superman’s
    holiday asteroid, has radius 0.3, and is traveling on the line given by the parametric
    equation
    x(t) = 2 + t, y(t) = −1 − t, z(t) = t.
    The second rock, made of kryptonite and set in motion by one of Superman’s enemies,
    has radius 0.1 and is traveling on a line given by the parametric equation
    x(s) = 3 − s, y(s) = 1, z(s) = 1 + s.
    Calculate the distance between the two lines (8 marks) and use this distance to prove
    that Superman has nothing to worry about. (2 marks)
    lol, this is a unique one. I'll work on some of it. Since the equation of the plane is x+y+z=18 it is easy to find points. We know that the point P(6,6,6) is on the plane. The armageddon point' A = (1,1,1) in space is on the terminal end of the vector n=<1,1,1> normal to the plane. I think the melbourne point should be the point through which the normal vector will pass through the plane, since this would correspond to the shortest distance between melbourne and "A," Ths distance turns out to be:

    D=\frac{\mid21\mid}{\sqrt{3}}

    I got this from an elementry formula, I will not derive it here. I'll just write it down:

    For the distance D between a point P_o(x_1,y_1,z_1) and a plane ax_1+by_1+cz_1+d =0

    D=\frac{\mid ax_1+by_1+cz_1+d \mid}{\sqrt{a^2+b^2+c^2}} , where, in our case, {a,b,c}=1 are the coefficients of the terms of the plane x+y+z -18=0

    The distance between A and melbourne was calculated by the formula I wrote above. The distance between A and Earth is the same as the distance between A and melbourne since A(1,1,1) can be expressed as a normal vector n=<1,1,1>

    To find the melbourne point we can draw from the origin of our coordinate system, a vector r_o with the terminal end meeting the plain at the melbourne point  P_o. and therefore, also meeting the tail of n=<1,1,1>. We can draw another vector r_1(6,6,6) from the origin to the point P(6,6,6). From P_o we draw a vector r_{\perp}. to P(6,6,6). Since r_{\perp}=r_1 -r_o is on the surface of the plane, it is perpendicular to n=<1,1,1>. We may write:

    n\cdot r_{\perp}=n\cdot (r_1-r_o)=0 Now all we have to do is solve the vector equation for r_o. I will leave that to you.
    Last edited by adkinsjr; August 26th 2009 at 08:14 PM.
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  3. #3
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    this is a unique one
    Hello milkyway
    The teacher who created the scenarios is brilliant.
    It just makes it more appealing.
    Very interesting.
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  4. #4
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    thanks! i figured it all out
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