A similar question was discussed before but it was in a 2-dimensional space. Perpendicular form of the straight line equation. So how can the derivation in a 2D space be generalized to an n-dimensional space? In other words, how can we derive the perpendicular form of a specific line perpendicular to the hyperplane defined as L = A.X1 + B.X2+ C.X3 + ... + K.Xn in an n-dimensional space?

  • $\begingroup$ In spaces with more than two dimensions what you call perpendicular form becomes a system of equations. $\endgroup$ – N74 Dec 19 '18 at 8:42
  • $\begingroup$ A line in more than two dimensions can’t be represented by a single implicit linear Cartesian equation. $\endgroup$ – amd Dec 19 '18 at 9:42
  • $\begingroup$ Thanks a lot. So how about the directions of the hyperplane? What do the cosine directions derived from L's equation represent in the n-dimensional space? And can those directions be used to derive a new coordinate system which is parallel to the hyperplane L in some direction (similar to a rotation in 2D)? I would also appreciate if you recommend me some references on this regard. $\endgroup$ – Mr. Gulliver Dec 19 '18 at 13:07

The vector ${\bf A}=(A_1,A_2, \cdots, A_n)$ is a vector normal to the plane.

Denoting by $\bf X$ the generic point of coordinates $(x_1,\cdots, x_n)$ and by $\bf P$ a given point, then $${\bf(X-P)}=\lambda \bf A$$ is the parametric equation of a line normal to the plane, and passing through $\bf P$.

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  • $\begingroup$ Thanks for your comment. In this sense, can we derive the angles between the normal vector A and the axes of the coordinate system using the above equation similarly to the 2D case (alpha in this answer: math.stackexchange.com/a/740382/627145)? $\endgroup$ – Mr. Gulliver Dec 19 '18 at 12:58
  • $\begingroup$ @Mr.Gulliver: Yes, provided that you take the normalized vector ${\bf A}/{|\bf A|}$, whose components will be the cosines of the angles that the line makes with each axis. $\endgroup$ – G Cab Dec 19 '18 at 16:36
  • $\begingroup$ @G Cab Thank you! $\endgroup$ – Mr. Gulliver Dec 20 '18 at 10:55

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