#### Chapter 10 Basic Concepts of Descriptive Vector Geometry

Section 10.2 Lines and Planes

# 10.2.4 Relative Positions of Lines and Planes in Space

While two lines in the plane can only have three different relative positions with respect to each other (lines are parallel, coincide, or intersect, see Section 9.2.3), two lines in space can have four different relative positions with respect to each other. These will be outlined in the Info Box below.
##### Info 10.2.13

Let two lines in space be given by vector equations. The line $g$ has the reference vector $\stackrel{\to }{a}$ and the direction vector $\stackrel{\to }{u}$, and the line $h$ has the reference vector $\stackrel{\to }{b}$ and the direction vector $\stackrel{\to }{v}$:

$g:\stackrel{\to }{r}=\stackrel{\to }{a}+s\stackrel{\to }{u} ; s\in ℝ ,$

$h:\stackrel{\to }{r}=\stackrel{\to }{b}+t\stackrel{\to }{v} ; t\in ℝ .$

The two lines $g$ and $h$ can have four different relative positions:
1. The lines are identical. In this case, the lines $g$ and $h$ have all their points in common, they coincide. This is the case if and only if the two direction vectors $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ are collinear and the lines have any one point in common.
2. The lines are parallel. This is the case if and only if the two direction vectors $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ are collinear and the two lines do not have any points in common.
3. The lines intersect. In this case, the lines $g$ and $h$ have exactly one point in common. This point is called the intersection point. This is the case if and only if the two direction vectors $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ are not collinear and the two lines have exactly one point in common.
4. Lines that are neither identical nor parallel and do not intersect are called skew. This is the case if and only if the two direction vectors are not collinear, and the two lines do not have any points in common.

(This figure will be released shortly.)

In practice, the relative position of two lines in space is investigated according as follows: first, we examine two direction vectors for collinearity, then we check whether the two lines have points in common. This uniquely identifies one of the four cases. The example below illustrates this approach for all four cases.
##### Example 10.2.14
Let the four lines $g$, $h$, $i$, and $j$ be given in parametric form by

$g:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill -4\hfill \end{array}\right) ; s\in ℝ$

$h:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 7\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right) ; t\in ℝ$

$i:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 8\hfill \end{array}\right)+u\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 3\hfill \\ \hfill -6\hfill \end{array}\right) ; u\in ℝ$

$j:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill 2\hfill \end{array}\right)+v\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill -3\hfill \end{array}\right) ; v\in ℝ .$

• The lines $g$ and $h$ are identical. The two direction vectors $\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill -4\hfill \end{array}\right)$ of $g$ and $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right)$ of $h$ are collinear. We have

$\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill -4\hfill \end{array}\right)=-2·\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right) .$

The point described by the position vector $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 7\hfill \end{array}\right)$ lies both on the line $h$ (as reference point) and on the line $g$ since we have for the line $g$:

$\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 7\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill -4\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1-2s\hfill \\ \hfill 2s\hfill \\ \hfill 3-4s\hfill \end{array}\right) ⇔ s=-1 .$

Thus, the vector $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 7\hfill \end{array}\right)$ results from the equation of $g$ for the parameter value $s=-1$.
• The lines $h$ and $i$ (and hence the lines $g$ and $i$) are parallel. The two direction vectors $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right)$ of $h$ and $\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 3\hfill \\ \hfill -6\hfill \end{array}\right)$ of $i$ are collinear.

$\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 3\hfill \\ \hfill -6\hfill \end{array}\right)=-3·\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right) .$

However, the lines $h$ and $i$ do not have any points in common: the reference point of one of the two lines is not a point on the other line. Here, we can check whether the reference vector $\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 8\hfill \end{array}\right)$ of the line $i$ can result as a position vector of the line $h$:

$\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 8\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 7\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \\ \hfill 2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1+t\hfill \\ \hfill -2-t\hfill \\ \hfill 7+2t\hfill \end{array}\right) .$

In this vector equation, $t=3$ results from the first component and $t=-2$ from the second, which is a contradiction. Hence, the two lines do not have any points in common.
• The lines $i$ and $j$ intersect. First we see that for these two lines the two direction vectors $\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 3\hfill \\ \hfill -6\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill -3\hfill \end{array}\right)$ are not collinear. There is no number $a\in ℝ$ such that

