Angle between skew lines

s_ingram · #1 $Old$ November 3rd, 2009, 09:33 AM

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

ukorov · #2 $Old$ November 3rd, 2009, 10:51 AM

The angle you are looking for should be angle HBC.