OBJECTIVES

When you finish this section, you should be able to:

1. Find a vector equation of a line in space (p. 733)
2. Find parametric equations of a line in space (p. 734)
3. Find symmetric equations of a line in space (p. 734)
4. Determine whether two distinct lines are skew, parallel, or intersecting (p. 736)
5. Find an equation of a plane (p. 737)
6. Determine whether two distinct planes are parallel or intersecting (p. 738)
7. Find the distance from a point to a plane (p. 739)

THEOREM Vector Equation of a line

A vector equation of a line in the direction $\mathbf{D}\neq \mathbf{0 }$ and containing the point $P_{0}=(x_{0},y_{0},z_{0})$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{r}=\mathbf{r}_{0}+t\mathbf{D} }$$

where $\mathbf{r}_{0}$ is the vector $\overrightarrow{{\it OP}_{0}}$ , $\mathbf{r}$ is the vector $\overrightarrow{\it OP}$, and $P= ( x,y,z)$ is any point on the line.

Find a vector equation of the line containing the points $P_{0}\,\,{=}\,\,(1,2,-1)$ and $P_{1}=(4,3,-2)$.

Solution The vector $\mathbf{D}$ in the direction from $P_{0}=(1,2,-1)$ to $P_{1}=(4,3,-2)$ is $$\begin{equation*} \mathbf{D}=\overrightarrow{P_{0}P_{1}}=3\mathbf{i+j-k} \end{equation*}$$

If we let $\mathbf{r}_{0}=\mathbf{i}+2\mathbf{j-k}$, then a vector equation of the line is $$\mathbf{r}=\mathbf{r}_{0}+t \mathbf{D}=\mathbf{i}+2\mathbf{j}-\mathbf{k} +t (3\mathbf{i}+\mathbf{j}-\mathbf{k})=(1+3t)\mathbf{i}+(2+t)\mathbf{j} +(-1-t)\mathbf{k}$$

NOW WORK

Problem 11(a).

NEED TO REVIEW?

Parametric equations are discussed in Section 9.1, pp. 637-642.

Find parametric equations of the line containing the point $(2,-3,1)$ and in the direction of the vector $4\mathbf{i}+\dfrac{3}{4}\mathbf{j}-\mathbf{k}$.

Solution Let $(x_{0},y_{0},z_{0})=(2,-3,1)$ and $a\mathbf{i}+b \mathbf{j}+c\mathbf{k}=4\mathbf{i}+\dfrac{3}{4}\mathbf{j}-\mathbf{k}$. Then, the parametric equations of the line are $$x=x_{0}+at=2+4t\qquad y=y_{0}+bt=-3+\frac{3}{4}t\qquad z=z_{0}+ct=1-t$$

NOW WORK

Problem 9(b).

NOTE

The symmetric equations actually represent a pair of equations

$\dfrac{x-x_0}{a}=\dfrac{y-y_{0}}{b}$

and

$\dfrac{x-x_{0}}{a} = \dfrac{z-z_0}{c}.$

Find symmetric equations of the line containing the point $(1,-1,2)$ and in the direction of the vector $5\mathbf{i}-2\mathbf{j}+3\mathbf{k}$.

Solution The components of the vector $5\mathbf{i}-2\mathbf{j}+3 \mathbf{k}$ are all nonzero. So, we use $a=5$, $b=-2$, and $c=3$ and the coordinates of the point $(1,-1,2)$ to obtain the symmetric equations $$\frac{x-1}{5}=\frac{y+1}{-2}=\frac{z-2}{3}$$

NOW WORK

Problem 9(c).

Find symmetric equations of the line that contains the point $(5,-2, 3)$ and is in the direction of the vector $\mathbf{D=}$ $3\mathbf{i}-2\mathbf{k}$.

Solution For $\mathbf{D=}$ $3\mathbf{i}-2\mathbf{k}$, $a=3$, $b=0$, and $c=-2.$ So, symmetric equations in the direction $\mathbf{D}$ are $$\dfrac{x-5}{3}=\frac{z-3}{-2}\qquad y=-2$$

NOW WORK

Problem 17.

A line is defined by the symmetric equations $\dfrac{x-6}{3}=\dfrac{y+2}{1}=\dfrac{z+3}{-2}$.

Solution (a) From the denominators of the symmetric equations, we find $a=3$, $b=1$, and $c=-2$. So, ${\bf D}= 3\mathbf{i}+\mathbf{j} -2\mathbf{k}$ is a vector in the direction of the line.

