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Homework Help: Points, lines and planes

  1. Aug 17, 2007 #1
    The problem statement, all variables and given/known data

    In this question, Earth is a plane described by the equation
    x + y + z = 18.

    Earth will be destroyed by an explosion that occurs at the point A = (1, 1, 1), also known as the “armageddon point”. It so happens that the school (considered as a point) will be the first place on Earth that is destroyed by this explosion.

    (a) Calculate the coordinates of the school.


    The attempt at a solution
    I don't get what the question is asking, if the school is first place to be destroyed, isn't it simply located at point A (1,1,1) ?
     
  2. jcsd
  3. Aug 17, 2007 #2
    (1,1,16) Is it right?
     
  4. Aug 17, 2007 #3
    @t_n_p
    if u read the lat line then u will note that the question is asking u the point on [y]earth[/u] that will be destroyed first.and note that [tex](1,1,1)[/tex] is not on earth.so the poin on earth taht will be destroyed firsyt should be nearest to [tex](1,1,1)[/tex] ,hence should be on the perpendicular from that point to the plane of earth and must lie on the plane of earth
     
  5. Aug 18, 2007 #4
    Not sure what the answer is yet..:cool:

    Ok, that's cleared up now.
    I'm still unclear on how to find the coordinates of the school though..:confused:
     
  6. Aug 18, 2007 #5

    learningphysics

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    I recommend drawing a sketch...

    You need to find the intersection of the line through (1,1,1) perpendicular to the given plane, and the plane... itself.

    What is the equation of the line?
     
  7. Aug 18, 2007 #6
    Not sure how to find the line that is perpendicular through (1,1,1) and the plane x + y + z = 18. :confused:
     
  8. Aug 18, 2007 #7

    learningphysics

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    What is the normal to the plane? You can get this directly from the equation of the plane.

    Remember that the normal is a vector perpendicular to the plane.
     
  9. Aug 18, 2007 #8
    I know the answer :the coordinates of the school is (6,6,6);
    The direction vector of earth n= (1,1,1) ,and A=(1,1,1) ;Suppose the school P=(x,y,z); the line perpendicular to the plane and through points A and P is :
    x-1=y-1=z-1, denoted as equation (1). In addition, we know x+y+z=18;
    so we can get the coordinates (6,6,6).
    Because of the particularity of the number , in this case, we only need two equations. In general, we can get another equation reflecting the shortest distance from A to the plane.
     
  10. Aug 19, 2007 #9
    Well if ax + by + cz = d is the equation of a plane, then the vector (a, b, c) is a normal vector of the plane.

    In this case, the vector is (1,1,1) which is point A :bugeye:

    What now to find the coordinates of the school? :confused::confused:
     
  11. Aug 19, 2007 #10

    HallsofIvy

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    Find the equations of the line through (1,1,1) with direction vector the same as the normal vector to the plane x+ y+ z= 18. That should be easy and that line will, of course, be perpendicular to the plane. All you have to now is find where that line hits the plane x+ y+ z= 18
     
  12. Aug 19, 2007 #11
    I'm bloody confused at the moment. Can you go through it real slow?
     
  13. Aug 19, 2007 #12
    the point A = (1, 1, 1), also known as the “armageddon point”.
     
  14. Aug 19, 2007 #13
    ?

    learningphysics said "You need to find the intersection of the line through (1,1,1) perpendicular to the given plane, and the plane... itself"

    But I don't know what to do! :bugeye:
     
  15. Aug 19, 2007 #14
    When you know the direction of a line and one point on this line, you can write
    an equation for that line . Then the intersection both belong to the line and the
    plane. you can get that point's coordinates
     
  16. Aug 19, 2007 #15
    Ok, so step by step.
    I know the one point on this line (1,1,1) how do I find the direction of the line?
     
  17. Aug 19, 2007 #16
    Suppose the coordinate of one point ,on a line ,is (a,b,c), and the direction
    vector of the line is n=(A,B,C). Then that line's expression is ((x-a)/A)=((y-b)/B)=((z-c)/C);
     
  18. Aug 19, 2007 #17
    When you know the expression for a plane ,Ax+By+Cz=D. Then the direction vector of this plane is (A,B,C), which also is the direction vector of the line
    perpendicularing to that plane .
     
  19. Aug 19, 2007 #18
    Just see from the expression for the plane
     
  20. Aug 19, 2007 #19

    learningphysics

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    You know that the normal to the plane is (1,1,1)... so that is a vector perpendicular to the plane. So a line that has the direction vector of the normal will be perpendicular to the plane.

    So what is the equation of the line that has the point (1,1,1) and has the direction vector (1,1,1) It is : (x,y,z) = (1,1,1) + (1,1,1)t .

    Find the intersection of this line with the given plane... You'll be solving for t. Then you can get x,y,z.
     
  21. Aug 20, 2007 #20
    So equation of the line is (1+1t)x + (1+1t)y+ (1+1t)z?
    How do I find intersection?

    So unbeleivably confused by this topic at the moment...
     
  22. Aug 20, 2007 #21

    learningphysics

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    No, the parametric equations for the line are
    x = 1 + t
    y = 1 + t
    z = 1 + t

    substitute these values for into your plane equation:

    x + y + z = 18

    then solve for t.
     
    Last edited: Aug 20, 2007
  23. Aug 20, 2007 #22
    I solved for t (t=3). How do I now find the coordinates of the school given I now know t?
     
  24. Aug 20, 2007 #23

    learningphysics

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    Check your algebra. It's not t=3.

    You're trying to find the intersection of the line and the plane. In other words you're trying to find x,y,z that satisfy the equations for the line:

    x = 1 + t
    y = 1 + t
    z = 1 + t

    AND the equation for the plane

    x + y + z = 18

    This is a system of 4 equations and 4 unknowns. So you're trying to find x, y and z.

    Once you've found t, just calculate x, y and z. Just use the equations.
     
  25. Aug 20, 2007 #24
    my bad, that was a typo and meant to be 5! Hence x=6, y=6 and z=6 and coordinates of school are (6,6,6). :devil:

    I know it may seem a stupid question, but I need to refresh my memory. How do I find the distance between 2 parallel lines (specifically, point (1,1,1) and (6,6,6))?
     
  26. Aug 20, 2007 #25

    learningphysics

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    :cool:

    Two lines or two points? The distance between two points (x1,y1,z1) and (x2,y2,z2) is [tex]\sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2}[/tex]

    The distance between two parallel lines is a more complex formula... your text might have it...

    Here's a link:
    http://www.owlnet.rice.edu/~comp360/lectures/ApplicationsofVG.pdf [Broken]

    There's a section that talks about distance between two parallel lines...
     
    Last edited by a moderator: May 3, 2017
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