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The direction of a two-dimensional(2D) plane in three-dimensional space(Euclidean space) is defined by the normal vector to this plane. And any (N-1)-dimensional hypersurface in N-dimensional can use the similar definition.

However, what is the best definition for the direction of a 2D plane in a high dimensional space? For example, a 2D plane in the four-dimensional space. The "normal vector" for this plane is not a vector, but also a "plane". The reason that I want to ask this question is I want to calculate the "angle" between two 2D planes in high dimensional space. Is there any definition for this? For example, one plane is constructed by v1 and v2, and another plane is constructed by v3 and v4, what is the "angle" between these two planes in N-dimensional?

Qi Zhong
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  • If the planes intersect, then you can just deal with the three-dimensional subspace spanned by them and compute the angle as you would in $\mathbb R^3$. However, just as you can have skew lines in $\mathbb R^3$, you can have skew planes in $\mathbb R^4$. How would you define the angle between such lines and planes? – amd Mar 26 '18 at 20:49
  • For skew lines in R3, we can use dot product OR move/translate one line to intersect with another to find the angle between them. For skew planes in R4, we need to find the directional vector or normal vector of planes, then use the dot product. But now we do not have the definition for them. Or, we can translate one plane to intersect with another, then deal with the three-dimensional subspace spanned by them as you said. But is this a standard method? And what if I only know the vectors span the two planes as I mentioned in the question? Is there any equation to find the angle directly? @amd – Qi Zhong Mar 26 '18 at 21:00
  • I will add that three-dimensional space is special in that you can identify a “direction” vector for any plane. As you’ve noted, the orthogonal complement of a plane in higher-dimensional spaces is not one-dimensional, which is why one can’t in general speak of a rotation axis but only of a plane of rotation when moving beyond 3D. – amd Mar 26 '18 at 21:13
  • I don’t know if there’s any “standard” since one could also argue that the angle between two skew lines or planes is undefined. Translating to force an intersection so that there is a dihedral angle is reasonable, but I would translate both flats so that they include the origin. – amd Mar 26 '18 at 21:14

1 Answers1

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Exterior algebra fulfills this need. If two planes passing through the origin are generated by vectors $\boldsymbol{v}_1,\boldsymbol{v}_2$ and $\boldsymbol{v}_3,\boldsymbol{v}_4$, then $\boldsymbol{v}_1\wedge\boldsymbol{v}_2$ and $\boldsymbol{v}_3\wedge\boldsymbol{v}_4$ are blades representing the planes that they generate, $$\langle \boldsymbol{v}_1\wedge \boldsymbol{v}_2 ,\boldsymbol{v}_3\wedge \boldsymbol{v}_4\rangle=\begin{vmatrix} \langle\boldsymbol{v}_1,\boldsymbol{v}_3\rangle &\langle\boldsymbol{v}_1,\boldsymbol{v}_4\rangle\\ \langle\boldsymbol{v}_2,\boldsymbol{v}_3\rangle &\langle\boldsymbol{v}_2,\boldsymbol{v}_4\rangle \end{vmatrix}$$ is an inner product, and $$\cos\theta=\frac{\langle \boldsymbol{v}_1\wedge \boldsymbol{v}_2 ,\boldsymbol{v}_3\wedge \boldsymbol{v}_4\rangle}{\sqrt{\langle \boldsymbol{v}_1\wedge \boldsymbol{v}_2 ,\boldsymbol{v}_1\wedge \boldsymbol{v}_2\rangle \langle \boldsymbol{v}_3\wedge \boldsymbol{v}_4 ,\boldsymbol{v}_3\wedge \boldsymbol{v}_4\rangle}}\text{.}$$ gives the angle between the planes.

K B Dave
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  • Thanks. This is exactly what I want. Additional question, if the vectors are complex vectors, i.e., in complex space, then should I any modification on the above equation? – Qi Zhong Mar 26 '18 at 21:26
  • I believe you take the absolute value of the right side of the second equation. More generally, there's a "standard" metric on $k$-dimensional subspaces of $n$-dimensional space, real or complex. – K B Dave Mar 26 '18 at 21:34
  • Dave, do you know any references(books or articles) that have these two equations? – Qi Zhong May 01 '18 at 16:11