## Line Constructions

1. We will use our knowledge of vectors to construct lines in 3-space, i.e., three-dimensional Euclidean space.

2. Constructing Lines. Suppose we have two points
(1)$\displaystyle A = (4,2,-1)\quad\mbox{and}\quad B = (3,5,7).$
We want to find a line ${\ell}$ passing through these points. What to do?

First we form the vector
(2)\displaystyle \begin{aligned} \vec{v}=\overrightarrow{AB} &= (3-4)\widehat{\textbf{\i}}+(5-2)\widehat{\textbf{\j}}+(7+1)\widehat{\textbf{k}}\\ &=-\widehat{\textbf{\i}}+3\widehat{\textbf{\j}}+8\widehat{\textbf{k}} \end{aligned}
This vector is parallel to ${\ell}$; the numbers given by this vector’s components (i.e., -1, 3, 8) are called the “Direction Numbers” of ${\ell}$.

In general, we have two distinct points ${P_{0}=(x_0,y_0,z_0)}$ and ${P=(x,y,z)}$ on the line ${\ell}$, then we construct the vector
(3)$\displaystyle \overrightarrow{P_{0}P} = (x-x_{0})\widehat{\textbf{\i}}+ (y-y_{0})\widehat{\textbf{\j}}+(z-z_{0})\widehat{\textbf{k}}$
and this is equal to some scalar multiple of ${\vec{v}}$ (i.e., it’s a dilation of the vector). We write
(4)$\displaystyle (x-x_{0})\widehat{\textbf{\i}}+ (y-y_{0})\widehat{\textbf{\j}}+(z-z_{0})\widehat{\textbf{k}} =t\vec{v}$
which lets us write
(5)$\displaystyle \left.\begin{array}{rl} x-x_{0} &=tv_{1}\\ y-y_{0} &=tv_{2}\\ z-z_{0} &=tv_{3} \end{array} \right\}\implies \left\{\begin{array}{rl} x &=x_{0}+tv_{1}\\ y &=y_{0}+tv_{2}\\ z &=z_{0}+tv_{3} \end{array}\right.$
This is the parametric equations of ${\ell}$. So returning to our example, we have
(6)\displaystyle \begin{aligned} x &= 4 - t\\ y &= 2 + 3t\\ z &= -1 + 8t \end{aligned}
where the constant terms are precisely the values of the components of ${A}$, and the coefficients of ${t}$ are the components of the vector ${\overrightarrow{AB}}$.

3. Distance From a Point to a Line What’s the distance from any point ${S}$ in 3-space to a given line ${\ell}$?

We pick a point ${P}$ on ${\ell}$ and form a vector ${\overrightarrow{PS}}$. The distance from ${S}$ to ${\ell}$ can be given as ${d}$, as in the following diagram:

We see ${d=\|\overrightarrow{PS}\|\sin(\theta)}$. If ${\vec{v}}$ is a vector parallel to ${\ell}$, then we have
(7)$\displaystyle \|\overrightarrow{PS}\times\widehat{v}\|=\|\overrightarrow{PS}\|\sin(\theta)$
This is all abstract, lets consider an example.

Example 1. Find the distance from ${S=(2,1,3)}$ to the line given by
(8)\displaystyle \begin{aligned} x &= -1+t\\ y &= 2+t\\ z &= 1+2t \end{aligned}
First we pick the point when ${t=0}$, we call it
(9)$\displaystyle P = (-1,2,1).$
Observe
(10)$\displaystyle \overrightarrow{PS} = 3\widehat{\textbf{\i}}-\widehat{\textbf{\j}}+2\widehat{\textbf{k}}.$
Now we need to find a vector parallel to the line. What to do? Construct a vector by considering the point when ${t=1}$, which would be ${P_{1}=(0,3,3)}$. Thus
(11)\displaystyle \begin{aligned} \vec{v} &=\overrightarrow{PP_{1}}\\ &=(-1-0)\widehat{\textbf{\i}}+(2-3)\widehat{\textbf{\j}}+(1-3)\widehat{\textbf{k}}\\ &=-\widehat{\textbf{\i}}-\widehat{\textbf{\j}}-2\widehat{\textbf{k}} \end{aligned}
Its unit vector is
(12)\displaystyle \begin{aligned} \widehat{v} &= \vec{v}/\|\vec{v}\|\\ &= \vec{v}/\sqrt{1+1+4}\\ &= \vec{v}/\sqrt{6} \end{aligned}
We have
(13)\displaystyle \begin{aligned} \overrightarrow{PS}\times\widehat{v} &= \frac{1}{\sqrt{6}}\begin{vmatrix} \widehat{\textbf{\i}} & \widehat{\textbf{\j}} & \widehat{\textbf{k}}\\ 3 & -1 & 2\\ -1 & -1 & -2 \end{vmatrix}\\ &= \frac{4\widehat{\textbf{\i}}+4\widehat{\textbf{\j}}-4\widehat{\textbf{k}}}{\sqrt{6}} \end{aligned}
This has its magnitude be
(14)$\displaystyle \|\overrightarrow{PS}\times\widehat{v}\| = \frac{4\sqrt{3}}{\sqrt{6}}=2\sqrt{2},$
which describes the distance between our point ${S}$ and the given line.