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--- lecture_22 [2016/04/21 12:31] – rupert
+++ lecture_22 [2017/04/20 09:02] (current) – rupert
@@ Line 1: / Line 1: @@
+====== The distance to a plane ======
+===== The distance from a point to a plane =====
+Let $\Pi$ be a plane in $\def\rt{\mathbb{R}^3}\rt$ with equation $ax+by+cz=d$, so that $\def\nn{\vec n}\nn=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c abc$ is a normal vector to $\Pi$. Also let $A$ be any point in $\rt$.
+The shortest path from $A$ to a point in $\Pi$ goes in the same direction as $\nn$. Let $B$ be any point in the plane $\Pi$.
+{{ :dpp.jpg?nolink&600 |}}
+From the diagram, we see that the shortest distance from $A$ to $\Pi$ is given by
+\[ \text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|\]
+where
+\[ \pp=\text{proj}_{\nn}{\vec{AB}}.\]
+Using the formula for $\text{proj}_{\vec w}\vec v$ and the fact that $\|c\vec v\|=|c|\,\|\vec v\|$ where $c$ is a scalar and $\vec v$ is a vector, we obtain the formula
+\[ \text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}.\]
+==== Example ====
+To find the distance from  $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$, choose any point $B$ in $\Pi$; for example, let $B=(2,1,0)$. Then $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$, so
+\[ \def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7.\]
 ==== Remark: the distance from the origin to a plane ====
@@ Line 35: / Line 57: @@
 \[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]
-===== The distance from a point to a line =====
+====== Lines in $\mathbb{R}^3$ ======
-Suppose $L$ is a line in $\def\rt{\mathbb{R}^3}\def\rn{\mathbb{R}^n}\rt$. Let $A$ be a point on $L$ and let $\def\vv{\vec v}\vv$ be a direction vector along $L$.
+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.
-Given a point $B$, how can we find $d=\text{dist}(B,L)$, the (shortest) distance from the point $B$ to the line $L$?
+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.
-{{ :dl.jpg?nolink&600 |}}
+The equation above is called a **parametric equation** for the line $L$, because of the free parameter $t$.
-Let $A$ be any point in $L$ and let $\theta$ be the angle between $AB$ and $\vv$. We have
+What do $a,b,c,d,e,f$ mean?
-\[ d=\|\vec{AB}\|\,\sin \theta = \frac{\|\vec{AB}\|\,\|\vv\|\sin \theta}{\|\vv\|} = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}.\]
+    * Set $t=0$: **$A=(a,b,c)$ is a point in $L$**
-So
+    * Set $t=1$: $B=(a+d,b+e,c+f)$ is another
-\[ \text{dist}(B,L) = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}\]
+    * So **$\vec {AB}=\c def$ is a direction along $L$**.
-where $A$ is any point in $L$.
-=== Example ===
+==== Example ====
-To find the distance from the point $B=(1,2,3)$ to the line \[L:\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}\]
+Find the parametric equation of the line $L$ in $\rt$ which passes through $A=(2,1,-3)$ and $B=(4,-1,5)$.
-we can choose $A=\c10{-1}$ so that $\vec{AB}=\c024$ and taking $\vv=\c41{-5}$, we obtain
+  * Note that $\vec{AB}=\c 2{-2}{8}$
-\[ \vec{AB}\times \vv = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\0&2&4\\4&1&-5\end{vmatrix}=\c{-14}{16}{-8}=2\c{-7}{8}{-4}\]
+  * So this is a direction vector along the line $L$
-so
+  * So the equation is $ L: \c xyz=\c 21{-3}+t\c 2{-2}8,\quad t\in \mathbb{R}$.
-\[ \def\dist{\text{dist}}\dist(B,L)=\frac{\|\vec{AB}\times\vv\|}{\|\vv\|}=\frac{2\sqrt{7^2+8^2+4^2}}{\sqrt{4^2+1^2+5^2}} = \frac{2\sqrt{129}}{\sqrt{42}} \approx 3.5051.\]
+  * (Same as $ L: \c xyz=\c 4{-1}5+t\c 2{-2}8,\quad t\in \mathbb{R}$).