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).
(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.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Geometry, Vector Calculus and tagged . Bookmark the permalink.

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