Learning Objectives
- Write the vector, parametric, and symmetric equations of a line through a given point in a given direction, and a line through two given points.
- Find the distance from a point to a given line.
Equations for a Line in Space
Let’s first explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or opposite directions. If two nonzero vectors, [latex]\textbf u[/latex] and [latex]\textbf v[/latex], are parallel, we claim there must be a scalar, [latex]k[/latex], such that [latex]{\bf{u}}=k{\bf{v}}[/latex]. If [latex]\textbf u[/latex] and [latex]\textbf v[/latex] have the same direction, simply choose [latex]k=\frac{||{\bf{u}}||}{||{\bf{v}}||}[/latex]. If [latex]\textbf u[/latex] and [latex]\textbf v[/latex] have opposite directions, choose [latex]k=-\frac{||{\bf{u}}||}{||{\bf{v}}||}[/latex]. Note that the converse holds as well. If [latex]{\bf{u}}=k{\bf{v}}[/latex] for some scalar [latex]k[/latex], then either [latex]\textbf u[/latex] and [latex]\textbf v[/latex] have the same direction [latex](k>0)[/latex] or opposite directions [latex](k<0)[/latex], so [latex]\textbf u[/latex] and [latex]\textbf v[/latex] are parallel. Therefore, two nonzero vectors [latex]\textbf u[/latex] and [latex]\textbf v[/latex] are parallel if and only if [latex]{\bf{u}}=k{\bf{v}}[/latex] for some scalar [latex]k[/latex]. By convention, the zero vector [latex]\textbf 0[/latex] is considered to be parallel to all vectors.
As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the direction vector (Figure 1). Let [latex]L[/latex] be a line in space passing through point [latex]P(x_0, y_0, z_0)[/latex]. Let [latex]{\bf{v}}=\langle a,b,c\rangle[/latex] be a vector parallel to [latex]L[/latex]. Then, for any point on line [latex]Q(x, y, z)[/latex], we know that [latex]\overrightarrow{PQ}[/latex] is parallel to [latex]\textbf v[/latex]. Thus, as we just discussed, there is a scalar, [latex]t[/latex], such that [latex]\overrightarrow{PQ}=t{\bf{v}}[/latex], which gives
Using vector operations, we can rewrite this equation as
[latex]\begin{aligned} \langle x-x_0,y-y_0,z-z_0\rangle&=\langle ta,tb,tc\rangle \\ \langle x,y,z\rangle -\langle x_0,y_0,z_0\rangle&=t\langle a,b,c\rangle \\ \langle x,y,z\rangle&=\langle x_0,y_0,z_0\rangle+t\langle a,b,c\rangle \end{aligned}[/latex].
Setting [latex]{\bf{r}}=\langle x,y,z\rangle[/latex] and [latex]{\bf{r}}_0=\langle x_0,y_0,z_0\rangle[/latex], we now have the vector equation of a line:
[latex]{\bf{r}}={\bf{r}}_0+t{\bf{v}}[/latex].
Equating components, the Vector Equation of a Line shows that the following equations are simultaneously true: [latex]x-x_0=ta[/latex], [latex]y-y_0=tb[/latex], and [latex]z-z_0=tc[/latex]. If we solve each of these equations for the component variables [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex], we get a set of equations in which each variable is defined in terms of the parameter [latex]t[/latex] and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:
[latex]x=x_0+ta \quad y=y_0+tb \quad z=z_0+tc[/latex].
If we solve each of the equations for [latex]t[/latex] assuming [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are nonzero, we get a different description of the same line:
[latex]\frac{x-x_0}a=t \quad \frac{y-y_o}b=t \quad \frac{z-z_0}c=t[/latex].
Because each expression equals [latex]t[/latex], they all have the same value. We can set them equal to each other to create symmetric equations of a line:
We summarize the results in the following theorem.
THEOREM: parametric and symmetric equations of a line
A line [latex]L[/latex] parallel to vector [latex]{\bf{v}}=\langle a,b,c\rangle[/latex] and passing through point [latex]P(x_0, y_0, z_0)[/latex] can be described by the following parametric equations:
[latex]x=x_0+ta, \ y=y_0+tb, \ z=z_0+tc[/latex].