$\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 3\hfill \\ \hfill -6\hfill \end{array}\right)=a\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill -3\hfill \end{array}\right)$

since the equality of the first component results in $a=-3$ and equality of the second component results in $a=1$, which is a contradiction. However, these two lines have a point in common that can be found by equating the position vectors for $i$ and $j$:

$\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 8\hfill \end{array}\right)+u\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 3\hfill \\ \hfill -6\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 4-3u\hfill \\ \hfill 3u\hfill \\ \hfill 8-6u\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1+v\hfill \\ \hfill 3+3v\hfill \\ \hfill 2-3v\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill 2\hfill \end{array}\right)+v\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill -3\hfill \end{array}\right) .$

Equating the first two components results in the two equations

$3-3u=v \text{and} u=1+v$

in the variables $u$ and $v$ with the solution $v=0$, $u=1$. Substituting these values into the equation for the third component results in

$8-6·1=2-3·0 ⇔ 2=2 .$

Thus, the vector equation for the position vectors is satisfied for the parameter values $u=1$ and $v=0$. Hence, the position vector of the intersection point results from substituting the parameter value $u=1$ into the equation of the line $i$ or from substituting the parameter value $v=0$ into the equation of the line $j$. For the intersection point of the lines we have $\left(1;3;2\right)$.
• The lines $g$ and $j$ (and hence the lines $h$ and $j$) are skew. As in the previous case of the intersecting lines it can be easily seen that the two direction vectors $\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill -4\hfill \end{array}\right)$ of $g$ and $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill -3\hfill \end{array}\right)$ of $j$ are not collinear. However, in this case the lines do not have any point in common which again can be found by equating the position vectors:

$\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 2\hfill \\ \hfill -4\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1-2s\hfill \\ \hfill 2s\hfill \\ \hfill 3-4s\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1+v\hfill \\ \hfill 3+3v\hfill \\ \hfill 2-3v\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill 2\hfill \end{array}\right)+v\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill -3\hfill \end{array}\right) .$

This vector equation involves a contradiction; there are no pairs of parameter values of $s$ and $v$ such that the equation is satisfied, and hence $g$ and $j$ do not have any points in common. Considering the first and the second components results in the two equations

$-2s=2+v \text{and} 2s=3+3v$

with the solution $v=-\frac{5}{4}$, $s=-\frac{3}{8}$. However, substituting this into the equation for the third component results in the contradiction

$3-4\left(-\frac{3}{8}\right)=2-3\left(-\frac{5}{4}\right) ⇔ \frac{9}{2}=\frac{23}{4} .$

##### Exercise 10.2.15
Tick the true statements:
The two lines given by the equations

$g:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 4\hfill \end{array}\right)+x\left(\begin{array}{c}\hfill -5\hfill \\ \hfill 10\hfill \\ \hfill -15\hfill \end{array}\right) ; x\in ℝ$

and

$h:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 7\hfill \end{array}\right)+y\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill 3\hfill \end{array}\right) ; y\in ℝ$

intersect since
 the two direction vectors are collinear, the two direction vectors are not collinear, the two direction vectors are collinear and the lines have a point in common, the two direction vectors are not collinear and the lines have a point in common, the two direction vectors are not collinear and the lines do not have any points in common.

Find the intersection point $S$ of the two lines $g$ and $h$.
$S=$
The position vector of the intersection point $\stackrel{\to }{S}$ results from the lines $g$ and $h$ for the parameter values
$x=$
and
$y=$ .

##### Exercise 10.2.16
The two lines given by the equations

$\gamma :\stackrel{\to }{r}=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill -2\hfill \end{array}\right) ; s\in ℝ$

and

$\kappa :\stackrel{\to }{r}=\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill 4\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill -\frac{3}{2}\hfill \\ \hfill c\hfill \\ \hfill 1\hfill \end{array}\right) ; t\in ℝ$

are parallel. Find the value of $c$ and specify which values the parameters $a$ and $b$ must not take simultaneously to ensure that the two lines are parallel.
$a\ne$

$b\ne$

$c=$

Consider two lines in space be given that are truly parallel or intersecting. Then these two lines uniquely define a plane (see figure below).
(This figure will be released shortly.)
This is always the plane that contains both lines.
The example below shows how to derive the parametric equation of the plane from two truly parallel or intersecting lines.
##### Example 10.2.17
• The two lines given by the equations

