Math 2.0

Geometry Proposition

Distance from a point to a plane

Builds on Euclidean distanceEquation of a planeNormal vector

Statement

Let aa, bb, cc and dd be real numbers with (a,b,c)≠(0,0,0)(a, b, c) \neq (0, 0, 0), and let P={(x,y,z)∈ℝ3:ax+by+cz=d}P = \{(x, y, z) \in \mathbb{R}^3 : a x + b y + c z = d\} be a plane. Let X=(x0,y0,z0)X = (x_0, y_0, z_0) be a point of ℝ3\mathbb{R}^3. The distance from XX to PP is the smallest Euclidean distance d(X,Q)d(X, Q) for Q∈PQ \in P. Then dist⁡(X,P)=∣ax0+by0+cz0−d∣a2+b2+c2.\operatorname{dist}(X, P) = \frac{|a x_0 + b y_0 + c z_0 - d|}{\sqrt{a^2 + b^2 + c^2}}.

Remarks

The smallest distance is reached at exactly one point of PP, the orthogonal projection of XX on PP. With n=(a,b,c)n = (a, b, c) this point is X−ax0+by0+cz0−da2+b2+c2 nX - \frac{a x_0 + b y_0 + c z_0 - d}{a^2 + b^2 + c^2} \, n. The number (ax0+by0+cz0−d)/∥n∥(a x_0 + b y_0 + c z_0 - d) / \|n\| is the signed distance. Its sign tells on which side of the plane the point is. The same formula holds for a hyperplane {x:a⋅x=d}\{x : a \cdot x = d\} with a≠0a \neq 0 in ℝn\mathbb{R}^n: the distance from x0x_0 is ∣a⋅x0−d∣/∥a∥|a \cdot x_0 - d| / \|a\|.

Reading path

To understand this page from the ground up, read these 4 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
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  1. Equation of a plane concept
  2. Euclidean distance concept
  3. Normal vector concept
  4. Distance from a point to a plane statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane, Converting general problem to distance-from-origin problem
About this page
Tag0K1E2S6Z69
URIm2://0K1E2S6Z69
Kindstatement, proposition
WritingW1 stated
MSC51N20
StatusThis copy of the site has no build data. Run m2 status 0K1E2S6Z69.
UsesEuclidean distance, Equation of a plane, Normal vector
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