(b) Since the line is defined by the symmetric equations $\dfrac{x-6}{3}=\dfrac{y+2}{1}=\dfrac{z+3}{-2},$ one point on the line is $( 6,-2,-3)$. To find a second point, we assign a value to $x$ say, $x=0.$ Then $$\begin{eqnarray*} \frac{0-6}{3} &=&\frac{y+2}{1}=\frac{z+3}{-2} \\[4pt] -2 &=&\frac{y+2}{1}=\frac{z+3}{-2} \end{eqnarray*}$$

Now we solve for $y$ and $z,$ and find $y=-4$ and $z=1$. So, another point on the line is $(0,-4,1)$.

NOW WORK

Problem 15.

Determine whether the lines given below intersect, are parallel, or are skew.

Solution (a) The line $l_{1}$ is in the direction of the vector $\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and the line $l_{2}$ is in the direction of the vector $\mathbf{i}-4\mathbf{j}+3\mathbf{k}$. Since these vectors are not parallel (do you know why?), $l_{1}$ and $l_{2}$ either intersect or are skew.

Suppose the lines intersect. Then there is some value of the parameter $t_{1}$ and some value of the parameter $t_{2}$ for which $\mathbf{r}_{1}= \mathbf{r}_{2}.$ Since $~\mathbf{r}_{1}=(3+t_{1})\mathbf{i}+(3-2t_{1}) \mathbf{j}+(1+t_{1})\mathbf{k}$ and $\mathbf{r}_{2}=(5+t_{2})\mathbf{i} +(1-4t_{2})\mathbf{j}+(1+3t_{2})\mathbf{k}$, for $\mathbf{r}_{1}$ to equal $\mathbf{r}_{2},$ we have $$\begin{equation*} (3+t_{1})\mathbf{i}+(3-2t_{1})\mathbf{j}+(1+t_{1})\mathbf{k}=(5+t_{2}) \mathbf{i}+(1-4t_{2})\mathbf{j}+(1+3t_{2})\mathbf{k}\qquad {\color{#0066A7}{\hbox{\(\mathbf{r}_{1}=\mathbf{r}_{2}\)}}} \end{equation*}$$

We equate the components of each vector to obtain $$3+t_{1}=5+t_{2} \qquad 3-2t_{1}=1-4t_{2} \qquad 1+t_{1}=1+3t_{2}$$

The result is a system of three equations containing two variables. From the third equation, $t_{1}=3t_{2}$. Now substitute $t_{1}=3t_{2}$ into each of the first two equations. $$\begin{equation*} \begin{array}{cllll} \begin{array}{rcl} 3+t_{1}&=&5+t_{2}\\ 3+3t_{2}&=&5+t_{2}\\ t_{2}&=&1 \end{array} & \ \ \ \ & \begin{array}{rcl} 3-2t_{1}&=&1-4t_{2}\\ 3-6t_{2}&=&1-4t_{2}\\ t_{2}&=&1 \end{array} & \ \ \ \ & {\color{#0066A7}{\hbox{\(t_{1}=3t_{2}\)}}} \end{array} \end{equation*}$$

Backsubstituting, we conclude $t_{1}=3$ and $t_{2}=1$. The point of intersection is found by substituting $t_{1}=3$ in $l_{1}$ or $t_{2}=1$ in $l_{2}$ . The result is $\mathbf{r}_{1}=\mathbf{r}_{2}=6\mathbf{i}-3\mathbf{j}+4 \mathbf{k}$. The point of intersection is $(6,-3,4)$.

(b) The line $l_{1}$ is in the direction of the vector $\mathbf{i}- \mathbf{j}+\mathbf{k}$, and the line $l_{2}$ is in the direction of the vector $\mathbf{i}-\mathbf{j}+2\mathbf{k}$. These vectors are not parallel, so the lines $l_{1}$ and $l_{2}$ either intersect or are skew. As before, suppose they intersect. Then $\mathbf{r}_{1}=\mathbf{r}_{2}\mathbf{\ }$so that $$\begin{eqnarray*} &&( 3+t_{1}) \mathbf{i}+\left( 2-t_{1}\right)\! \mathbf{j}+( 1+t_{1}) \mathbf{k}=( 5+t_{2})\mathbf{i}+( 6-t_{2})\mathbf{j}+( 1+2t_{2}) \mathbf{k}\\[4pt] &&3+t_{1}=5+t_{2} \qquad 2-t_{1}=6-t_{2} \qquad 1+t_{1}=1+2t_{2} \qquad {\color{#0066A7}{\hbox{Equate components.}}} \end{eqnarray*}$$

