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--- lecture_22 [2017/04/20 08:57] – rupert
+++ lecture_22 [2017/04/20 09:02] (current) – rupert
@@ Line 57: / Line 57: @@
 \[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]
+====== Lines in $\mathbb{R}^3$ ======
+A line $L$ in $\rt$ has an equation of the form
+\[ L: \c xyz=\c abc+t\c def, \quad t\in \mathbb R\]
+where $a,b,c,d,e,f$ are fixed numbers.
+The variable $t$ is called a "free parameter": it's "free" because it can take any value, and it's a "parameter" because this is just another name for a variable.
+The equation above is called a **parametric equation** for the line $L$, because of the free parameter $t$.
+What do $a,b,c,d,e,f$ mean?
+    * Set $t=0$: **$A=(a,b,c)$ is a point in $L$**
+    * Set $t=1$: $B=(a+d,b+e,c+f)$ is another
+    * So **$\vec {AB}=\c def$ is a direction along $L$**.
+==== Example ====
+Find the parametric equation of the line $L$ in $\rt$ which passes through $A=(2,1,-3)$ and $B=(4,-1,5)$.
+  * Note that $\vec{AB}=\c 2{-2}{8}$
+  * So this is a direction vector along the line $L$
+  * So the equation is $ L: \c xyz=\c 21{-3}+t\c 2{-2}8,\quad t\in \mathbb{R}$.
+  * (Same as $ L: \c xyz=\c 4{-1}5+t\c 2{-2}8,\quad t\in \mathbb{R}$).