Parametric Lines from a Point and Direction

Describing a Line by a Point and a Direction

A line in three-dimensional space can be specified by giving one point on it and a nonzero vector pointing along it. The point tells us where the line is anchored; the direction vector tells us how its points are arranged. Together, they provide a compact way to describe every point on the line.

The parameter in a parametric equation is a real number that determines how far to move from the given point in the direction of the vector. It is not necessarily a distance: its scale depends on the length of the chosen direction vector.

Definition: Let \(P\) be a point with position vector \(\mathbf{p}\), and let \(\mathbf{v}\ne\mathbf{0}\) be a direction vector. The line through \(P\) in direction \(\mathbf{v}\) consists of the points whose position vectors are \[ \mathbf{r}=\mathbf{p}+t\mathbf{v},\qquad t\in\mathbb{R}. \] This is the vector parametric equation of the line.

The condition \(\mathbf{v}\ne\mathbf{0}\) matters. If \(\mathbf{v}=\mathbf{0}\), then \(\mathbf{p}+t\mathbf{v}=\mathbf{p}\) for every \(t\), which describes one point rather than a line.

From the Vector Equation to Components

In three dimensions, write the point and direction vector as \(\mathbf{p}=\langle x_0,y_0,z_0\rangle\) and \(\mathbf{v}=\langle a,b,c\rangle\), where at least one of \(a,b,c\) is nonzero. If \(\mathbf{r}=\langle x,y,z\rangle\), then the vector equation says that the position vector of a point on the line is obtained by adding \(t\mathbf{v}\) to \(\mathbf{p}\).

\begin{aligned} \langle x,y,z\rangle &=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle\\ &=\langle x_0+ta,\ y_0+tb,\ z_0+tc\rangle. \end{aligned}

Equating corresponding components gives the three parametric equations.

Component form: The line through \((x_0,y_0,z_0)\) with direction vector \(\langle a,b,c\rangle\ne\mathbf{0}\) has parametric equations \[ x=x_0+ta,\qquad y=y_0+tb,\qquad z=z_0+tc,\qquad t\in\mathbb{R}. \] The same parameter \(t\) must be used in all three equations.

Each coordinate changes linearly with \(t\). At \(t=0\), the point is the given point. Increasing or decreasing \(t\) moves along the line in one of the two directions parallel to \(\mathbf{v}\). Since \(t\) ranges over all real numbers, the equations describe the entire line, not only a ray or a segment.

Why the Parametric Form Describes Exactly a Line

To see why the formula works, consider any real value of \(t\). The difference between the position vector it produces and the starting point is

\mathbf{r}-\mathbf{p}=t\mathbf{v}.

This difference is a scalar multiple of \(\mathbf{v}\), so the resulting point lies along the direction of the line through \(P\). Conversely, any point on that line has a displacement from \(P\) parallel to \(\mathbf{v}\). Such a displacement is \(t\mathbf{v}\) for some real number \(t\), so that point is included in the equation. Thus the equation gives both points on the line and every point on the line.

Worked Example: Write a Line in Vector and Component Form

Find parametric equations for the line through \(P=(2,-1,3)\) with direction vector \(\mathbf{v}=\langle 4,2,-2\rangle\). Then find the point corresponding to \(t=1\) and check it.

The position vector of \(P\) is \(\mathbf{p}=\langle 2,-1,3\rangle\). Since \(\mathbf{v}\ne\mathbf{0}\), the line is given by

\mathbf{r}=\langle 2,-1,3\rangle+t\langle 4,2,-2\rangle,\qquad t\in\mathbb{R}.

Multiplying the direction vector by \(t\) and adding componentwise gives

x=2+4t,\qquad y=-1+2t,\qquad z=3-2t.

At \(t=1\), the coordinates are \(x=2+4=6\), \(y=-1+2=1\), and \(z=3-2=1\). The resulting point is \((6,1,1)\). To check, subtract the starting point: \((6,1,1)-(2,-1,3)=(4,2,-2)\), exactly the direction vector used for one unit of parameter.

