• Write different types of equations (vector, parametric, symmetric) to describe lines in space
• Find the distance from a point to a given line
• Write equations for planes using vectors and scalars
• Find the angle between two planes

Equations for a Line in Space

You’re already comfortable writing equations for lines in two dimensions. In three dimensions, the concept is similar, but instead of using slope to describe direction, we use a direction vector parallel to the line.

Two vectors are parallel when they point in the same or opposite directions. For nonzero vectors [latex]\textbf{u}[/latex] and [latex]\textbf{v}[/latex], this means there exists a scalar [latex]k[/latex] such that [latex]\textbf{u} = k\textbf{v}[/latex].

parallel vectors

Two nonzero vectors [latex]\textbf{u}[/latex] and [latex]\textbf{v}[/latex] are parallel if and only if [latex]\textbf{u} = k\textbf{v}[/latex] for some scalar [latex]k[/latex]. When [latex]k > 0[/latex], the vectors have the same direction. When [latex]k < 0[/latex], they have opposite directions.

The zero vector [latex]\textbf 0[/latex] is considered to be parallel to all vectors.

Vector Equation of a Line

Consider a line [latex]L[/latex] passing through point [latex]P(x_0, y_0, z_0)[/latex] with direction vector [latex]\textbf{v} = \langle a, b, c \rangle[/latex]. For any point [latex]Q(x, y, z)[/latex] on this line, the vector [latex]\overrightarrow{PQ}[/latex] must be parallel to [latex]\textbf{v}[/latex].

This means [latex]\overrightarrow{PQ} = t\textbf{v}[/latex] for some scalar [latex]t[/latex]. Since [latex]\overrightarrow{PQ} = \langle x - x_0, y - y_0, z - z_0 \rangle[/latex], we have:

[latex]\langle x - x_0, y - y_0, z - z_0 \rangle = t\langle a, b, c \rangle[/latex]

Using vector addition, we can rewrite this as:

[latex]\langle x, y, z \rangle = \langle x_0, y_0, z_0 \rangle + t\langle a, b, c \rangle[/latex]

Setting [latex]\textbf{r} = \langle x, y, z \rangle[/latex] and [latex]\textbf{r}_0 = \langle x_0, y_0, z_0 \rangle[/latex], we get the vector equation of a line:

[latex]\textbf{r} = \textbf{r}_0 + t\textbf{v}[/latex]

Parametric and Symmetric Equations

From the vector equation, we can extract three component equations:

• [latex]x - x_0 = ta[/latex]
• [latex]y - y_0 = tb[/latex]
• [latex]z - z_0 = tc[/latex]

Solving for [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex] gives us the parametric equations:

[latex]x = x_0 + ta, \quad y = y_0 + tb, \quad z = z_0 + tc[/latex]

If [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are all nonzero, we can solve each equation for [latex]t[/latex]:

[latex]\frac{x - x_0}{a} = t, \quad \frac{y - y_0}{b} = t, \quad \frac{z - z_0}{c} = t[/latex]

Since all expressions equal [latex]t[/latex], we can set them equal to create symmetric equations:

[latex]\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}[/latex]

line equations in three dimensions

A line [latex]L[/latex] through point [latex]P(x_0, y_0, z_0)[/latex] parallel to vector [latex]\textbf{v} = \langle a, b, c \rangle[/latex] can be described by:

• Vector form: [latex]\textbf{r} = \textbf{r}_0 + t\textbf{v}[/latex]
• Parametric form: [latex]x = x_0 + ta, \quad y = y_0 + tb, \quad z = z_0 + tc[/latex]
• Symmetric form: [latex]\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}[/latex] (when [latex]a, b, c \neq 0[/latex])

 The parametric and symmetric equations of a line are not unique. Using a different direction vector or different point on the line creates equivalent but different-looking equations that describe the same line.

Find parametric and symmetric equations of the line passing through points [latex](1, 4, -2)[/latex] and [latex](-3, 5, 0)[/latex].

Show Solution

First, identify a vector parallel to the line:

[latex]{\bf{v}}=\langle-3-1,5-4,0-(-2)\rangle=\langle-4,1,2\rangle[/latex].

Use either of the given points on the line to complete the parametric equations:

[latex]x=1-4t[/latex], [latex]y=4+t[/latex], and [latex]z=-2+2t[/latex].

Solve each equation for [latex]t[/latex] to create the symmetric equation of the line:

[latex]\frac{x-1}{-4}=y-4=\frac{z+2}2[/latex].

Line Segments

Sometimes you need equations for just a portion of a line, not the entire line. This requires restricting the parameter [latex]t[/latex] to a specific interval.

Consider two points [latex]P(x_0, y_0, z_0)[/latex] and [latex]Q(x_1, y_1, z_1)[/latex] with position vectors [latex]\textbf{p} = \langle x_0, y_0, z_0 \rangle[/latex] and [latex]\textbf{q} = \langle x_1, y_1, z_1 \rangle[/latex]. To find the equation of the line segment between these points, we use [latex]P[/latex] as our reference point and [latex]\overrightarrow{PQ} = \langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \rangle[/latex] as our direction vector.

Starting with the vector equation [latex]\textbf{r} = \textbf{p} + t(\overrightarrow{PQ})[/latex], we can manipulate this algebraically:

[latex]\begin{aligned} \textbf{r} &= \textbf{p} + t(\overrightarrow{PQ}) \\ &= \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)\textbf{p} + t\textbf{q} \end{aligned}[/latex]

This gives us the vector equation for a line segment: [latex]\textbf{r} = (1-t)\textbf{p} + t\textbf{q}[/latex]

We can also derive the parametric equations directly:

[latex]\begin{aligned} \textbf{r} &= \textbf{p} + t(\overrightarrow{PQ}) \\ \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]

Notice what happens at the endpoints:

• When [latex]t = 0[/latex]: [latex]\textbf{r} = \textbf{p}[/latex] (we’re at point [latex]P[/latex])
• When [latex]t = 1[/latex]: [latex]\textbf{r} = \textbf{q}[/latex] (we’re at point [latex]Q[/latex])

line segment equation

The line segment between points [latex]P[/latex] and [latex]Q[/latex] is described by:

• Vector form: [latex]\textbf{r} = (1-t)\textbf{p} + t\textbf{q}, \quad 0 \leq t \leq 1[/latex]
• Parametric form: [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 \leq 1[/latex]

The key difference from a full line is the restriction [latex]0 \leq t \leq 1[/latex], which limits us to just the segment between the two endpoints.

Find parametric equations of the line segment between the points [latex]P(2, 1, 4)[/latex] and [latex]Q(3, -1 ,3)[/latex].

Show Solution

By the Parametric Equations of a Line, we have

[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].

Working with each component separately, we get

[latex]\begin{aligned} x&=x_0+t(x_1-x_0) \\ &=2+t(3-2) \\ &=2+t, \end{aligned}[/latex]

[latex]\begin{aligned} y&=y_0+t(y_1-y_0) \\ &=1+t(-1-1) \\ &=1-2t, \end{aligned}[/latex]

and

[latex]\begin{aligned} z&=z_0+t(z_1-z_0) \\ &=4+t(3-4) \\ &=4-t. \end{aligned}[/latex]

Therefore, the parametric equations for the line segment are

[latex]x=2+t,y=1-2t,z=4-t, \ 0\leq t\leq1[/latex].