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Math Help - tetrahedron volume

  1. #1
    n22
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    tetrahedron volume

    Hello,

    I am having a bit of trouble-Most of the working out has been done .My problem is in green text.Thanks.
    Find the volume of the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through (2,0,0),(0,2,0),(0,0,3).


    first z was integrated first in this question
    but what I dont understand is what is meant by "shadow" ..Please see below.
    we know that the equation of the plane is :
    6x-12+6y+4z Here is a diagram:insert pic.tetrahedron volume-tetrahedron-volume.png


    the normal found was 6i+6j+4k


    now after the z limits were found the y limits were found .
    The solutions I have state that where z=0 a line parallel to the y axis will enter the "shadow" at y=0 and exits throu point on line passing thru the two points (2,0) and (0,2),then equation of this line is y=-x+2.
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  2. #2
    MHF Contributor

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    Re: tetrahedron volume

    It's not standard terminology but I think the "shadow", at z= 0, is the projection of the tetrahedron to the xy-plane, z= 0.

    But the equation of the plane is NOT "6x- 12+ 6y+ 12z"- that's not even an equation! The plane that crosses the x-axis at (2, 0, 0), the y-axis at (0, 2, 0), and the z-axis at (0, 0, 3) has equation 3x+ 3y+ 2z= 6. It projection onto z= 0 is the triangle with vertices at (0, 0, 0), (2, 0, 0), and (0, 2, 0). The three lines of that triangle are x= 0, y= 0, and x+ y= 2 or y= -x+ 2.
    Thanks from n22
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  3. #3
    n22
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    Re: tetrahedron volume

    lol..i forgot the zero..!..

    does one usually think of the shadow projection for the two inner variables?why? does it make it easier to visualizze...should I do this for every question?
    Quote Originally Posted by HallsofIvy View Post
    It's not standard terminology but I think the "shadow", at z= 0, is the projection of the tetrahedron to the xy-plane, z= 0.

    But the equation of the plane is NOT "6x- 12+ 6y+ 12z"- that's not even an equation! The plane that crosses the x-axis at (2, 0, 0), the y-axis at (0, 2, 0), and the z-axis at (0, 0, 3) has equation 3x+ 3y+ 2z= 6. It projection onto z= 0 is the triangle with vertices at (0, 0, 0), (2, 0, 0), and (0, 2, 0). The three lines of that triangle are x= 0, y= 0, and x+ y= 2 or y= -x+ 2.
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