Let $\theta$ denote the angle between the planes $2x+2y+z = 1$ and $-x + 2y - 2z = 2$. The planes intersect in a line that we call L. Find the plane that contains L and forms acute angle $\theta/2$ with both planes.
First of all, $n_1 = <2, 2, 1>$ and $n_2 = <-1, 2, -2>$. Then $\cos \theta$ gives that $\theta = 90°$, using the dot product.Thus the angle the plane in question forms with the two planes would be $45°$ each.
I'm unsure about what to do next, however. Let $n_3$ be the normal to the plane the problem asks for—plane 3. Depending on how I draw the 3 planes, it seems that I could either get $n_3 = n_2 - n_1$ (or $n_3 = n_1 - n_2$; the key is $n_3$ is the difference of the normals of two given planes) or $n_3 = n_1 + n_2$. And computing in these two different ways give different planes. Which one would I know is the correct approach (without using any of the graphing calculators etc.)?
Or would both planes be accepted as the answer, since in this case $\theta = 90°$? What if $\theta$ was an acute or obtuse angle, however? How would my problem-solving approach change, to still get the plane that contains L and forms $\theta/2$ (acute) angle with the two planes?