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The equation for the angle between two planes, given their normals are $\vec n_1$ and $\vec n_2$, is $$\theta=cos^{-1}\frac{\vec n_1.\vec n_2}{|\vec n_1||\vec n_2|}$$ Now I got the angle to be $\pi -\theta$, because I considered the direction of one normal to be opposite to the correct direction, as is shown here;

I had been thinking of, say $\vec n_Q$, as pointing in the opposite direction.

Looking this up on Wikipedia, apparently there's something called orientability that can help you choose which normal you need to consider. But there's nothing there about planes; just "real projective planes" and other stuff of that sort I don't know anything about right now.

So how do you choose which normal you use in case of simple 3D planes?

harry
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  • If orientation is not a consideration, you can choose a unique angle $\theta\in[0,\frac12\pi]$ given by $\theta=\arccos\lvert\hat{\mathbf{n}}_1\cdot\hat{\mathbf{n}}_2\rvert$. – user10354138 Apr 23 '21 at 07:02
  • @user10354138: so is the angle between two planes defined to be always the acute angle? – harry Apr 23 '21 at 07:49
  • The angle between two unoriented planes in $\mathbb{R}^3$ is always acute or right, similar to angle between two unoriented lines in $\mathbb{R}^2$. – user10354138 Apr 23 '21 at 07:51
  • user10354138: thanks, got that. But just this; you keep specifying the orientation of planes and lines. Do these simple 3D shapes have orientation, too? And for lines; since there would otherwise be ambiguity, unlike with vectors, we again define the angle to be the acute/right angle. Have I got that right? – harry Apr 23 '21 at 08:01

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The normal of a plane, and the direction of a line, are each unique up to non-zero scalar multiplication.

The angle between two lines, between two planes, and between a line & a plane are each (defined to be) between $0$ and $\frac{\pi}2$ (inclusive); on the other hand, the angle between two vectors is (defined to be) between $0$ and $\pi$ (inclusive).

ryang
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  • Okay, so we just use the acute angle between the normal vectors to describe the angle between the planes, which can be done by taking the absolute value as commented by user10354138. This seems to imply I can get the inter-plane angle using the normal to either side. But if the normal's unique, how do I decide which one the actual, unique 'normal to the plane' is? – harry Apr 23 '21 at 11:34
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    @HarryHolmes If a plane has equation x+y+z=2, it can equivalently be specified as -3x-3y-3z=-6, right? This is because its normal is unique up to non-zero scalar multiplication; here I’d chosen (1 1 1) and (-3 -3 -3) respectively. – ryang Apr 23 '21 at 11:58
  • Oh, okay, got it. So no, planes aren't really orientable, and there are two unit normals that can be specified(and infinitely more with scalar multiplication), and here you can use any pair and just use the acute angle. Have I got it right? – harry Apr 23 '21 at 12:13
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    @HarryHolmes Yes, your given formula should be modified to enclose the fraction between ||, then it will automatically give the non-obtuse angle, as desired. – ryang Apr 23 '21 at 12:36