# Angle between a line on a plane and its projection on to a different plans

Two planes π1 and π2 intersect in a line l. The angle between planes π1 and π2 is 45◦.Let A be a point on l, and let m be the line in plane π1 passing through A that is perpendicular to l. Let B be a point in plane π1 so that the angle between line m and line AB is 45◦,and let P be the projection of B onto plane π2. Find the angle between lines AB and AP.

I know that the cosine of the angle is the dot product of the directional vectors divided by the magnitude of the vectors, but I don't know how to get the directional vectors. I'm pretty sure that the angle APB is equal to 90◦, but how would I either find the directional vectors or find angle ABP.

The angle $$BAP$$ is $$\pi/6$$. One way to see it is to find a frame where $$A$$ is the point $$(0,0,0)$$, $$\pi_1$$ is the plane $$z=0$$, $$\pi_2$$ is the plane $$y=z$$. $$B$$ is the point $$(1,1,0)$$. Then $$P = (1, 1/2, 1/2)$$ and $$AP \cdot AB = \|AP\|\|AB\|\sqrt{3}/2$$
Let the direction of the line L is $$\vec L$$ the vector $$\vec P$$ representing the projection line of L on the the plane with nornal vector as $$\vec N$$ will be such that: $$\vec P$$ is perpendicular to $$\vec N$$ and co-planar with $$\vec L$$ and $$\vec N$$ . Thus, $$\vec P= \hat N \times (\vec L \times \hat N)$$. Next, $$|\vec P|=|\hat N \times \vec L|, \vec P. \vec L= \vec L^2-(\vec L. \hat N)^2=(\vec L \times \hat N)^2.$$ So the angle berween $$\vec L$$ and $$\vec P$$ is given by: $$\cos \theta= \frac{ \vec P. \vec L}{|\vec P|~ |\vec L|}=|\hat L \times \hat N|= \sin \phi \implies \theta=\pi/2- \phi,$$ where $$\phi$$ is angle between the line L and the second plane.