Now substitute $t_{1}=2t_{2}$ (from the third equation) into each of the first two equations. $$\begin{equation*} \begin{array}{cllll} \begin{array}{rcl} 3+t_{1}&=&5+t_{2} \\[6pt] 3+2t_{2}&=&5+t_{2} \\[6pt] t_{2}&=&2 \end{array} & \ \ \ \ & \begin{array}{rcl} 2-t_{1}&=&6-t_{2} \\[6pt] 2-2t_{2}&=&6-t_{2} \\[6pt] t_{2}&=&-4 \end{array} & \ \ \ \ & {\color{#0066A7}{\hbox{\(t_{1}=2t_{2}\)}}} \end{array} \end{equation*}$$

The system of equations has no solution, so the assumption that $\mathbf{r} _{1}=\mathbf{r}_{2}\mathbf{,}$ for some $t_{1}$ and some $t_{2}$ is false. The lines are neither parallel nor intersecting, so they are skew.

CAUTION

When working with pairs of lines, be sure to use a different parameter for each line.

NOW WORK

Problems 23(a) and 29(a).

THEOREM Vector Equation of a Plane

The vector equation of the plane containing the point $P_{0}$ is $$(\mathbf{r}-\mathbf{r}_{0})\,{\cdot}\, \mathbf{N}=0$$

where $\mathbf{r}_{0}$ is the vector $\overrightarrow{{\it OP}_{0}},$ $\mathbf{r }$ is the vector $\overrightarrow{\it OP}$ of any other point $P$ on the plane, and $\mathbf{N}$ is normal to the plane.

THEOREM General Equation of a Plane

An equation of the form $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox { Ax+By+Cz=D }}$$

with at least one of the numbers $A, B, C$ not $0$, is called the general equation of a plane, and the vector $A\mathbf{i}+B\mathbf{j}+C \mathbf{k}$ is normal to this plane.

Find the general equation of the plane containing the point $( 1,2,-1)$ if the vector $\mathbf{N=\,}2\mathbf{i}+3\mathbf{j}-4 \mathbf{k}$ is normal to the plane. Then use the intercepts to graph the plane.

Solution The general equation of a plane is $Ax+By+Cz=D,$ where $A,$ $B,$ and $C$ are the components of a normal vector to the plane. Since the normal is $\mathbf{N=\,}2\mathbf{i}+3\mathbf{j}-4\mathbf{k}$, we have $$2x+3y-4z=D$$

To find $D$, we use the point $(1,2,-1)$. Then $D=2(1) +3(2) -4(-1) =12.$ The general equation of the plane is $$2x+3y-4z=12$$

Figure 54 $2x+3y-4z=12$

We find the intercepts by letting two variables equal zero and then solving for the third variable. For example, to find the $x$-intercept, we let $y=0$ and $z=0.$ Then $2x=12$ or $x=6.$ Similarly, the $y$-intercept is $4$ and the $z$-intercept is $-3.$ Figure 54 illustrates how the intercepts are used to graph the plane.

NOW WORK

Problem 31.

Find the general equation of the plane containing the points $$P_{1}=(1,-1,2)\quad P_{2}=( 3,0,0)\quad P_{3}=(4,2,1)$$

Solution The vectors $\mathbf{v}=\overrightarrow{P_{1}~P_{2}}=2 \mathbf{i}+\mathbf{j}-2\mathbf{k}$ and $\mathbf{w}=\overrightarrow{P_{1}~P_{3} }=3\mathbf{i}+3\mathbf{j}-\mathbf{k}$ lie in the plane. The vector $$\begin{equation*} \mathbf{N}=\mathbf{v}\times \mathbf{w}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 2 & 1 & -2 \\[3pt] 3 & 3 & -1 \end{array} \right\vert =\left\vert \begin{array}{r@{\quad}r} 1 & -2 \\[3pt] 3 & -1 \end{array} \right\vert \mathbf{i}-\left\vert \begin{array}{r@{\quad}r} 2 & -2 \\[3pt] 3 & -1 \end{array} \right\vert \mathbf{j}+\left\vert \begin{array}{r@{\quad}r} 2 & 1 \\[3pt] 3 & 3 \end{array} \right\vert \mathbf{k}=5\mathbf{i}-4\mathbf{j}+3\mathbf{k} \end{equation*}$$

is orthogonal to both $\mathbf{v}$ and $\mathbf{w}$ and so is normal to the plane. Using the point $(1,-1,2)$ and the vector $\mathbf{N}$, the general equation of the plane is $$\begin{eqnarray*} 5(x-1)-4(y+1)+3(z-2)& =&0 \\[2pt] 5x-5-4y-4+3z-6& =&0 \\[2pt] 5x-4y+3z& =&15 \end{eqnarray*}$$

NOW WORK

Problem 43(a).

