Kuzee
Vector Algebra and Analytic Geometry of Space

Distance Formulas

The Distance from a Point to a Plane

The distance formula from a point to a plane

The distance from the point P(x0,y0,z0)P(x_0, y_0, z_0) to the plane Ax+By+Cz+D=0Ax + By + Cz + D = 0 is:

d=∣Ax0+By0+Cz0+D∣A2+B2+C2d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}

The Geometric Meaning

The distance from a point to a plane is the length of the perpendicular from the point to the plane.

Special Cases

  • when the point lies on the plane, the distance is 0
  • the distance is always a non-negative number

The Distance from a Point to a Line

The distance formula from a point to a line

The distance from the point P(x0,y0,z0)P(x_0, y_0, z_0) to the line x−x1l=y−y1m=z−z1n\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} is:

d=∣PP1⃗×s⃗∣∣s⃗∣d = \frac{|\vec{PP_1} \times \vec{s}|}{|\vec{s}|}

where PP1⃗=(x0−x1,y0−y1,z0−z1)\vec{PP_1} = (x_0 - x_1, y_0 - y_1, z_0 - z_1) and s⃗=(l,m,n)\vec{s} = (l, m, n).

A Vector Derivation

This uses the geometric meaning of the cross product: ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta

The distance from a point to a line equals the length of the perpendicular from the point to the line.

The Distance between Two Parallel Planes

The distance formula between two parallel planes

The distance between the parallel planes Ax+By+Cz+D1=0Ax + By + Cz + D_1 = 0 and Ax+By+Cz+D2=0Ax + By + Cz + D_2 = 0 is:

d=∣D1−D2∣A2+B2+C2d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}

The Basis of the Derivation

The distance between two parallel planes equals the absolute difference of the constant terms divided by the magnitude of the normal vector.

Techniques for Applying Them

Simplifying the Computation

  1. Unify the form: turn the various equations into a standard form
  2. Choose a coordinate system wisely: pick one that makes the computation easier
  3. Use geometric relationships: use symmetry and perpendicularity to simplify the computation

Common Mistakes

  1. Sign errors: the absolute value in the distance formula cannot be omitted
  2. The unit vector: the normal vector has to be normalised
  3. Substituting the coordinates: watch that the coordinates of the point correspond to the variables of the equation

Exercises

Exercise 1

Find the distance from the point P(2,1,−1)P(2, 1, -1) to the plane x−y+2z−4=0x - y + 2z - 4 = 0.

the distance from a point to a planethe distance formulacoordinate computation

Approach:

Use the distance formula from a point to a plane.

Steps:

  1. d=∣2−1+2×(−1)−4∣12+12+22=∣2−1−2−4∣6=56=566d = \frac{|2 - 1 + 2 \times (-1) - 4|}{\sqrt{1^2 + 1^2 + 2^2}} = \frac{|2 - 1 - 2 - 4|}{\sqrt{6}} = \frac{5}{\sqrt{6}} = \frac{5\sqrt{6}}{6}

Answer: the distance is 566\frac{5\sqrt{6}}{6}

Exercise 2

Find the distance from the point P(1,1,1)P(1, 1, 1) to the plane x+y+z=6x + y + z = 6.

the distance from a point to a planecomputing a distancesimplifying the computation

Approach:

Compute it directly with the distance formula.

Steps:

  1. d=∣1+1+1−6∣12+12+12=33=3d = \frac{|1 + 1 + 1 - 6|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}

Answer: the distance is 3\sqrt{3}

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
ddDistancedistanceThe distance between geometric objects
PP1⃗\vec{PP_1}Vectorvector from P to P1The vector from the point P to the point P1
s⃗\vec{s}Direction vectordirection vectorThe direction vector of the line
×\timesCross productcross productThe cross product of vectors
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