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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.

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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 $AP \cdot AB = \|AP\|\|AB\|\sqrt{3}/2$

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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.

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