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# 3.2: Lines and Planes in Eⁿ

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To obtain a line in $$E^{2}$$ or $$E^{3}$$ passing through two points $$\overline{a}$$ and $$\overline{b},$$ we take the vector

$\vec{u}=\overrightarrow{a b}=\overline{b}-\overline{a}$

and, so to say, "stretch" it indefinitely in both directions, i.e., multiply $$\vec{u}$$ by all possible scalars $$t \in E^{1} .$$ Then the set of all points $$\overline{x}$$ of the form

$\overline{x}=\overline{a}+t \vec{u}$

is the required line. It is natural to adopt this as a definition in $$E^{n}$$ as well.

Below, $$\overline{a} \neq \overline{b}$$.

Definition: parametric equation of the line

The line $$\overline{a b}$$ through the points $$\overline{a}, \overline{b} \in E^{n}$$ (also called the line through $$\overline{a},$$ in the direction of the vector $$\vec{u}=\overline{b}-\overline{a}$$ ) is the set of all points $$\overline{x} \in E^{n}$$ of the form

$\overline{x}=\overline{a}+t \vec{u}=\overline{a}+t(\overline{b}-\overline{a}),$

where $$t$$ varies over $$E^{1} .$$ We call $$t$$ a variable real parameter and $$\vec{u}$$ a direction vector for $$\overline{a b} .$$ Thus

$\text{Line } \overline{a b}=\left\{\overline{x} \in E^{n} | \overline{x}=\overline{a}+t \vec{u} \text{ for some } t \in E^{1}\right\}, \quad \vec{u}=\overline{b}-\overline{a} \neq \overline{0}.$

The formula

$\overline{x}=\overline{a}+t \vec{u}, \text{ or } \overline{x}=\overline{a}+t(\overline{b}-\overline{a}),$

is called the parametric equation of the line. (We briefly say "the line $$\overline{x}=\overline{a}+t \vec{u} . "$$ ) It is equivalent to $$n$$ simultaneous equations in terms of coordinates, namely,

$x_{k}=a_{k}+t u_{k}=a_{k}+t\left(b_{k}-a_{k}\right), \quad k=1,2, \ldots, n.$

Note 1. As the vector $$\vec{u}$$ is anyway being multiplied by all real numbers $$t$$, the line (as a set of points) will not change if $$\vec{u}$$ is replaced by some $$c \vec{u}\left(c \in E^{1}\right,$$ $$c \neq 0 ) .$$ In particular, taking $$c=1 /|\vec{u}|,$$ we may replace $$\vec{u}$$ by $$\vec{u} /|\vec{u}|, \quad a$$ unit vector.$$.$$ We may as well assume that $$\vec{u}$$ is a unit vector itself.

If we let $$t$$ vary not over all of $$E^{\perp}$$ but only over some interval in $$E^{\perp},$$ we obtain what is called a line segment. In particular, we define the open line segment $$L(\overline{a}, \overline{b}),$$ the closed line segment $$L[\overline{a}, \overline{b}],$$ the half-open line segment $$L(\overline{a}, \overline{b}],$$ and the half-closed line segment $$L[\overline{a}, \overline{b}),$$ as we did for $$E^{1}$$.

Definition: endpoints of the segment

Given $$\vec{u}=\overline{b}-\overline{a},$$ we set

\begin{aligned} \text { (i) } L(\overline{a}, \overline{b})=\{\overline{a}+t \vec{u} | 0<t<1\} ; & \quad \text { (ii) } L[\overline{a}, \overline{b}]=\{\overline{a}+t \vec{u} | 0 \leq t \leq 1\} \\ \text { (iii) } L(\overline{a}, \overline{b}]=\{\overline{a}+t \vec{u} | 0<t \leq 1\} ; & \quad \text { (iv) } L[\overline{a}, \overline{b})=\{\overline{a}+t \vec{u} | 0 \leq t<1\} \end{aligned}

In all cases, $$\overline{a}$$ and $$\overline{b}$$ are called the endpoints of the segment; $$\rho(\overline{a}, \overline{b})=|\overline{b}-\overline{a}|$$ is its length; and $$\frac{1}{2}(\overline{a}+\overline{b})$$ is its midpoint.