Obtaining a Direction from Two Points

If two distinct points are given, their difference provides a direction vector for the line through them. If their position vectors are \(\mathbf{a}\) and \(\mathbf{b}\), then \(\mathbf{b}-\mathbf{a}\) points from the first point to the second. Because the points are distinct, this difference is nonzero and can be used in the line equation. Either point may serve as the starting point.

Worked Example: Parametrize the Line Through Two Points

Find vector and component parametric equations for the line through \(A=(-2,1,0)\) and \(B=(1,5,-2)\). Check that both points are included.

Subtract the coordinates of \(A\) from those of \(B\) to obtain a direction vector:

\overrightarrow{AB} =\langle 1-(-2),\ 5-1,\ -2-0\rangle =\langle 3,4,-2\rangle.

Using \(A\) as the starting point, the vector equation is

\mathbf{r}=\langle -2,1,0\rangle+t\langle 3,4,-2\rangle,\qquad t\in\mathbb{R}.

Therefore, the component equations are \(x=-2+3t\), \(y=1+4t\), and \(z=-2t\). At \(t=0\), these give \((-2,1,0)=A\). At \(t=1\), they give \((-2+3,1+4,-2)=(1,5,-2)=B\). Thus both specified points lie on the parametrized line.

Interpreting the Parameter

The parameter records a signed multiple of the direction vector added to the starting point. For example, \(t=0\) gives \(\mathbf{p}\), \(t=1\) gives \(\mathbf{p}+\mathbf{v}\), and \(t=-1\) gives \(\mathbf{p}-\mathbf{v}\). A value such as \(t=2\) gives \(\mathbf{p}+2\mathbf{v}\).

In the two-point example, the direction vector has length \(\sqrt{3^2+4^2+(-2)^2}=\sqrt{29}\). Therefore, increasing \(t\) by \(1\) moves a distance \(\sqrt{29}\), not a distance of \(1\). More generally, a change of parameter from \(t_1\) to \(t_2\) changes the position vector by \((t_2-t_1)\mathbf{v}\); the distance moved is \(|t_2-t_1|\|\mathbf{v}\|\). The parameter itself represents distance only if the direction vector has length \(1\), with its sign indicating orientation along the line.

Equivalent Parametrizations

A geometric line does not depend on which point on it is chosen as the starting point or which nonzero scalar multiple of its direction vector is used. These choices change the equation’s appearance, but not the set of points it describes.

Theorem: Two parametric equations \(\mathbf{r}=\mathbf{p}+t\mathbf{v}\) and \(\mathbf{r}=\mathbf{q}+s\mathbf{w}\), with nonzero direction vectors \(\mathbf{v}\) and \(\mathbf{w}\), describe the same line exactly when there are real numbers \(a\) and \(c\), with \(c\ne 0\), such that \[ \mathbf{q}=\mathbf{p}+a\mathbf{v} \qquad\text{and}\qquad \mathbf{w}=c\mathbf{v}. \]

For the forward direction, suppose these conditions hold. Substituting them into the second equation gives

\mathbf{r} =\mathbf{q}+s\mathbf{w} =\mathbf{p}+a\mathbf{v}+sc\mathbf{v} =\mathbf{p}+(a+cs)\mathbf{v}.

As \(s\) ranges over all real numbers, \(a+cs\) also ranges over all real numbers because \(c\ne0\). Hence the second equation traces exactly the same points as the first.

For the reverse direction, suppose the two equations describe the same line. The point \(\mathbf{q}\), obtained by setting \(s=0\), must then be on the first line, so \(\mathbf{q}=\mathbf{p}+a\mathbf{v}\) for some real \(a\). The point obtained by setting \(s=1\) is also on that line. Its displacement from \(\mathbf{q}\) is \(\mathbf{w}\), and its displacement from \(\mathbf{p}\) must be some scalar multiple \(b\mathbf{v}\). Therefore, \(\mathbf{q}+\mathbf{w}=\mathbf{p}+b\mathbf{v}\). Using \(\mathbf{q}=\mathbf{p}+a\mathbf{v}\), we get \(\mathbf{w}=(b-a)\mathbf{v}\). Since \(\mathbf{w}\ne\mathbf{0}\), \(b-a\ne0\), as required.

