plane intersection?

Question

Find the angle of intersection of the plane 3x + 3y -4z=-3 with the plane 3x-2y+3z=4

i've already found
n1 = <3,3,-4> & n2 = <3,-2,3>
cos(theta) = (n1*n2)/(|n1| |n2|)
theta = cos ^-1 (-9/sqrt(748))
in radians i get 1.9 and degrees i get 109
but when i type them in as answers i get it wrong.. what did i do wrong??

Answer 1

Where two planes intersect, they create two sets of vertical angles - either both of these angles are right, or one is obtuse and one is acute. In general, where the problem asks for the angle of intersection, it is asking for the _acute_ angle. You returned the obtuse angle. The correct answer would be the supplement of that angle, which is 180°-109.2° or 70.8°.