
# 1.5: Lines and Planes


Now that we know how to perform some operations on vectors, we can start to deal with some familiar geometric objects, like lines and planes, in the language of vectors. The reason for doing this is simple: using vectors makes it easier to study objects in 3-dimensional Euclidean space. We will first consider lines.

## Line through a point, parallel to a vector

Let $$P = (x_{0}, y_{0}, z_{0})$$ be a point in $$\mathbb{R}^{3}$$, let $$\textbf{v} = (a,b,c)$$ be a nonzero vector, and let $$L$$ be the line through $$P$$ which is parallel to $$\textbf{v}$$ (see Figure 1.5.1).

Figure 1.5.1

Let $$\textbf{r} = (x_{0}, y_{0}, z_{0})$$ be the $$\textit{vector}$$ pointing from the origin to $$P$$. Since multiplying the vector $$\textbf{v}$$ by a scalar $$t$$ lengthens or shrinks $$\textbf{v}$$ while preserving its direction if $$t > 0$$, and reversing its direction if $$t < 0$$, then we see from Figure 1.5.1 that every point on the line $$L$$ can be obtained by adding the vector $$t \textbf{v}$$ to the vector $$\textbf{r}$$ for some scalar $$t$$. That is, as $$t$$ varies over all real numbers, the vector $$\textbf{r} + t \textbf{v}$$ will point to every point on $$L$$. We can summarize the $$\textit{vector representation of \(L$$}\) as follows:

For a point $$P = (x_{0}, y_{0}, z_{0})$$ and nonzero vector $$\textbf{v}$$ in $$\mathbb{R}^{3}$$, the line $$L$$ through $$P$$ parallel to $$\textbf{v}$$ is given by
$$\textbf{r} + t \textbf{v}, \text{for} -\infty < t < \infty \label{Eq1.16}$$
where $$\textbf{r} = (x_{0}, y_{0}, z_{0})$$ is the vector pointing to $$P$$.

Note that we used the correspondence between a vector and its terminal point. Since $$\textbf{v} = (a,b,c)$$, then the terminal point of the vector $$\textbf{r} + t \textbf{v}$$ is $$(x_{0} + at, y_{0} + bt, z_{0} + ct)$$. We then get the $$\textit{parametric representation of L}$$ with the $$\textit{parameter}$$ $$t$$:

For a point $$P = (x_{0}, y_{0}, z_{0})$$ and nonzero vector $$\textbf{v} = (a,b,c)$$ in $$\mathbb{R}^{3}$$, the line $$L$$ through $$P$$ parallel to $$\textbf{v}$$ consists of all points $$(x,y,z)$$ given by
$$x = x_{0} + at,\quad y = y_{0} + bt,\quad z = z_{0} + ct, \text{for} -\infty < t < \infty \label{Eq1.17}\] Note that in both representations we get the point $$P$$ on $$L$$ by letting $$t = 0$$. In Equation \ref{Eq1.17}, if $$a \ne 0$$, then we can solve for the parameter $$t$$: $$t = (x - x_{0})/a$$. We can also solve for $$t$$ in terms of $$y$$ and in terms of $$z$$ if neither $$b$$ nor $$c$$, respectively, is zero: $$t = (y - y_{0})/b$$ and $$t = (z - z_{0})/c$$. These three values all equal the same value $$t$$, so we can write the following system of equalities, called the $$\textit{symmetric representation of L}$$: For a point $$P = (x_{0}, y_{0}, z_{0})$$ and vector $$\textbf{v} = (a,b,c)$$ in $$\mathbb{R}^{3}$$ with $$a$$, $$b$$ and $$c$$ all nonzero, the line $$L$$ through $$P$$ parallel to $$\textbf{v}$$ consists of all points $$(x,y,z)$$ given by the equations $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ What if, say, $$a = 0$$ in the above scenario? We can not divide by zero, but we do know that $$x = x_{0} + at$$, and so $$x = x_{0} + 0t = x_{0}$$. Then the symmetric representation of $$L$$ would be: $x = x_{0}, \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ Note that this says that the line $$L$$ lies in the $$\textit{plane}$$ $$x = x_{0}$$, which is parallel to the $$yz$$-plane (Figure 1.5.2). Similar equations can be derived for the cases when $$b = 0$$ or $$c = 0$$. Figure 1.5.2 You may have noticed that the vector representation of $$L$$ in Equation \ref{Eq1.16} is more compact than the parametric and symmetric formulas. That is an advantage of using vector notation. Technically, though, the vector representation gives us the $$\textit{vectors}$$ whose terminal points make up the line $$L$$, not just $$L$$ itself. So you have to remember to identify the vectors $$\textbf{r} + t \textbf{v}$$ with their terminal points. On the other hand, the parametric representation $$\textit{always}$$ gives just the points on $$L$$ and nothing else. Solution (a) Let $$\textbf{r} = (2,3,5)$$. Then by Equation \ref{Eq1.16}, $$L$$ is given by:$$\nonumber \textbf{r} + t \textbf{v} = (2,3,5) + t(4,-1,6), \text{for} -\infty < t < \infty\]

