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Thread: 3D prism problem

  1. #1
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    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 $\displaystyle x+y+5z=\sqrt{2}$, where A and B are the points $\displaystyle (0,\sqrt{2},0)$ and $\displaystyle (\sqrt{2},0,0)$ 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
    $\displaystyle 5x+5y-2z=5\sqrt{2}$.
    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.
    F(x,y,z)
    AF=BF from this x=y
    AB=AF=BF=2
    Perpendicular distance of F from plane ABCD=$\displaystyle \sqrt 3=AF(\cos \theta)$
    Where $\displaystyle \theta$ is the angle of elevation of F from A.
    $\displaystyle \cos\theta=[(\frac{1}{3\sqrt 3})(\frac{x-\sqrt 2}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)})+(\frac{1}{3\sqrt 3})(\frac{y}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)})+(\frac{5}{3\sqrt 3})(\frac{z}{\surd(x^2-2\sqrt 2x+2+y^2+z^2)})]$
    Fill in the values for the perpendicular distance.
    Then we have $\displaystyle 9=\frac{2x-2\sqrt 2+2y+10z}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)}$
    BF=$\displaystyle \surd (x^2-2\sqrt 2x+2+y^2+z^2)=2$
    So
    $\displaystyle 9=\frac{2x-2\sqrt 2+2y+10z}{2}$
    $\displaystyle 9=x-\sqrt 2+y+5z$
    $\displaystyle 5x+5y=45+5\sqrt 2-25z=5\sqrt 2+2z$ from the plane ABF
    $\displaystyle z=\frac{5}{3}$
    now x=y
    $\displaystyle 2x=9+\sqrt 2-5(\frac{5}{3})$
    $\displaystyle x=y=\frac{1}{3}+\frac{\sqrt 2}{2}$
    So $\displaystyle F(\frac{1}{3}+\frac{\sqrt 2}{2}, \frac{1}{3}+\frac{\sqrt 2}{2}, \frac{5}{3})$
    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
    $\displaystyle -2x=-2y=5z-9-\sqrt 2$

    the answer is supposed to be $\displaystyle -2x=-2y=5z+9-\sqrt 2$
    I can't find where I'm wrong.
    Thanks!
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  2. #2
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    A unit vector perpendicular to ABCD is $\displaystyle \frac1{3\sqrt3}(1,1,5)$. The midpoint of AB is $\displaystyle P = \bigl(\frac{\sqrt2}2,\frac{\sqrt2}2,0\bigr)$. The point F is at a distance $\displaystyle \sqrt3$ from P along this vector. Therefore F is the point $\displaystyle \bigl(\frac{\sqrt2}2\pm\frac13,\frac{\sqrt2}2\pm\f rac13,\pm\frac53\bigr)$. 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 $\displaystyle x+y+5z<\sqrt{2}$ or $\displaystyle x+y+5z>\sqrt{2}$. The origin clearly lies in the first of these, hence so does F, which means that we must take the minus signs. Thus $\displaystyle F = \bigl(\frac{\sqrt2}2-\frac13,\frac{\sqrt2}2-\frac13,-\frac53\bigr)$.

    That should account for the change of sign from –9 to 9 in the final result.
    Last edited by Opalg; Jun 21st 2010 at 10:06 AM.
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