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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?

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  • $\begingroup$ in which form are the planes given? $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 17:18
  • $\begingroup$ and the direction vector of the line? $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 17:19
  • $\begingroup$ @Dr.SonnhardGraubner planes are not specified in the original question. I suppose the only relevant information would be their normal vectors. The direction of all these vectors (normal, directional) are not important because we only care about non-obtuse angles. So they can be either way. $\endgroup$ – Vim Nov 11 '17 at 17:27
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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?

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  • $\begingroup$ i think you can complete your proof now $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 17:39
  • $\begingroup$ Thanks for your effort typesetting it... I'm afraid I can't. I could see that these inverse trig functions will lead to a horrible mess. (The goal is prove $|\alpha - \beta|\le\gamma$. $\endgroup$ – Vim Nov 11 '17 at 17:39
  • $\begingroup$ but i have found no better approach $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 17:40
  • $\begingroup$ i think it is also possible to prove that $$\sin(\alpha)+\sin(\beta)\le cos(\gamma)$$ and now substitute the given terms $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 17:42
  • $\begingroup$ no. Take $\alpha=\beta=\pi/4,\gamma=\pi/2$ for example, which is easy to draw. $\endgroup$ – Vim Nov 11 '17 at 17:46

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