(b) $$L$$ consists of the points $$(x,y,z)$$ such that
$$\nonumber x = 2 + 4t, y = 3 - t, z = 5 + 6t, \text{for} -\infty < t < \infty\] (c) $$L$$ consists of the points $$(x,y,z)$$ such that$$\nonumber \frac{x - 2}{4} = \frac{y - 3}{-1} = \frac{z - 5}{6}\]

(d) Letting $$t=1$$ and $$t=2$$ in part(b) yields the points $$(6,2,11)$$ and $$(10,1,17)$$ on $$L$$.

## Line through two points

Figure 1.5.3

Let $$P_{1} = (x_{1}, y_{1}, z_{1})$$ and P_{2} = (x_{2}, y_{2}, z_{2})\) be distinct points in $$\mathbb{R}^{3}$$, and let $$L$$ be the line through $$P_{1}$$ and $$P_{2}$$. Let $$\textbf{r}_{1} = (x_{1}, y_{1}, z_{1})$$ and $$\textbf{r}_{2} = (x_{2}, y_{2}, z_{2})$$ be the vectors pointing to $$P_{1}$$ and $$P_{2}$$, respectively. Then as we can see from Figure 1.5.3, $$\textbf{r}_{2} - \textbf{r}_{1}$$ is the vector from $$P_{1}$$ to $$P_{2}$$. So if we multiply the vector $$\textbf{r}_{2} - \textbf{r}_{1}$$ by a scalar $$t$$ and add it to the vector $$\textbf{r}_{1}$$, we will get the entire line $$L$$ as $$t$$ varies over all real numbers. The following is a summary of the vector, parametric, and symmetric forms for the line $$L$$:

Let $$P_{1} = (x_{1}, y_{1}, z_{1})$$, $$P_{2} = (x_{2}, y_{2}, z_{2})$$ be distinct points in $$\mathbb{R}^{3}$$, and let $$\textbf{r}_{1} = (x_{1}, y_{1}, z_{1})$$, $$\textbf{r}_{2} = (x_{2}, y_{2}, z_{2})$$. Then the line $$L$$ through $$P_{1}$$ and $$P_{2}$$ has the following representations:

$$\textit{Vector:}$$
$$\textbf{r}_{1} + t(\textbf{r}_{2} - \textbf{r}_{1}) \text{, for} -\infty < t < \infty\] $$\textit{Parametric:}$$$$x = x_{1} + (x_{2} - x_{1})t, y = y_{1} + (y_{2} - y_{1})t, z = z_{1} + (z_{2} - z_{1})t, \text{for} -\infty < t < \infty \label{Eq1.21}\]

$$\textit{Symmetric:}$$
$$\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} \text{if $$x_{1} \ne x_{2}$$, $$y_{1} \ne y_{2}$$, and $$z_{1} \ne z_{2}$$}\] Solution By Equation \ref{Eq1.21}, $$L$$ consists of the points $$(x,y,z)$$ such that$$\nonumber x = -3 + 7t, y = 1 +3t, z = -4 -2t, \text{for} -\infty < t < \infty\]

