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Let $H_1$, $H_2$ be two hyperplanes of $\Bbb{R}^n$. Their normal vectors and bias terms are, respectively, given by $\mathbf{w}_1$, $\mathbf{w}_2$, $b_1$, and $b_2$. That is, they are given as $$H_1: \mathbf{w}_1^\top\mathbf{x}+b_1=0 $$ and $$ H_2: \mathbf{w}_2^\top\mathbf{x}+b_2=0. $$ I am looking for a way of "comparing" the above hyperplanes. More specifically, I need to quantify their similarity using some function of their parameters ($\mathbf{w}_i$, $b_i$, $i=1,2$). For instance, a desired function $q(\mathbf{w}_1,\mathbf{w}_2,b_1,b_2)$ would be zero when $H_1$ and $H_2$ coincide.

One such quantity could be the Euclidean norm of the difference between $\mathbf{w}_1$ and $\mathbf{w}_2$, i.e., $q=\lVert\mathbf{w}_1-\mathbf{w}_2\rVert$, but obviously this is not a good choice, since every pair of parallel hyperplanes would lead to $q=0$ (which would mean "absolutely similar").

I am absolutely unaware of such issues; is there any way of quantifying such similarity?

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    $|\langle w_1,w_2 \rangle| +|b_2-b_1|$? – 2017-02-01
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    @DanRust thanks for your comment! But, wouldn't that mean that two hyperplanes that include the origin are "identical" if $\mathbf{w}_1\bot\mathbf{w}_2$? – 2017-02-01
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    Sorry, replace the first term with $\|w_1\|\|w_2\|-|\langle w_1,w_2 \rangle|$, so the full thing is $\|w_1\|\|w_2\|-|\langle w_1,w_2 \rangle| +|b_1-b_2|$. This uses the fact that the Cauchy-Schwarz inequality is only an equality when $w_1$ and $w_2$ are linearly dependent. – 2017-02-01
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    @DanRust this makes sense, indeed. Thanks for your time! I wonder if there is any other approach that uses different norms. I'll wait for some answers. Thanks again :) – 2017-02-01
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    You might want to normalize the vectors $w_i$ before you take the above actually. So $1-|\langle w_1,w_2 \rangle|/\|w_1\|\|w_2\|+|b_1-b_2|$ might be a better behaved measure. – 2017-02-01

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