Looks good to me, though I've been having trouble since my hari-kari.
I need to find the angle between the planes x-z=1 and 2y+z=1
So the normal vectors for the planes would be
<1,0,-1) and <0,2,1) Correct?
And the magnitudes of those vectors would be
(2)^(1/2) and (5)^(1/2) Correct?
Thus Cos(Theta)= -1/((2^.5)*(5^.5))
Leaving me with an angle of 1.8925 Radians?
Is this all correct?
When two planes (or two lines) intersect they make four angles but only two distinct angles- the "opposite" (vertical) angles are congruent. The two distinct angles between two planes add to radians. radians.
As to why that formula did not immediately give you the correct answer, note that <-1, 0, 1> is also normal to the first plane and if you had used that, you would have got . I.e., to get the smaller angle, use the absolute value.