If the constants [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are all nonzero, then [latex]L[/latex] can be described by the symmetric equation of the line:
[latex]\frac{x-x_0}a=\frac{y-y_o}b=\frac{z-z_0}c[/latex].
The parametric equations of a line are not unique. Using a different parallel vector or a different point on the line leads to a different, equivalent representation. Each set of parametric equations leads to a related set of symmetric equations, so it follows that a symmetric equation of a line is not unique either.
Example: equations of a line in space
Find parametric and symmetric equations of the line passing through points [latex](1, 4, -2)[/latex] and [latex](-3, 5, 0)[/latex].
try it
Find parametric and symmetric equations of the line passing through points [latex](1, -3, 2)[/latex] and [latex](5, -2 ,8)[/latex].
Sometimes we don’t want the equation of a whole line, just a line segment. In this case, we limit the values of our parameter [latex]t[/latex]. For example, let [latex]P(x_0, y_0, z_0)[/latex] and [latex]Q(x_1, y_1, z_1)[/latex] be points on a line, and let [latex]{\bf{p}}=\langle x_0,y_0,z_0\rangle[/latex] and [latex]{\bf{q}}=\langle x_1,y_1,z_1\rangle[/latex] be the associated position vectors. In addition, let [latex]{\bf{r}}=\langle x,y,z\rangle[/latex]. We want to find a vector equation for the line segment between [latex]P[/latex] and [latex]Q[/latex]. Using [latex]P[/latex] as our known point on the line, and [latex]\overrightarrow{PQ}=\langle x_1-x_0,y_1-y_0,z_1-z_0\rangle[/latex] as the direction vector equation, the Vector Equation of a Line gives
[latex]{\bf{r}}={\bf{p}}+t \left( \overrightarrow{PQ}\right)[/latex].
Using properties of vectors, then
[latex]\begin{aligned} {\bf{r}}&={\bf{p}}+ t \left(\overrightarrow{PQ}\right) \\ &=\langle x_0,y_0,z_0\rangle+t\langle x_1-x_0,y_1-y_0,z_1-z_0\rangle \\ &=\langle x_0,y_0,z_0\rangle+t(\langle x_1,y_1,z_1\rangle-\langle x_0,y_0,z_0\rangle) \\ &=\langle x_0,y_0,z_0\rangle+t\langle x_1,y_1,z_1\rangle-t\langle x_0,y_0,z_0\rangle \\ &=(1-t)\langle x_0,y_0,z_0\rangle+t\langle x_1,y_1,z_1\rangle \\ &=(1-t){\bf{p}}+t{\bf{q}} \end{aligned}[/latex].
Thus, the vector equation of the line passing through [latex]P[/latex] and [latex]Q[/latex] is
[latex]{\bf{r}}=(1-t){\bf{p}}+t{\bf{q}}[/latex].
Remember that we didn’t want the equation of the whole line, just the line segment between [latex]P[/latex] and [latex]Q[/latex]. Notice that when [latex]t=0[/latex], we have [latex]{\bf{r}}={\bf{p}}[/latex], and when [latex]t=1[/latex], we have [latex]{\bf{r}}={\bf{q}}[/latex]. Therefore, the vector equation of the line segment between [latex]P[/latex] and [latex]Q[/latex] is
[latex]{\bf{r}}=(1-t){\bf{p}}+t{\bf{q}}[/latex], [latex]0\leq t\leq1[/latex].
Going back to the Vector Equation of a Line, we can also find parametric equations for this line segment. We have
[latex]\begin{aligned} {\bf{r}}&={\bf{p}}+\left(\overrightarrow{PQ}\right) \\ \langle x,y,z\rangle &=\langle x_0,y_0,z_0\rangle+t\langle x_1-x_0,y_1-y_0,z_1-z_0\rangle \\ &=\langle x_0+t(x_1-x_0),y_0+t(y_1-y_0),z_0+t(z_1-z_0)\rangle \end{aligned}[/latex].