$g:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 4\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill -5\hfill \\ \hfill 10\hfill \\ \hfill -15\hfill \end{array}\right) ; s\in ℝ$

and

$h:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 7\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill 3\hfill \end{array}\right) ; t\in ℝ$

intersect at the point $S=\left(1;2;4\right)$ (see Exercise 10.2.15). Thus, they uniquely define a plane $E$ that contains both $g$ and $h$. For the equation of the plane $E$ in parametric form, the position vectors of three appropriate points are used resulting from the given equations of the lines $g$ and $h$. The intersection point is a suitable reference point of $E$, with the reference vector

$\stackrel{\to }{S}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 4\hfill \end{array}\right).$

Then, the position vectors of two points $P$ and $Q$ result, for example, from substituting the parameter values $s=1$ and $t=1$ into the equations of the lines in parametric form:

$\stackrel{\to }{P}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 4\hfill \end{array}\right)+1·\left(\begin{array}{c}\hfill -5\hfill \\ \hfill 10\hfill \\ \hfill -15\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 12\hfill \\ \hfill -11\hfill \end{array}\right) ,$

$\stackrel{\to }{Q}=\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 7\hfill \end{array}\right)+1·\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill 3\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 7\hfill \\ \hfill -2\hfill \\ \hfill 10\hfill \end{array}\right) .$

Thus, we have the direction vectors

$\stackrel{\to }{SP}=\stackrel{\to }{P}-\stackrel{\to }{S}=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 12\hfill \\ \hfill -11\hfill \end{array}\right)-\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 4\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -5\hfill \\ \hfill 10\hfill \\ \hfill -15\hfill \end{array}\right)$

and

$\stackrel{\to }{SQ}=\stackrel{\to }{Q}-\stackrel{\to }{S}=\left(\begin{array}{c}\hfill 7\hfill \\ \hfill -2\hfill \\ \hfill 10\hfill \end{array}\right)-\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 4\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 6\hfill \\ \hfill -4\hfill \\ \hfill 11\hfill \end{array}\right) .$

Hence, a possible equation of the plane $E$ in parametric form is given by

$E:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 4\hfill \end{array}\right)+\mu \left(\begin{array}{c}\hfill -5\hfill \\ \hfill 10\hfill \\ \hfill -15\hfill \end{array}\right)+\nu \left(\begin{array}{c}\hfill 6\hfill \\ \hfill -4\hfill \\ \hfill 11\hfill \end{array}\right) ; \mu ,\nu \in ℝ .$

• The two lines $g$ and $h$ given by the equations

$g:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill -2\hfill \end{array}\right) ; s\in ℝ$

and

$h:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill -\frac{3}{2}\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right) ; t\in ℝ$

are parallel (see Exercise 10.2.16). This uniquely defines a plane $F$ that contains both $g$ and $h$. The parametric equation of the plane $F$ is derived from the position vectors of three appropriate points resulting from the equations of the lines $g$ and $h$. The position vectors of points on $g$ or $h$ that result, for example, from substituting the parameter values $s=0$, $s=1$, and $t=0$ are suitable. The first vector

$\stackrel{\to }{A}=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \end{array}\right)+0·\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill -2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \end{array}\right)$

can be used as a reference vector. From

$\stackrel{\to }{B}=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \end{array}\right)+1·\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill -2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 4\hfill \\ \hfill -2\hfill \end{array}\right)$

it immediately results that as the first direction vector $\stackrel{\to }{AB}$ of the plane $F$ the direction vector $\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill -2\hfill \end{array}\right)$ of $g$ can be used. The second direction vector of the plane results from the position vector of a point $B$ on $h$ for the parameter value $t=0$, i.e. the reference point of $h$:

$\stackrel{\to }{B}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right) .$

Hence,

$\stackrel{\to }{AB}=\stackrel{\to }{B}-\stackrel{\to }{A}=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right)-\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill -5\hfill \\ \hfill 4\hfill \end{array}\right)$

is the second direction vector of $F$. Thus, a parameter form of the equation of $F$ is