Note that in $$E^{1},$$ line segments simply become intervals, $$(a, b),[a, b],$$ etc.

To describe a plane in $$E^{3},$$ we fix one of its points, $$\overline{a},$$ and a vector $$\vec{u}=\overrightarrow{a b}$$ perpendicular to the plane (imagine a vertical pencil standing at $$\overline{a}$$ on the horizontal plane of the table). Then a point $$\overline{x}$$ lies on the plane iff $$\vec{u} \perp \overrightarrow{a x}$$ . It is natural to accept this as a definition in $$E^{n}$$ as well.

Definition: plane

Given a point $$\overline{a} \in E^{n}$$ and a vector $$u \neq \overrightarrow{0},$$ we define the plane (also called hyperplane if $$n>3$$) through $$\overline{a}$$, orthogonal to $$\vec{u},$$ to be the set of all $$\overline{x} \in E^{n}$$ such that $$\vec{u} \perp \overrightarrow{a x},$$ i.e., $$\vec{u} \cdot(\overline{x}-\overline{a})=0,$$ or, in terms of components,

$\sum_{k=1}^{n} u_{k}\left(x_{k}-a_{k}\right)=0,\text{ where } \vec{u} \neq \overrightarrow{0}\text{ (i.e., not all values } \ u_{k}\text{ are 0). }$

We briefly say

$\text{ "the plane } \vec{u} \cdot(\overline{x}-\overline{a})=0\text{" or "the plane } \sum_{k=1}^{n} u_{k}\left(x_{k}-a_{k}\right)=0\text{"}$

(this being the equation of the plane). Removing brackets in (3), we have

$u_{1} x_{2}+u_{2} x_{2}+\cdots+u_{n} x_{n}=c,\text{ or } \vec{u} \cdot \overline{x}=c,\text{ where } c=\sum_{k-1}^{n} u_{k} a_{k}, \vec{u} \neq \overrightarrow{0}.$

An equation of this form is said to be linear in $$x_{1}, x_{2}, \ldots, x_{n}.$$

Theorem $$\PageIndex{1}$$

A set $$A \subseteq E^{n}$$ is a plane (hyperplane) iff $$A$$ is exactly the set of all $$\overline{x} \in E^{n}$$ satisfying $$(4)$$ for some fixed $$c \in E^{1}$$ and $$\vec{u}=\left(u_{1}, \ldots, u_{n}\right) \neq \overline{0}.$$

Proof

Add proof here and it will automatically be hidden Indeed, as we saw above, each plane has an equation of the form (4).

Conversely, any equation of that form (with, say, $$u_{1} \neq 0 )$$ can be written as

$u_{1}\left(x_{1}-\frac{c}{u_{1}}\right)+u_{2} x_{2}+u_{3} x_{3}+\cdots+u_{n} x_{n}=0.$

Then, setting $$a_{1}=c / u_{1}$$ and $$a_{k}=0$$ for $$k \geq 2,$$ we transform it into $$(3),$$ which is, by definition, the equation of a plane through $$\overline{a}=\left(c / u_{1}, 0, \ldots, 0\right),$$ orthogonal to $$u=\left(u_{1}, \ldots, u_{n}\right). \square$$

Thus, briefly, planes are exactly all sets with linear equations (4). In this connection, equation (4) is called the general equation of a plane. The vector $$\vec{u}$$ is said to be normal to the plane. Clearly, if both sides of (4) are multiplied by a nonzero scalar $$q,$$ one obtains an equivalent equation (representing the same set). Thus we may replace $$u_{k}$$ by $$q u_{k},$$ i.e., $$\vec{u}$$ by $$q \vec{u},$$ without affecting the plane. In particular, we may replace $$\vec{u}$$ by the unit vector $$\vec{u} /|\vec{u}|,$$ as in lines is called the normalization of the equation). Thus

$\frac{\vec{u}}{|\vec{u}|} \cdot(\overline{x}-\overline{a})=0$

and

$\overline{x}=\overline{a}+t \frac{\vec{u}}{|\vec{u}|}$

are the normalized (or normal) equations of the plane (3) and line (1), respectively.

