From fffdaa2fec6445dc4a2da6eb19186a71c8f9d00a Mon Sep 17 00:00:00 2001
From: nicebyte <12521759+nicebyte@users.noreply.github.com>
Date: Mon, 20 May 2024 21:30:11 -0700
Subject: [PATCH] Create Plane.md

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+# Plane Equation
+
+A plane with a normal vector $\boldsymbol{\overline{N}}$ that is at the distance $D$ from the origin is the set of all points whose projection onto the 
+normal $\boldsymbol{\overline{N}}$ is $D$ units long. In other words, it is the set of points $P$ for which
+$P \cdot \boldsymbol{\overline{N}} = D$, or, written out in coordinates: $P_{x}N_{x} + P_{y}N_{y} + P_{z}N_{z} - D = 0$.
+
+To test whether a particular point $P$ is on the plane, it is enough to compute
+$\langle P_{x}, P_{y}, P_{z}, 1 \rangle \cdot \langle N_{x}, N_{y}, N_{z}, -D \rangle$. The result is the distance from the point to the plane,
+which is $0$ when the point lies exactly on the plane, positive when the point is in front of the plane and negative when 
+the point is behind the plane.
+
+# Transforming Planes
+
+Assume we have a plane in coordinate space $A$ with a normal vector $\boldsymbol{\overline{N}}$, $D$ units asway from the origin. We
+want to transform this representation into its equivalent for coordinate space $B$.
+
+Assume $M_{AB}$ is the matrix representing the transformation from $A$ to $B$. $M_{AB}\boldsymbol{\overline{N}}$ would be the normal in
+the new coordinate space.
+The distance from the new origin will be $M_{AB}\boldsymbol{\overline{N}} \cdot M_{AB}D\boldsymbol{\overline{N}}$
+(because $D\boldsymbol{\overline{N}}$ lies exactly on the plane).
+
+
+Categories: [[:Mathematics]], [[:Geometry]]