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Write vector equations of lines

A line is represented in vector form as r=a+td\mathbf r=\mathbf a+t\mathbf d, where a\mathbf a is the position vector of a known point, d0\mathbf d\ne\mathbf0 is a direction vector, and tt ranges over all real numbers; from two points, a valid direction vector is their difference. Equivalent equations may use different points or scalar multiples of the direction vector, while restricting tt would describe a ray or segment rather than the entire line. The treatment excludes abstract vector spaces and higher-dimensional generalizations.

Detailed Explanation: Write vector equations of lines

A vector equation for a line has the form

r=a+td,\mathbf r=\mathbf a+t\mathbf d,

where:

  • r\mathbf r is the position vector of any point on the line,
  • a\mathbf a is the position vector of a known point on the line,
  • d\mathbf d is a nonzero direction vector,
  • tRt\in\mathbb R.

When two points are given, subtract their coordinates to find a direction vector.

Example

Find a vector equation for the line through

A(2,1,3)andB(5,1,3).A(2,-1,3) \quad\text{and}\quad B(5,1,-3).

First, use one of the points as the known point. Take AA:

a=(213).\mathbf a= \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix}.

Next, find a direction vector by subtracting the coordinates:

AB=BA=(513)(213)=(326).\overrightarrow{AB}=B-A = \begin{pmatrix} 5\\ 1\\ -3 \end{pmatrix} - \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} = \begin{pmatrix} 3\\ 2\\ -6 \end{pmatrix}.

Therefore, the vector equation of the line is

r=(213)+t(326),tR\boxed{ \mathbf r= \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} +t \begin{pmatrix} 3\\ 2\\ -6 \end{pmatrix}, \qquad t\in\mathbb R }

The value t=0t=0 gives point AA. The value t=1t=1 gives

(213)+(326)=(513),\begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} + \begin{pmatrix} 3\\ 2\\ -6 \end{pmatrix} = \begin{pmatrix} 5\\ 1\\ -3 \end{pmatrix},

which is point BB. Since tt can be any real number, the equation describes the entire line.

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3D Space - Line Vector Form - Two Points to Vector Form


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