# 3.2: Lines and Planes in Eⁿ

- Page ID
- 19036

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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.