Angle Between Two Planes (No Vector and Coordinates Allowed, Only Allowed to Use Non-Coordinate Geometry Tricks)

3dgeometrysolid-geometry

Cube $ABCD.EFGH$ have side length 6 units. P is the midpoint of $EH$, Q is the midpoint of $AD$, the angle between $BFPQ$ and $BDG$ is $\lambda$.

My attempt :

1) Because the line of intersection is outside the cube, i translated $BFPQ$ 3 units forward (from my point of view) so it become $DHP'Q'$ and the line intersection after translation is $DO$.

2) I calculated the distance from $B$ to $DO$ and distance from $Q'$ to $DO$, because in order to find the angle between two plane (without relying on normal vector) you must find 2 lines perpendicular to the line of intersection, Those 2 perpendicular lines in this case are $BO$ and $Q'U$

3) When i checked in GeoGebra it turns out that $BO$ and $Q'U$ are skew to each other. That's what caused my answer to be wrong.

My question is : How do i avoid the 2 lines perpendicular to the lines of intersection being skew to each other (avoid $BO$ and $Q'U$ being skew) ? and if i have to translate the skew lines how do i figure out how far do i need to translate one of the line so that the two lines aren't skew anymore ? is there any other easier way to do this ? of course without vector and coordinate.

Cube

Best Answer

The two lines must be perpendicular to the line of intersection at the same point. In your case, the simplest thing to do is drawing a perpendicular $UV$ on plane $BDG$ (I don't know if point $V$ in your figure is the one I mean) and compute $\lambda=\angle Q'UV$.

EDIT.

Computing $UV$ and $Q'V$ is not difficult:

  • in rectangle $DQ'P'H$ you have $UQ'=3\sqrt{\frac{5}{6}}$ and $DU=\frac{5\sqrt6}{2}=\frac{5}{6}DO$;

  • in equilateral triangle $DBG$ line $UV$ is parallel to $BG$, hence $DV=\frac{5}{6}DB=5\sqrt2$ and $UV=\frac{5}{2}\sqrt2$.

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