Acute angle between two planes (shortcut not working)

Question

The point A, B, C, D have co-ordinates as follows $$\ A: (2,1,-2) ~~~~~~B: (4,1,-1)~~~~ ~C: (3,-2,-1)~~~~~~~ D: (3,6,2) $$

The plane R passes through A,B and C and plane Q passes through A,B and D. Find the acute angle between the two planes.

Attempt

I am aware of the fool proof method of finding the acute angle between the normal vectors of the plane. However, I tried finding the angle between AC and AD. Since one plane contains the line AC and the second contains the line AD, finding the angle between these lines should provide me the angle between the planes. However, the answer is not coming correct. Can someone point out the flaw in my method?

Answer 1

The definition of the dihedral angle between the planes $\alpha$ an $\beta$ is as follows: Take a plane $\gamma$ perpendicular to $\alpha$ and $\beta$. $\gamma$ will intersect $\alpha$ and $\beta$ in two lines. The angle of these two lines is the dihedral angle as shown below.

The short cut doesn't work because the lines $AC$ and $AD$ are not (necessarily) in a perpendicular plane.

It is an absolute theorem that the angle between lines belonging to a plane not perpendicular to $\alpha$ and $\beta$ is less than the dihedral angle.

Answer 2

The angle between vectors in a pair of intersecting planes bears little to no relation to the angle between the planes. Consider, for example, the $x$-$y$ and $x$-$z$ planes. The angle between the planes is of course $\pi/2$, but if you take vectors nearly parallel to the $x$-axis, the angle between them is either near zero or near $\pi$.