$F:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \end{array}\right)+\lambda \left(\begin{array}{c}\hfill 3\hfill \\ \hfill -2\hfill \\ \hfill -2\hfill \end{array}\right)+\mu \left(\begin{array}{c}\hfill 5\hfill \\ \hfill -5\hfill \\ \hfill 4\hfill \end{array}\right) ; \lambda ,\mu \in ℝ .$

A line and a plane in space can only have three different relative positions with respect to each other, as outlined in the Info Box below.
##### Info 10.2.18

Let a line $g$ with reference point $\stackrel{\to }{a}$ and direction vector $\stackrel{\to }{u}$ and a plane $E$ with the reference vector $\stackrel{\to }{b}$ and the direction vectors $\stackrel{\to }{v}$ and $\stackrel{\to }{w}$ in space be given in parametric form by the equations

$g:\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{u} ; \lambda \in ℝ$

and

$E:\stackrel{\to }{r}=\stackrel{\to }{b}+\mu \stackrel{\to }{v}+\nu \stackrel{\to }{w} ; \mu ,\nu \in ℝ .$

Then the line $g$ and the plane $E$ can have three different relative positions:
1. The line $g$ lies in the plane $E$. This is the case if and only if the three direction vectors $\stackrel{\to }{u}$, $\stackrel{\to }{v}$, and $\stackrel{\to }{w}$ are coplanar, and the reference point of the line lies in the plane.
2. The line $g$ is parallel to the plane $E$. This is the case if and only if the three direction vectors $\stackrel{\to }{u}$, $\stackrel{\to }{v}$, and $\stackrel{\to }{w}$ are coplanar and the reference point of the line does not lie in the plane.
3. The lines $g$ and the plane $E$ intersect. This is the case if and only if the three direction vectors $\stackrel{\to }{u}$, $\stackrel{\to }{v}$, and $\stackrel{\to }{w}$ are not coplanar.

(This figure will be released shortly.)

To investigate the relative position of a given line and plane, we first examine the three direction vectors for collinearity, then we check whether the reference point of the line lies in the plane. This uniquely identifies one of the three possible cases. If the line and the plane intersect, we can calculate the intersection point. The example below illustrates a few approaches.
##### Example 10.2.19
Let the plane $E$ be given by the parametric equation

$E:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right) ; s,t\in ℝ .$

• A line with the vector $\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill -4\hfill \end{array}\right)$ as its direction vector either lies in the plane or is parallel to the plane $E$, since $\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill -4\hfill \end{array}\right)$ is coplanar to the two direction vectors of $E$. From the condition

$\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill -4\hfill \end{array}\right)=s\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right) ,$

we have $s=1$, $t=-2$. Hence, the line

$g:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 3\hfill \\ \hfill 0\hfill \end{array}\right)+x\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill -4\hfill \end{array}\right) ; x\in ℝ$

lies in the plane $E$, since the reference point $\left(-1;3;0\right)$ lies in $E$.

$\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 3\hfill \\ \hfill 0\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 2+3s\hfill \\ \hfill 2-s\hfill \\ \hfill 2+2t\hfill \end{array}\right) ⇔ s=t=-1 .$

Hence, the position vector $\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 3\hfill \\ \hfill 0\hfill \end{array}\right)$ of the line results from the parametric equation of the plane for the parameter values $s=t=-1$. In contrast, the line

$h:\stackrel{\to }{r}=y\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill -4\hfill \end{array}\right) ; y\in ℝ$

is parallel to the plane $E$ since $h$ has the origin $\left(0;0;0\right)$ as its reference point. The origin does not lie in the plane $E$ since there are no parameter values $s$ and $t$ such that the vector equation

$\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 2+3s\hfill \\ \hfill 2-s\hfill \\ \hfill 2+2t\hfill \end{array}\right)$

is satisfied. Considering the first component implies $s=-\frac{2}{3}$, and $s=2$ results from the second component; this is a contradiction.
• Every line with a direction vector that is not coplanar to the two direction vectors $\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right)$ of $E$ intersect the plane $E$ at exactly one point. An example of such a line is

$k:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)+\mu \left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right) ; \mu \in ℝ .$

The direction vector $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)$ is not coplanar to $\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right)$ since the condition