## Distance between a point and a line

Figure 1.5.4

Let $$L$$ be a line in $$\mathbb{R}^{3}$$ in vector form as $$\textbf{r} + t \textbf{v}$$ (for $$-\infty < t < \infty$$), and let $$P$$ be a point not on $$L$$. The distance $$d$$ from $$P$$ to $$L$$ is the length of the line segment from $$P$$ to $$L$$ which is perpendicular to $$L$$ (see Figure 1.5.4). Pick a point $$Q$$ on $$L$$, and let $$\textbf{w}$$ be the vector from $$Q$$ to $$P$$. If $$\theta$$ is the angle between $$\textbf{w}$$ and $$\textbf{v}$$, then $$d = \norm{\textbf{w}}\,\sin \theta$$. So since $$\norm{\textbf{v} \times \textbf{w}} = \norm{\textbf{v}}\,\norm{\textbf{w}}\,\sin \theta$$ and $$\textbf{v} \ne \textbf{0}$$, then:

$d = \frac{\norm{\textbf{v} \times \textbf{w}}}{\norm{\textbf{v}}}$

Find the distance $$d$$ from the point $$P = (1,1,1)$$ to the line $$L$$ in Example 1.20.

Solution

From Example 1.20, we see that we can represent $$L$$ in vector form as: $$\textbf{r} + t \textbf{v}$$, for $$\textbf{r} = (-3,1,-4)$$ and $$\textbf{v} = (7,3,-2)$$. Since the point $$Q = (-3,1,-4)$$ is on $$L$$, then for $$\textbf{w} = \overrightarrow{QP} = (1,1,1) - (-3,1,-4) = (4,0,5)$$, we have:

$\nonumber \textbf{v} \times \textbf{w} = \left|\begin{array}{rrr}\textbf{i} & \textbf{j} & \textbf{k}\\7 & 3 & -2\\ 4 & 0 & 5\end{array}\right| = \left|\begin{array}{rr}3 & -2\\0 & 5\end{array}\right| \textbf{i} \;-\; \left|\begin{array}{rr}7 & -2\\4 & 5\end{array}\right| \textbf{j} \;+\; \left|\begin{array}{rr}7 & 3\\4 & 0\end{array}\right| \textbf{k} = 15\,\textbf{i} - 43\,\textbf{j} - 12\,\textbf{k} \text{ , so}\\ \nonumber d = \frac{\norm{\textbf{v} \times \textbf{w}}}{\norm{\textbf{v}}} = \frac{\norm{15\,\textbf{i} - 43\,\textbf{j} - 12\,\textbf{k}}}{\norm{(7,3,-2)}} = \frac{\sqrt{15^{2} + (-43)^{2} + (-12)^{2}}}{\sqrt{7^{2} + 3^{2} + (-2)^{2}}} = \frac{\sqrt{2218}}{\sqrt{62}} = 5.98$

It is clear that two lines $$L_{1}$$ and $$L_{2}$$, represented in vector form as $$\textbf{r}_{1} + s \textbf{v}_{1}$$ and $$\textbf{r}_{2} + t \textbf{v}_{2}$$, respectively, are parallel (denoted as $$L_{1} \parallel L_{2}$$) if $$\textbf{v}_{1}$$ and $$\textbf{v}_{2}$$ are parallel. Also, $$L_{1}$$ and $$L_{2}$$ are perpendicular (denoted as $$L_{1} \perp L_{2}$$) if $$\textbf{v}_{1}$$ and $$\textbf{v}_{2}$$ are perpendicular.

Figure 1.5.5

In 2-dimensional space, two lines are either identical, parallel, or they intersect. In 3-dimensional space, there is an additional possibility: two lines can be $$\textbf{skew}$$, that is, they do not intersect but they are not parallel. However, even though they are not parallel, skew lines are on parallel planes (see Figure 1.5.5).

To determine whether two lines in $$\mathbb{R}^{3}$$ intersect, it is often easier to use the parametric representation of the lines. In this case, you should use different parameter variables (usually $$s$$ and $$t$$) for the lines, since the values of the parameters may not be the same at the point of intersection. Setting the two $$(x,y,z)$$ triples equal will result in a system of 3 equations in 2 unknowns ($$s$$ and $$t$$).