NOTE

The planes $p_{1}$: $2x+y-z=3$ and $p_{2}$: $4x+2y-2z=6$ are identical. Do you see why?

Find an equation of the line of intersection of the two planes $$x-2y+z=-1\quad \hbox{and}\quad 3x+y-z=4$$

Solution First notice that the normals $\mathbf{N}_{1} = \mathbf{i}\, -\, 2\mathbf{j}\, +\, \mathbf{k}$ and $\mathbf{N}_{2} = 3\mathbf{i}\, +\, \mathbf{j}\, -\, \mathbf{k}$ of the two planes are not parallel. To find an equation of the line of intersection, we need a point on the line and the direction of the line. Since the line of intersection is perpendicular to both $\mathbf{N}_{1}$ and $\mathbf{N}_{2}$, it is parallel to the vector $\mathbf{D}= \mathbf{N} _{1}\times \mathbf{N}_{2}$. $$\mathbf{D=N}_{1}\times \mathbf{N}_{2}= \left|\begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 1 & -2 & 1 \\[3pt] 3 & 1 & -1 \end{array}\right| =\mathbf{i}+4\mathbf{j}+7\mathbf{k}$$

The vector $\mathbf{D}$ gives the direction of the line. We can find a point on the line by locating any point common to both planes. For example, if $z=0$, then $x-2y=-1$ and $3x+y=4$. Solving these equations simultaneously, we find $x=1$ and $y=1$. So, the point $(1,1,0)$ is on the line, and symmetric equations of the line of intersection are $$\dfrac{x-1}{1}=\dfrac{y-1}{4}=\dfrac{z}{7}$$

NOTE

We did not have to choose $z=0$; a value could have been assigned to any of the variables and the resulting two equations solved. Different choices would give equivalent equations of the line.

NOW WORK

Problem 47.

THEOREM Distance from a Point to a Plane

The distance $d$ from the point $P_{0}=(x_{0},y_{0},z_{0})$ to the plane $Ax+By+Cz=D$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ d=\dfrac{\left\vert \,Ax_{0}+By_{0}+Cz_{0}-D\,\right\vert }{\sqrt{A^{2}+B^{2}+C^{2}}}}\tag {1}$$

Find the distance $d$ from the point $( 2,3,-1)$ to the plane $x+4y+z=5$.

Solution We use (1) with $A=1$, $B=4$, $C=1$, $D=5$, and $P_{0}=(2,3,-1)$. Then $$d=\frac{\left\vert (1)(2)+(4)(3)+(1)(-1)-5\right\vert }{\sqrt{1+16+1}}= \frac{8}{3\sqrt{2}}=\frac{4\sqrt{2}}{3}$$

NOW WORK

Problem 51.

Find the distance $d$ between the parallel planes $$p_{1}:\ 2x+3y-z=3 \quad \hbox{and}\quad p_{2}: \ 2x+3y-z=4$$

Solution The planes $p_{1}$ and $p_{2}$ are parallel because $\mathbf{N}_{1}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}=\mathbf{N}_{2}$ and $D_{1}=3$ is not equal to $D_{2}=4$.

We find the distance between the planes by choosing a point on one plane and using the formula for distance from that point to the other plane. A point on $p_{1}$ can be found by letting $x=0$ and $y=0.$ Then $z=-3,$ and $( 0,0,-3)$ is a point on $p_{1}.$ The distance $d$ from the point $( 0,0,-3)$ to the plane $p_{2}$ is $$d=\dfrac{\left\vert \,2(0) +3(0) -1 (-3) -4\,\right\vert }{\sqrt{2^{2}+3^{2}+(-1) ^{2}}}=\dfrac{1}{\sqrt{ 14}}=\dfrac{\sqrt{14}}{14}$$

NOW WORK

Problems 55.