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Math Help - shortest diatance between plane and tower

  1. #1
    Eater of Worlds
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    shortest diatance between plane and tower

    Here's a calc III problem(that was my approach) if anyone would like a shot. Not particularly difficult but kind of interesting.

    "An airplane approaches the end of a runway from the east with an angle of 20 degrees. A tower 50 feet high is located 400 feet south and 800 feet east of the runway. How close does the plane come to the top of the tower?".

    Here's a 'paint' drawing. It's a little tilted, but it should suffice.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  2. #2
    Eater of Worlds
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    No one wanted to tackle the problem, huh?.

    Well, here's is the way I done it. I was wondering if someone else had a different method or verification of my solution.

    We want the distance of the point P_{1}(400,800,50) from the line of descent of the plane. The point P_{2}(0,0,0) is on the line and the line has direction vector V=cos(20)j+sin(20)k

    Then we have P_{2}P_{1}=400i+800j+50k

    P_{1}P_{2}\times{V}=\begin{vmatrix}i&j&k\\400&800&  50\\0&cos(20)&sin(20)\end{vmatrix}

    =(800sin(20)-50cos(20))i-400sin(20)j+400cos(20)k

    The distance of P1 from the line using the distance from a point to a line formula.

    \frac{|P_{2}P_{1}\times{V}|}{|V|}

    \frac{|(800sin(20)-50cos(20))|i-400sin(20)j+400cos(20)k|}{|cos(20)j+sin(20)k|}

    =\frac{\sqrt{(800sin(20)-50cos(20))^{2}+(-400sin(20))^{2}+(400cos(20))^{2}}}{\sqrt{cos^{2}(2  0)+sin^{2}(20)}}

    \approx\boxed{459.74} \;\ feet
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