Example 1.22

Find the point of intersection (if any) of the following lines:
$$\nonumber \frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z - 1}{-1} \text{ and } x + 3 = \frac{y - 8}{-3} = \frac{z + 3}{2}\] Solution First we write the lines in parametric form, with parameters $$s$$ and $$t$$:$$\nonumber x = -1 + 3s, y = 2 + 2s, z = 1 - s \text{ and } x = -3 + t, y = 8 - 3t, z = -3 + 2tThe lines intersect when $$(-1 + 3s,2 + 2s,1 - s) = (-3 + t,8 - 3t,-3 + 2t)$$ for some $$s$$, $$t$$: \begin{align*} \nonumber -1 + 3s = -3 + t &: \Rightarrow t = 2 + 3s\\ \nonumber 2 + 2s = 8 - 3t &: \Rightarrow 2 + 2s = 8 - 3(2 + 3s) = 2 - 9s \Rightarrow 2s = -9s \Rightarrow s = 0 \Rightarrow t = 2 + 3(0) = 2\\ \nonumber 1 - s = -3 + 2t &: 1 - 0 = -3 + 2(2) \Rightarrow 1 = 1 \checkmark \text{(Note that we had to check this.)} \end{align*} Letting $$s = 0$$ in the equations for the first line, or letting $$t = 2$$ in the equations for the second line, gives the point of intersection $$(-1,2,1)$$. We will now consider planes in 3-dimensional Euclidean space. ## Plane through a point, perpendicular to a vector Let $$P$$ be a plane in $$\mathbb{R}^{3}$$, and suppose it contains a point $$P_{0} = (x_{0}, y_{0}, z_{0})$$. Let $$\textbf{n} = (a,b,c)$$ be a nonzero vector which isper pendicular to the plane $$P$$. Such a vector is called a $$\textbf{normal vector}$$ (or just a $$\textit{normal}$$) to the plane. Now let $$(x,y,z)$$ be any point in the plane $$P$$. Then the vector $$\textbf{r} = (x - x_{0}, y - y_{0}, z - z_{0})$$ lies in the plane $$P$$ (Figure 1.5.6). So if $$\textbf{r} \ne \textbf{0}$$, then $$\textbf{r} \perp \textbf{n}$$ and hence $$\textbf{n} \cdot \textbf{r} = 0$$. And if $$\textbf{r} = \textbf{0}$$ then we still have $$\textbf{n} \cdot \textbf{r} = 0$$. Figure 1.5.6 The plane $$P$$ Conversely, if $$(x,y,z)$$ is any point in $$\mathbb{R}^{3}$$ such that $$\textbf{r} = (x - x_{0}, y - y_{0}, z - z_{0}) \ne \textbf{0}$$ and $$\textbf{n} \cdot \textbf{r} = 0$$, then $$\textbf{r} \perp \textbf{n}$$ and so $$(x,y,z)$$ lies in $$P$$. This proves the following theorem: Theorem 1.18 Let $$P$$ be a plane in $$\mathbb{R}^{3}$$, let $$(x_{0}, y_{0}, z_{0})$$ be a point in $$P$$, and let $$\textbf{n} = (a,b,c)$$ be a nonzero vector which is perpendicular to $$P$$. Then $$P$$ consists of the points $$(x,y,z)$$ satisfying the vector equation:\textbf{n} \cdot \textbf{r} = 0\]

where $$\textbf{r} = (x - x_{0}, y - y_{0}, z - z_{0})$$, or equivalently:

$$a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0 \label{Eq1.25}\] The above equation is called the $$\textbf{point-normal form}$$ of the plane $$P$$. Solution By Equation \ref{Eq1.25}, the plane $$P$$ consists of all points $$(x,y,z)$$ such that:$$\nonumber 2(x + 3) + 4(y - 1) + 8(z - 3) = 0 \]

If we multiply out the terms in Equation \ref{Eq1.25} and combine the constant terms, we get an equation of the plane in $$\textbf{normal form}$$:

$ax + by + cz + d = 0$

For example, the normal form of the plane in Example 1.23 is $$2x + 4y + 8z - 22 = 0$$.