Note 2. The equation $$x_{k}=c$$ (for a fixed $$k )$$ represents a plane orthogonal to the basic unit vector $$\vec{e}_{k}$$ or, as we shall say, to the kth axis. The equation results from (4) if we take $$\vec{u}=\vec{e}_{k}$$ so that $$u_{k}=1,$$ while $$u_{i}=0$$ $$(i \neq k).$$ For example, $$x_{1}=c$$ is the equation of a plane orthogonal to $$\vec{e}_{1} ;$$ it consists of all $$\overline{x} \in E^{n},$$ with $$x_{1}=c$$ (while the other coordinates of $$\overline{x}$$ are arbitrary$$) .$$ In $$E^{2},$$ it is a line. In $$E^{1},$$ it consists of $$c$$ alone.

Two planes (respectively, two lines) are said to be perpendicular to each other iff their normal vectors (respectively, direction vectors) are orthogonal; similarly for parallelism. A plane $$\vec{u} \cdot \overline{x}=c$$ is said to be perpendicular to a line $$\overline{x}=\overline{a}+t \vec{v}$$ iff $$\vec{u} \| \vec{v} ;$$ the line and the plane are parallel iff $$\vec{u} \perp \vec{v}$$ .

Note 3. When normalizing, as in (5) or (6), we actually have two choices of a unit vector, namely, $$\pm \vec{u} /|\vec{u}| .$$ If one of them is prescribed, we speak of a directed plane (respectively, line).

Example $$\PageIndex{1}$$

(a) Let $$\overline{a}=(0,-1,2), \overline{b}=(1,1,1),$$ and $$\overline{c}=(3,1,-1)$$ in $$E^{3} .$$ Then the line $$\overline{a b}$$ has the parametric equation $$\overline{x}=\overline{a}+t(\overline{b}-\overline{a})$$ or, in coordinates, writing $$x, y, z$$ for $$x_{1}, x_{2}, x_{3},$$

$x=0+t(1-0)=t, y=-1+2 t, z=2-t.$

This may be rewritten

$t=\frac{x}{1}=\frac{y+1}{2}=\frac{z-2}{-1},$

where $$\vec{u}=(1,2,-1)$$ is the direction vector (composed of the denominators. Normalizing and dropping $$t,$$ we have

$\frac{x}{1 / \sqrt{6}}=\frac{y+1}{2 / \sqrt{6}}=\frac{z-2}{-1 / \sqrt{6}}$

(the so-called symmetric form of the normal equations).

Similarly, for the line $$\overline{b c},$$ we obtain

$t=\frac{x-1}{2}=\frac{y-1}{0}=\frac{z-1}{-2},$

where "$$t=(y-1) / 0$$" stands for "$$y-1=0$$." (It is customary to use this notation.)

(b) Let $$\overline{a}=(1,-2,0,3)$$ and $$\vec{u}=(1,1,1,1)$$ in $$E^{4} .$$ Then the plane normal to $$\vec{u}$$ through $$\overline{a}$$ has the equation $$(\overline{x}-\overline{a}) \cdot \vec{u}=0,$$ or

$\left(x_{1}-1\right) \cdot 1+\left(x_{2}+2\right) \cdot 1+\left(x_{3}-0\right) \cdot 1+\left(x_{4}-3\right) \cdot 1=0,$

or $$x_{1}+x_{2}+x_{3}+x_{4}=2 .$$ Observe that, by formula (4), the coefficients of $$x_{1}, x_{2}, x_{3}, x_{4}$$ are the components of the normal vector $$\vec{u}$$ (here $$(1,1,1,1) ).$$

Now define a map $$f : E^{4} \rightarrow E^{1}$$ setting $$f(\overline{x})=x_{1}+x_{2}+x_{3}+x_{4}$$ (the left-hand side of the equation). This map is called the linear functional corresponding to the given plane. (For another approach, see Problems 4-6 below.)

(c) The equation $$x+3 y-2 z=1$$ represents a plane in $$E^{3},$$ with $$\vec{u}=(1,3,-2)$$ . The point $$\overline{a}=(1,0,0)$$ lies on the plane (why?), so the plane equation may be written $$(\overline{x}-\overline{a}) \cdot \vec{u}=0$$ or $$\overline{x} \cdot \vec{u}=1,$$ where $$\overline{x}=(x, y, z)$$ and $$\overline{a}$$ and $$\vec{u}$$ are as above.