Then, the parametric equations are
[latex]x=x_0+t(x_1-x_0)[/latex], [latex]y=y_0+t(y_1-y_0)[/latex], [latex]z=z_0+t(z_1-z_0)[/latex], [latex]0\leq t\leq1[/latex].
Example: parametric equations of a line segment
Find parametric equations of the line segment between the points [latex]P(2, 1, 4)[/latex] and [latex]Q(3, -1 ,3)[/latex].
try it
Find parametric equations of the line segment between points [latex]P(-1, 3, 6)[/latex] and [latex]Q(-8, 2, 4)[/latex].
Distance between a Point and a Line
We already know how to calculate the distance between two points in space. We now expand this definition to describe the distance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculate these distances. When building a home, for example, builders must consider “setback” requirements, when structures or fixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned about the distances between populated areas and proposed flight paths.
Let [latex]L[/latex] be a line in the plane and let [latex]M[/latex] be any point not on the line. Then, we define distance [latex]d[/latex] from [latex]M[/latex] to [latex]L[/latex] as the length of line segment [latex]\overline{MP}[/latex], where [latex]P[/latex] is a point on [latex]L[/latex] such that [latex]\overline{MP}[/latex] is perpendicular to [latex]L[/latex] (Figure 2).
By Theorem: Area of a Parallelogram, vectors [latex]\overrightarrow{PM}[/latex] and [latex]\textbf v[/latex] form two sides of a parallelogram with area [latex]||\overrightarrow{PM}\times{\bf{v}}||[/latex]. Using a formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:
[latex]||\overrightarrow{PM}\times{\bf{v}}||=||{\bf{v}}||d[/latex].
We can use this formula to find a general formula for the distance between a line in space and any point not on the line.
THEOREM: distance from a point to a line
Let [latex]L[/latex] be a line in space passing through point [latex]P[/latex] with direction vector [latex]\textbf v[/latex]. If [latex]M[/latex] is any point not on [latex]L[/latex], then the distance from [latex]M[/latex] to [latex]L[/latex] is
[latex]d=\frac{||\overrightarrow{PM}\times{\bf{v}}||}{||{\bf{v}}||}[/latex].
Example: calculating the distance from a point to a line
Find the distance between t point [latex]M=(1, 1, 3)[/latex] and line [latex]\frac{x-3}4=\frac{y+1}2=z-3[/latex].
try it
Find the distance between point [latex](0, 3, 6)[/latex] and the line with parametric equations [latex]x=1-t[/latex], [latex]y=1+2t[/latex], [latex]z=5+3t[/latex].
Watch the following video to see the worked solution to the above Try IT.
Relationships between Lines
Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a single point. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the lines are said to be skew lines (Figure 5).
Example: classifying lines in space
For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.
- [latex]L_1[/latex]: [latex]x=2s-1[/latex], [latex]y=s-1[/latex], [latex]z=s-4[/latex]
[latex]L_2[/latex]: [latex]x=t-3[/latex], [latex]y=3t+8[/latex], [latex]z=5-2t[/latex] - [latex]L_1[/latex]: [latex]x=-y=z[/latex]
[latex]L_2[/latex]: [latex]\frac{x-3}2=y=z-2[/latex] - [latex]L_1[/latex]: [latex]x=6s-1[/latex], [latex]y=-2s[/latex], [latex]z=3s+1[/latex]
[latex]L_2[/latex]: [latex]\frac{x-4}6=\frac{y+3}{-2}=\frac{z-1}3[/latex]
try it
Describe the relationship between the lines with the following parametric equations:
[latex]x=1-4t[/latex], [latex]y=3+t[/latex], [latex]z=8-6t[/latex]
[latex]x=2+3s[/latex], [latex]y=2s[/latex], [latex]z=-1-3s[/latex].
Candela Citations
- CP 2.45. Authored by: Ryan Melton. License: CC BY: Attribution
- Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/books/calculus-volume-3/pages/1-introduction. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction