In $\Bbb R^3$, prove if a line intersects two non-parallel planes, the difference between the two angles it forms with the two planes is no greater than the angle formed by the two planes themselves.
Here, "angles" always mean the acute or right angle, i.e. non-obtuse angle.
It is quite intuitive. But when I tried formalising it using vectors, I encountered trouble. So I suppose the unit normal vectors of the two planes are $n_{1,2}$ and the unit directional vector of the line is $b$. I'm aware that the angles the line forms with the planes are $\arcsin |n_{1,2} \cdot b|$ respectively. And that formed by the two planes is $\arccos |n_1\cdot n_2|$. By using the formula $$\cos(x -y)=\cos x\cos y +\sin x \sin y$$ I was able to convert the problem to showing that $$|n_1\cdot n_2|\le|b\cdot n_1|+|b\cdot n_2|+\sqrt{(1-|b\cdot n_1|^2)(1-|b\cdot n_2|^2)}$$ The complexity is well beyond my expectations.
Is there any simple approach?
set $$\vec{n_1}=(A_1,B_1,C_1)$$ $$\vec{n_2}=(A_2.B_2,C_2)$$ $$\vec{a}=(a_1,b_1,c_1)$$ then $$\sin(\alpha)=\frac{|a_1A_1+B_1b_1+C_1c_1|}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{a_1^2+b_1^2+c_1^2}}$$ $$\sin(\beta)=\frac{|a_1A_2+B_2b_1+C_2c_1|}{\sqrt{A_2^2+B_2^2+C_2^2}\sqrt{a_1^2+b_1^2+c_1^2}}$$ $$\gamma=\arccos\left(\frac{|A_1A_2+B_1B_2+C_1C_2|}{\sqrt{A_1^2+B_1^2+C_2^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right)$$ Can you finish?