Worked Example: Verify Two Different Equations Give the Same Line

Show that \(\mathbf{r}=\langle 10,3,-1\rangle+s\langle -8,-4,4\rangle\) describes the same line as \(\mathbf{r}=\langle 2,-1,3\rangle+t\langle 4,2,-2\rangle\). Find the relationship between the parameters and check a point from each form.

Let \(\mathbf{p}=\langle 2,-1,3\rangle\) and \(\mathbf{v}=\langle 4,2,-2\rangle\). The second starting point satisfies \(\langle 10,3,-1\rangle=\mathbf{p}+2\mathbf{v}\), because \(\mathbf{p}+2\mathbf{v}=\langle 2+8,-1+4,3-4\rangle=\langle 10,3,-1\rangle\). Its direction vector is \(-2\mathbf{v}\), since \(-2\langle 4,2,-2\rangle=\langle -8,-4,4\rangle\).

Thus the second equation becomes \(\mathbf{r}=\mathbf{p}+2\mathbf{v}-2s\mathbf{v}=\mathbf{p}+(2-2s)\mathbf{v}\). The parameter relationship is \(t=2-2s\). As \(s\) ranges over all real values, \(t\) does too, so both equations describe the same complete line.

For a direct check, set \(s=0\). The second equation gives \(\langle 10,3,-1\rangle\); in the first equation this is \(t=2\), giving \(\langle 2,-1,3\rangle+2\langle 4,2,-2\rangle=\langle 10,3,-1\rangle\). Set \(s=1\) in the second equation: the point is \(\langle 10,3,-1\rangle+\langle -8,-4,4\rangle=\langle 2,-1,3\rangle\), which is \(t=0\) in the first equation.

A Common Mistake: Using Different Parameters in Different Coordinates

Each coordinate equation must use the same parameter value because all three coordinates describe one point reached by a single movement along the direction vector. Using separate, unrelated parameters generally produces points that do not lie on the intended line.

For the line through \((2,-1,3)\) with direction \(\langle 4,2,-2\rangle\), an incorrect setup might use \(x=2+4t\), \(y=-1+2s\), and \(z=3-2u\), with \(t,s,u\) independent. Taking \(t=1\), \(s=0\), and \(u=0\) produces \((6,-1,3)\). Its displacement from \((2,-1,3)\) is \((4,0,0)\), which is not a scalar multiple of \((4,2,-2)\). This point is not on the specified line.

The correction is to use one shared parameter: \(x=2+4t\), \(y=-1+2t\), and \(z=3-2t\). At \(t=1\), these give \((6,1,1)\), whose displacement \((4,2,-2)\) is exactly the direction vector.

Using the Form Carefully

When setting up a parametric line, identify a point on the line and a nonzero direction vector, then place them in \(\mathbf{r}=\mathbf{p}+t\mathbf{v}\). Expand componentwise only after the vector equation is correct. A quick check is to substitute \(t=0\): the result must be the chosen starting point. Substituting \(t=1\) should give that starting point plus one copy of the direction vector.

Changing the starting point or multiplying the direction vector by a nonzero scalar can produce an equivalent equation, but changing just one coordinate’s parameter cannot. Keep the parameter real and common to all components so the equation covers the entire line.

Check Your Understanding

Use the point-and-direction form to answer these questions.

Write vector and component parametric equations for the line through \((1,0,-2)\) with direction vector \(\langle -1,3,2\rangle\).
For the line in Question 1, what point corresponds to \(t=-2\)?
Find a direction vector for the line through \((0,2,1)\) and \((3,-1,5)\), and write a vector parametric equation for it.
Why must a direction vector in a line parametrization be nonzero?
What conditions on the starting points and direction vectors ensure two parametric equations describe the same line?