$\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)=a\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 3a\hfill \\ \hfill -a\hfill \\ \hfill 2b\hfill \end{array}\right)$

cannot be satisfied by any choice of $a,b\in ℝ$. Considering the first component would imply $a=\frac{1}{3}$ and the second would imply $a=-1$; this is a contradiction. From equating the position vectors of the line $k$ and the plane $E$, the intersection point can be calculated:

$\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)+\mu \left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -3+\mu \hfill \\ \hfill 1+\mu \hfill \\ \hfill \mu \hfill \end{array}\right)=\left(\begin{array}{c}\hfill 2+3s\hfill \\ \hfill 2-s\hfill \\ \hfill 2+2t\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 2\hfill \\ \hfill 2\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 2\hfill \end{array}\right) .$

If we are only interested in the intersection point, it is sufficient to determine the parameter value of the line for which this vector equation is satisfied. The position vector of the intersection point then results from substituting the determined parameter value into the equation of the line. Considering the first two components of this vector equation results in a system of two linear equations in the variables $\mu$ and $s$:

$\mu =5+3s \text{and} \mu =1-s ,$

with the solution $\mu =2$. Thus, the intersection point has the position vector

$\left(\begin{array}{c}\hfill -3\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)+2\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 3\hfill \\ \hfill 2\hfill \end{array}\right) .$

##### Exercise 10.2.20
Let the plane $E$ be given by the equation

$E:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 8\hfill \\ \hfill -2\hfill \\ \hfill 0\hfill \end{array}\right)+s\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill 2\hfill \end{array}\right)+t\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right) ; s,t\in ℝ$

and the line $g$ by the equation

$g:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right)+u\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 4\hfill \\ \hfill c\hfill \end{array}\right) ; u\in ℝ$

whose reference point does not lie in the plane $E$.
Find the missing component $c$ such that the line $g$ is parallel to the line $E$.
$c=$
For all other values of $c$, calculate the intersection point $S=\left(x;y;z\right)$ depending on $c$. Specify the three components of $S$ separately.
$x=$

$y=$

$z=$

##### Exercise 10.2.21
Let the line $h$ be given by the equation

$h:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 3\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right)+\rho \left(\begin{array}{c}\hfill -8\hfill \\ \hfill 9\hfill \\ \hfill 1\hfill \end{array}\right) ; \rho \in ℝ .$

Find the following:
1. Value of the parameter $\rho$ for which the line $h$ intersects the $xy$-plane: $\rho =$
2. Value of the parameter $\rho$ for which the line $h$ intersects the $yz$-plane: $\rho =$
3. Value of the parameter $\rho$ for which the line $h$ intersects the $xz$-plane: $\rho =$

If we consider two planes in space, we find that they can have three different relative positions with respect to each other which correspond to the three different relative positions of two lines described in Section 9.2.3.
These three cases are outlined in the Info Box below.
##### Info 10.2.22

Let the plane ${E}_{1}$ with the reference vector ${\stackrel{\to }{a}}_{1}$ and the two direction vectors ${\stackrel{\to }{u}}_{1}$ and ${\stackrel{\to }{v}}_{1}$ and the plane ${E}_{2}$ with the reference vector ${\stackrel{\to }{a}}_{2}$ and the two direction vectors ${\stackrel{\to }{u}}_{2}$ and ${\stackrel{\to }{v}}_{2}$ be given by the equations

${E}_{1}:\stackrel{\to }{r}={\stackrel{\to }{a}}_{1}+\mu {\stackrel{\to }{u}}_{1}+\nu {\stackrel{\to }{v}}_{1} ; \mu ,\nu \in ℝ$

${E}_{2}:\stackrel{\to }{r}={\stackrel{\to }{a}}_{2}+\rho {\stackrel{\to }{u}}_{2}+\sigma {\stackrel{\to }{v}}_{2} ; \rho ,\sigma \in ℝ .$

