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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 $\displaystyle \theta$ show that $\displaystyle y \sin\theta  5x \cos\theta = 0$.
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 $\displaystyle \theta$ will be $\displaystyle \angle ADB$.
$\displaystyle \tan\theta = \frac{4x}{y}$
$\displaystyle \frac{\sin\theta}{\cos\theta} = \frac{4x}{y}$
$\displaystyle y \sin\theta  4x \cos\theta = 0$
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 $\displaystyle \angle HAD$
This ends up giving me $\displaystyle y \sin\theta  3x\cos\theta = 0$
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

The angle you are looking for should be angle HBC.