Results 1 to 2 of 2

Thread: 3D prism problem

  1. #1
    Senior Member
    Jul 2009

    3D prism problem

    The cross section of a right prism is an equilateral triangle. The rectangular face ABCD of the prism lies on the plane x+y+5z=\sqrt{2}, where A and B are the points (0,\sqrt{2},0) and (\sqrt{2},0,0) respectively. EF is the edge in which the other two rectangular faces ADEF and BCEF meet.
    Prove that the equation of the plane containing A, B, F is
    If the origin lies inside the prism, determine the equations of the line EF.

    I have proven the first part, but I can't get the equations of EF.
    AF=BF from this x=y
    Perpendicular distance of F from plane ABCD= \sqrt 3=AF(\cos \theta)
    Where \theta is the angle of elevation of F from A.
    \cos\theta=[(\frac{1}{3\sqrt 3})(\frac{x-\sqrt 2}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)})+(\frac{1}{3\sqrt 3})(\frac{y}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)})+(\frac{5}{3\sqrt 3})(\frac{z}{\surd(x^2-2\sqrt 2x+2+y^2+z^2)})]
    Fill in the values for the perpendicular distance.
    Then we have 9=\frac{2x-2\sqrt 2+2y+10z}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)}
    BF= \surd (x^2-2\sqrt 2x+2+y^2+z^2)=2
    9=\frac{2x-2\sqrt 2+2y+10z}{2}
    9=x-\sqrt 2+y+5z
    5x+5y=45+5\sqrt 2-25z=5\sqrt 2+2z from the plane ABF
    now x=y
    2x=9+\sqrt 2-5(\frac{5}{3})
    x=y=\frac{1}{3}+\frac{\sqrt 2}{2}
    So F(\frac{1}{3}+\frac{\sqrt 2}{2}, \frac{1}{3}+\frac{\sqrt 2}{2}, \frac{5}{3})
    The direction ratios of the line are 5:5:-2 since it is perpendicular to the plane ABF
    the equations for the line I arrive at are
    -2x=-2y=5z-9-\sqrt 2

    the answer is supposed to be -2x=-2y=5z+9-\sqrt 2
    I can't find where I'm wrong.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    A unit vector perpendicular to ABCD is  \frac1{3\sqrt3}(1,1,5). The midpoint of AB is P = \bigl(\frac{\sqrt2}2,\frac{\sqrt2}2,0\bigr). The point F is at a distance \sqrt3 from P along this vector. Therefore F is the point \bigl(\frac{\sqrt2}2\pm\frac13,\frac{\sqrt2}2\pm\f  rac13,\pm\frac53\bigr). Notice the plus/minus signs. They are there because F could be on either side of the plane ABCD. But we are told that the origin lies inside the prism. Therefore F and the origin must lie in the same half-space x+y+5z<\sqrt{2} or x+y+5z>\sqrt{2}. The origin clearly lies in the first of these, hence so does F, which means that we must take the minus signs. Thus F = \bigl(\frac{\sqrt2}2-\frac13,\frac{\sqrt2}2-\frac13,-\frac53\bigr).

    That should account for the change of sign from –9 to 9 in the final result.
    Last edited by Opalg; Jun 21st 2010 at 11:06 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prism Problem...
    Posted in the Geometry Forum
    Replies: 0
    Last Post: Jan 13th 2010, 06:54 PM
  2. Right Prism
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Nov 22nd 2008, 10:29 PM
  3. Prism
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Aug 4th 2008, 05:10 AM
  4. rectangular prism problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Jul 29th 2008, 09:38 PM
  5. Prism
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 10th 2005, 09:16 AM

Search Tags

/mathhelpforum @mathhelpforum