The planes ${E}_{1}$ and ${E}_{2}$ can have three different possible relative positions with respect to each other:
1. The planes ${E}_{1}$ and ${E}_{2}$ are identical if they have all points in common. This is the case if and only if the three direction vectors ${\stackrel{\to }{u}}_{1}$, ${\stackrel{\to }{v}}_{1}$, ${\stackrel{\to }{u}}_{2}$ and the three direction vectors ${\stackrel{\to }{u}}_{1}$, ${\stackrel{\to }{v}}_{1}$, ${\stackrel{\to }{v}}_{2}$ are coplanar and the reference point of ${E}_{1}$ lies in ${E}_{2}$.
2. The planes ${E}_{1}$ and ${E}_{2}$ are parallel if they do not have any points in common. This is the case if and only if the three direction vectors ${\stackrel{\to }{u}}_{1}$, ${\stackrel{\to }{v}}_{1}$, ${\stackrel{\to }{u}}_{2}$ and the three direction vectors ${\stackrel{\to }{u}}_{1}$, ${\stackrel{\to }{v}}_{1}$, ${\stackrel{\to }{v}}_{2}$ are coplanar and the reference point of ${E}_{1}$ does not lie in ${E}_{2}$.
3. The planes ${E}_{1}$ and ${E}_{2}$ intersect if the points they have in common form a line. This is the case if and only if the three direction vectors ${\stackrel{\to }{u}}_{1}$, ${\stackrel{\to }{v}}_{1}$, ${\stackrel{\to }{u}}_{2}$ or the three direction vectors ${\stackrel{\to }{u}}_{1}$, ${\stackrel{\to }{v}}_{1}$, ${\stackrel{\to }{v}}_{2}$ are not coplanar.

(This figure will be released shortly.)

Of course, in the conditions of the three cases outlined in the Info Box above, the planes can be exchanged; it can also be checked whether the reference vector of ${E}_{2}$ lies in ${E}_{1}$; this makes no difference. If the planes intersect, the set of intersection points can be determined. Sets of intersection points were already discussed in Section 4.3, where the solvability of systems of linear equations in three variables was interpreted geometrically. A sound understanding of this interpretation is now presumed in this Module and a brief repetition of the material presented in Section 4.3 is highly recommended. The example below illustrates how the relative position of two planes is determined.
##### Example 10.2.23
Let the three planes $E$, $F$, and $G$ be given by the equations

$E:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \\ \hfill -2\hfill \end{array}\right)+a\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right) ; a,b\in ℝ ,$

$F:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)+c\left(\begin{array}{c}\hfill 5\hfill \\ \hfill -2\hfill \\ \hfill -1\hfill \end{array}\right)+d\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -2\hfill \\ \hfill 3\hfill \end{array}\right) ; c,d\in ℝ ,$

and

$G:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)+x\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 0\hfill \end{array}\right)+y\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right) ; x,y\in ℝ .$

• The planes $E$ and $F$ are parallel. The directions vectors $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)$ of $E$ and the first direction vector $\left(\begin{array}{c}\hfill 5\hfill \\ \hfill -2\hfill \\ \hfill -1\hfill \end{array}\right)$ of $F$ are coplanar since the condition

$a\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill -2\hfill \\ \hfill -1\hfill \end{array}\right)$

is satisfied for $a=b=1$. Likewise, the direction vectors $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)$ of $E$ and the second direction vector $\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -2\hfill \\ \hfill 3\hfill \end{array}\right)$ of $F$ are coplanar since the condition

$a\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -2\hfill \\ \hfill 3\hfill \end{array}\right)$

is satisfied for $a=1$ and $b=-1$. Moreover, the reference point of $F$ does not lie in $E$ since the condition

$\left(\begin{array}{c}\hfill 5\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \\ \hfill -2\hfill \end{array}\right)+a\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill a+4b\hfill \\ \hfill 2-2a\hfill \\ \hfill -2+a-2b\hfill \end{array}\right)$

cannot be satisfied for any value of $a$ and $b$. Considering the second component results in $a=\frac{1}{2}$. Substituting this value if $a$ into the equation for the first component results in $b=\frac{9}{8}$. Substituting these two values into the equation for the third component results in the contradiction $0=-2+\frac{1}{2}-\frac{9}{4}$. However, choosing another reference point for $F$, for example the same as for $E$, would result in an equation that describes a plane identical to the plane $E$, i.e. another equivalent parametric representation of one and the same plane. For example, the equation

${F}^{\text{'}}:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \\ \hfill -2\hfill \end{array}\right)+\alpha \left(\begin{array}{c}\hfill 5\hfill \\ \hfill -2\hfill \\ \hfill -1\hfill \end{array}\right)+\beta \left(\begin{array}{c}\hfill -3\hfill \\ \hfill -2\hfill \\ \hfill 3\hfill \end{array}\right) ; \alpha ,\beta \in ℝ$

represents such a plane.
• The planes $E$ and $G$ intersect. Both direction vectors $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 0\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right)$ of $G$ are not coplanar to the direction vectors $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)$ of $E$. For the second direction vector of $G$, we find that the condition

