# Intersecting Angles of Three Planes [closed]

There are three planes, A, B, and C, all of which intersect at a single point, P. The angles between the planes are given: $$\angle\mathbf{AB}=\alpha$$ $$\angle\mathbf{BC}=\beta$$ $$\angle\mathbf{CA}=\gamma$$ $$0\lt\alpha,\beta,\gamma\le\frac{\pi}{2}$$

The intersection of any two of these planes form lines. The intersection of AB is $\mathbf{\overline{AB}}$, the intersection of BC is $\mathbf{\overline{BC}}$, and the intersection of CA is $\mathbf{\overline{CA}}$. It is given that none of these lines are parallel to each other and that they all intersect at the same single point, P.

Please express the lesser of the two angles formed by the intersection of these lines in terms of $\alpha$, $\beta$, and $\gamma$: $$\angle\mathbf{\overline{AB}\;\overline{BC}}=?$$ $$\angle\mathbf{\overline{AB}\;\overline{CA}}=?$$ $$\angle\mathbf{\overline{BC}\;\overline{CA}}=?$$

## closed as off-topic by Dando18, Namaste, Leucippus, Moishe Kohan, RohanDec 21 '17 at 4:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

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If $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, are unit vectors perpendicular to $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, then $\mathbf{a}\times\mathbf{b}$ is a vector of magnitude $\sin\alpha$ directed along line $\overline{\mathbf{A}\mathbf{B}}$, and so on. The quantity $(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c})$ is then proportional to the cosine of $\angle\overline{\mathbf{A}\mathbf{B}}\ \overline{\mathbf{B}\mathbf{C}}$: $$(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c})= \sin\alpha\sin\beta \cos(\angle\overline{\mathbf{A}\mathbf{B}}\ \overline{\mathbf{B}\mathbf{C}}).$$ On the other hand, according to a well-know vector identity: $$(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c})= (\mathbf{a}\cdot\mathbf{b})\cdot(\mathbf{b}\cdot\mathbf{c})- (\mathbf{a}\cdot\mathbf{c})\cdot(\mathbf{b}\cdot\mathbf{b})= \cos\alpha\cos\beta-\cos\gamma.$$ It follows that $$\cos(\angle\overline{\mathbf{A}\mathbf{B}}\ \overline{\mathbf{B}\mathbf{C}})= \left|{\cos\alpha\cos\beta-\cos\gamma\over\sin\alpha\sin\beta}\right|,$$ where I inserted the absolute value in order to pick the lower angle.
• If I assign a coordinate system such that plane $\mathbf{A}$ is the $\mathbf{xy}$ plane and $\mathbf{B}$ intersects $\mathbf{A}$ along the $\mathbf{x}$ axis, I can get the normal vectors for $\mathbf{A}$ and $\mathbf{B}$, and I can make the origin point $\mathbf{P}$. The problem now is that I cannot see how to get the normal vector for plane $\mathbf{C}$ because I don't know where it is in relation to the other two planes. I only know the angles it makes with the other planes and that it contains the point $\mathbf{P}$, which is the origin. – TheNewGuy Dec 20 '17 at 20:18
If you call the last mentioned angles as $a,b,c$ and the spherical triangle $ABC$ then from Spherical Trigonometry (sphere has center at $P$,) the Law of Sines is valid:
$$\dfrac{\sin \alpha}{\sin a} = \dfrac{\sin \beta}{\sin b} =\dfrac{\sin \gamma}{\sin c}= \dfrac{6\, Volume\, PABC}{\sin a\sin b\sin c }$$