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Math Help - Angle between skew lines

  1. #1
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    Thumbs down Angle between skew lines

    Hi folks,

    I have a cuboid ABCDEFGH as shown in the attached diagram on which AB = 4x, AE = 3x and BC = y. If the angle between the skew lines BH and AD is \theta show that  y \sin\theta - 5x \cos\theta = 0.

    Now the angle between two skew lines is found by projecting one line onto a plane that contains the other. So, if I project BH onto the plane ABCD, the projection is line BD and therefore the angle \theta will be \angle ADB.

    \tan\theta = \frac{4x}{y}
    \frac{\sin\theta}{\cos\theta} = \frac{4x}{y}
    y \sin\theta - 4x \cos\theta = 0

    which is wrong. I tried projecting the other way, that is projecting BH onto the plane AEHD so that the projection becomes line AH and the required angle between the skew lines is \angle HAD

    This ends up giving me y \sin\theta - 3x\cos\theta = 0
    wrong again and not even the same as the first result. So, it seems I don't know how to calculate the angle between two skew lines! Can anyone explain?

    regards and thanks
    Attached Thumbnails Attached Thumbnails Angle between skew lines-m017.jpg  
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  2. #2
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    The angle you are looking for should be angle HBC.
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