$\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right)=a\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)$

cannot be satisfied for any value of $a$ and $b$. Considering the first two components would result in $a=b=0$. Substituting these values into the equation for the third equation would result in a contradiction. The intersection line of the two planes is calculated by equating the position vectors of the two planes. Here, we have

$\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 2\hfill \\ \hfill -2\hfill \end{array}\right)+a\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill a+4b\hfill \\ \hfill 2-2a\hfill \\ \hfill -2+a-2b\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 5+x\hfill \\ \hfill -2x\hfill \\ \hfill 1+3y\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)+x\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 0\hfill \end{array}\right)+y\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right) .$

This vector equation corresponds to a system of three linear equations in four variables $x$, $y$, $a$, and $b$. According to the methods described in Section 4.4, it is solved by taking one variable as a parameter and solving the system for the other variables as functions of this parameter. This left over parameter will be the parameter in the equation of the intersection line in vector form. Which of the variables is taken for the parameter doesn't matter. Here, we use $x$ as parameter. Considering the first two components of the vector equation result in the system of two linear equations

$a+4b=5+x \text{and} 2-2a=-2x ,$

with the solution $a=1+x$, $b=1$. Substituting this solution into the third component results in

$-2+\left(1+x\right)-2·1=1+3y ⇔ y=\frac{1}{3}x-\frac{4}{3} .$

Now, substituting $y=\frac{1}{3}x-\frac{4}{3}$ or $a=1+x$ and $b=1$ into the equation of the plane $G$ or the plane $E$ results - for the same parameter value - in the same parametric representation of the line $h$, namely the intersection line of the two planes. Specifically, substituting the parameters into the equation of $G$ results in:

$h:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)+x\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 0\hfill \end{array}\right)+\left(\frac{1}{3}x-\frac{4}{3}\right)\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 3\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill 0\hfill \\ \hfill -3\hfill \end{array}\right)+x\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 1\hfill \end{array}\right) ; x\in ℝ .$

##### Exercise 10.2.24
Let the two planes $E$ and $F$ be given by the equations

$E:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 3\hfill \\ \hfill 0\hfill \end{array}\right)+a\left(\begin{array}{c}\hfill 2\hfill \\ \hfill -5\hfill \\ \hfill 8\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 0\hfill \\ \hfill -1\hfill \\ \hfill 4\hfill \end{array}\right) ; a,b\in ℝ$

and

$F:\stackrel{\to }{r}=\left(\begin{array}{c}\hfill 5\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)+c\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 3\hfill \\ \hfill x\hfill \end{array}\right)+d\left(\begin{array}{c}\hfill 2\hfill \\ \hfill y\hfill \\ \hfill 12\hfill \end{array}\right) ; c,d\in ℝ ,$

where the reference point of $F$ does not lie in the plane $E$.
Find the values of the missing components $x$ and $y$ of $F$ such that the planes $F$ and $E$ are parallel.
$x=$

$y=$

##### Exercise 10.2.25
Let the two planes $E$ and $F$ be given by the equations

$E:\stackrel{\to }{r}=a\left(\begin{array}{c}\hfill 2\hfill \\ \hfill -5\hfill \\ \hfill 8\hfill \end{array}\right)+b\left(\begin{array}{c}\hfill 0\hfill \\ \hfill -1\hfill \\ \hfill 4\hfill \end{array}\right) ; a,b\in ℝ$

and

$F:\stackrel{\to }{r}=c\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 3\hfill \\ \hfill 4\hfill \end{array}\right)+d\left(\begin{array}{c}\hfill 2\hfill \\ \hfill -1\hfill \\ \hfill 0\hfill \end{array}\right) ; c,d\in ℝ .$

These planes intersect, and the intersection line is given by the equation

$g:\stackrel{\to }{r}=\xi \left(\begin{array}{c}\hfill 4\hfill \\ \hfill x\hfill \\ \hfill y\hfill \end{array}\right) ; \xi \in ℝ .$

Find the values of the missing components $x$ and $y$ of the direction vector of the intersection line.
$x=$

$y=$