## Plane containing three noncollinear points

In 2-dimensional and 3-dimensional space, two points determine a line. Two points do not determine a plane in $$\mathbb{R}^{3}$$. In fact, three $$\textit{collinear}$$ points (i.e. all on the same line) do not determine a plane; an infinite number of planes would contain the line on which those three points lie. However, three $$\textit{noncollinear}$$ points do determine a plane. For if $$Q$$, $$R$$ and $$S$$ are noncollinear points in $$\mathbb{R}^{3}$$, then $$\overrightarrow{QR}$$ and $$\overrightarrow{QS}$$ are nonzero vectors which are not parallel (by noncollinearity), and so their cross product $$\overrightarrow{QR} \times \overrightarrow{QS}$$ is perpendicular to both $$\overrightarrow{QR}$$ and $$\overrightarrow{QS}$$. So $$\overrightarrow{QR}$$ and $$\overrightarrow{QS}$$ (and hence $$Q$$, $$R$$ and $$S$$) lie in the plane through the point $$Q$$ with normal vector $$\textbf{n} = \overrightarrow{QR} \times \overrightarrow{QS}$$ (see Figure 1.5.7).

Figure 1.5.7: Noncollinear points $$Q, R, S$$

Solution

Let $$Q = (2,1,3)$$, $$R = (1,-1,2)$$ and $$S = (3,2,1)$$. Then for the vectors $$\overrightarrow{QR} = (-1,-2,-1)$$ and $$\overrightarrow{QS} = (1,1,-2)$$, the plane $$P$$ has a normal vector
$$\nonumber \textbf{n} = \overrightarrow{QR} \times \overrightarrow{QS} = (-1,-2,-1) \times (1,1,-2) = (5,-3,1)$$
So using Equation \ref{Eq1.25} with the point $$Q$$ (we could also use $$R$$ or $$S$$), the plane $$P$$ consists of all points $$(x,y,z)$$ such that:
$\nonumber 5(x - 2) - 3(y - 1) + (z - 3) = 0$

or in normal form,

$\nonumber 5x - 3y + z - 10 = 0$

We mentioned earlier that skew lines in $$\mathbb{R}^{3}$$ lie on separate, parallel planes. So two skew lines do not determine a plane. But two (nonidentical) lines which either intersect or are parallel do determine a plane. In both cases, to find the equation of the plane that contains those two lines, simply pick from the two lines a total of three noncollinear points (i.e. one point from one line and two points from the other), then use the technique above, as in Example 1.24, to write the equation. We will leave examples of this as exercises for the reader.

## Distance between a point and a plane

The distance between a point in $$\mathbb{R}^{3}$$ and a plane is the length of the line segment from that point to the plane which is perpendicular to the plane. The following theorem gives a formula for that distance.

Theorem 1.19

Let $$Q = (x_{0}, y_{0}, z_{0})$$ be a point in $$\mathbb{R}^{3}$$, and let $$P$$ be a plane with normal form $$ax + by + cz + d = 0$$ that does not contain $$Q$$. Then the distance $$D$$ from $$Q$$ to $$P$$ is:

D = \frac{|ax_{0} + by_{0} + cz_{0} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}}\] Proof: Let $$R = (x,y,z)$$ be any point in the plane $$P$$ (so that $$ax + by + cz + d = 0$$) and let $$\textbf{r} = \overrightarrow{RQ} = (x_{0} - x, y_{0} - y, z_{0} - z)$$. Then $$\textbf{r} \ne \textbf{0}$$ since $$Q$$ does not lie in $$P$$. From the normal form equation for $$P$$, we know that $$\textbf{n} = (a,b,c)$$ is a normal vector for $$P$$. Now, any plane divides $$\mathbb{R}^{3}$$ into two disjoint parts. Assume that $$\textbf{n}$$ points toward the side of $$P$$ where the point $$Q$$ is located. Place $$\textbf{n}$$ so that its initial point is at $$R$$, and let $$\theta$$ be the angle between $$\textbf{r}$$ and $$\textbf{n}$$. Then $$0^{\circ} < \theta < 90^{\circ}$$, so $$\cos \theta > 0$$. Thus, the distance $$D$$ is $$\cos \theta \,\norm{\textbf{r}} = |\cos \theta|\,\norm{\textbf{r}}$$ (see Figure 1.5.8). Figure 1.5.8 By Theorem 1.6 in Section 1.3, we know that $$\cos \theta = \dfrac{\textbf{n} \cdot \textbf{r}}{\norm{\textbf{n}} \norm{\textbf{r}}}$$, so \nonumber \begin{align}D &= |\cos \theta|\,\norm{\textbf{r}}= \dfrac{|\textbf{n} \cdot \textbf{r}|}{\norm{\textbf{n}} \norm{\textbf{r}}}\,\norm{\textbf{r}}= \dfrac{|\textbf{n} \cdot \textbf{r}|}{\norm{\textbf{n}}}= \dfrac{|a(x_{0} - x) + b(y_{0} - y) + c(z_{0} - z)|}{\sqrt{a^{2} + b^{2} + c^{2}}} \\ \nonumber &= \dfrac{ax_{0} + by_{0} + cz_{0} - (ax + by + cz)}{\sqrt{a^{2} + b^{2} + c^{2}}}= \dfrac{|ax_{0} + by_{0} + cz_{0} - (-d)|}{\sqrt{a^{2} + b^{2} + c^{2}}}= \dfrac{|ax_{0} + by_{0} + cz_{0} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}} \\ \end{align} If $$\textbf{n}$$ points away from the side of $$P$$ where the point $$Q$$ is located, then $$90^{\circ} < \theta < 180^{\circ}$$ and so $$\cos \theta < 0$$. The distance $$D$$ is then $$|\cos \theta| \, \norm{\textbf{r}}$$, and thus repeating the same argument as above still gives the same result.\nonumber D = \frac{|5(2) - 3(4) + 1(-5) - 10|}{\sqrt{5^{2} + (-3)^{2} + 1^{2}}} = \frac{|-17|}{\sqrt{35}} = \frac{17}{\sqrt{35}} \approx 2.87\]

## Line of intersection of two planes

Figure 1.5.9

Note that two planes are parallel if they have normal vectors that are parallel, and the planes are perpendicular if their normal vectors are perpendicular. If two planes do intersect, they do so in a line (see Figure 1.5.9). Suppose that two planes $$P_{1}$$ and $$P_{2}$$ with normal vectors $$\textbf{n}_{1}$$ and $$\textbf{n}_{2}$$, respectively, intersect in a line $$L$$. Since $$\textbf{n}_{1} \times \textbf{n}_{2} \perp \textbf{n}_{1}$$, then $$\textbf{n}_{1} \times \textbf{n}_{2}$$ is parallel to the plane $$P_{1}$$. Likewise, $$\textbf{n}_{1} \times \textbf{n}_{2} \perp \textbf{n}_{2}$$ means that $$\textbf{n}_{1} \times \textbf{n}_{2}$$ is also parallel to $$P_{2}$$. Thus, $$\textbf{n}_{1} \times \textbf{n}_{2}$$ is parallel to the intersection of $$P_{1}$$ and $$P_{2}$$, i.e. $$\textbf{n}_{1} \times \textbf{n}_{2}$$ is parallel to $$L$$. Thus, we can write $$L$$ in the following vector form:

$L: \textbf{r} + t(\textbf{n}_{1} \times \textbf{n}_{2}) \text{, for} -\infty < t < \infty$

where $$\textbf{r}$$ is any vector pointing to a point belonging to both planes. To find a point in both planes, find a common solution $$(x,y,z)$$ to the two normal form equations of the planes. This can often be made easier by setting one of the coordinate variables to zero, which leaves you to solve two equations in just two unknowns.

\nonumber \begin{align} -&3y + z - 10 = 0 \\ \nonumber &4y - z + 3 = 0 \\ \end{align}

The second equation gives $$z = 4y + 3$$, substituting that into the first equation gives $$y = 7$$. Then $$z = 31$$, and so the point $$(0,7,31)$$ is on $$L$$. Since $$\textbf{n}_{1} \times \textbf{n}_{2} = (-1,7,26)$$, then $$L$$ is given by:
$$\nonumber \textbf{r} + t(\textbf{n}_{1} \times \textbf{n}_{2}) = (0,7,31) + t(-1,7,26), \text{for} -\infty < t < \infty$$
or in parametric form:
\nonumber x = -t, y = 7 + 7t, z = 31 +26t, \text{for} -\infty < t < \infty\]