They're both fine! And yes the angle between two planes with normal vectors and can be calculated as:
1. The planes -x+2y+3z=4 and 2x+3y-z=4 intersect to form a line. Find an equation for the line.
I found two points that satisfy both equations.
(1,1,1) (2, 6/11, 18/11)
The equation is x-1 / 1 = y-2 / -5/11 = z-1 / 7/11
Is it correct?
2. To the nearest tenth of a degree, find the size of the acute dihedral angle formed by the intersecting planes. Hmmm....
I have an approach but I am not sure if it will work.
The perpendicular vector of the plane -x+2y +3z = 4 is
(-1, 2, 3) and the perpendicular vector for the plane 2x+3y-z=4 is (2, 3, -1)
cos(x) = 1/14
x=85.9
I think I read somewhere that you use the dot product of the perpendicular vectors
of a plane to get the dihedral angle.... ?