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--- lecture_22_slides [2016/04/20 12:06] – [Lines in $\mathbb{R}^3$] rupert
+++ lecture_22_slides [2017/04/18 09:33] (current) – rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
-==== Last time ====
+====== The distance to a plane ======
+===== Distance from $A$ to $\Pi$ =====
+  * $\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\def\nn{\vec n}\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\def\rt{\bR^3}A$: any point in $\def\rt{\mathbb R^3}\rt$. $\Pi$: a plane with normal vector $\nn$.
+  * $\nn$ is direction of shortest path from $A$ to $\Pi$
+  * Let $B$ be any point in the plane $\Pi$.{{ :dpp.jpg?nolink&600 |}}
+  * (shortest) distance from $A$ to $\Pi$ is $\text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|$
+  * where $\pp=\text{proj}_{\nn}{\vec{AB}}$.
+  * Do some algebra: we get $\text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}$.
+==== Example ====
+Find the distance from $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$.
+  * choose any point $B$ in $\Pi$, e.g. $B=(2,1,0)$
+  * $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$
+  * So $\def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7$.
-  * Take a plane $\Pi$ with normal vector $\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\def\nn{\vec n}\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\def\rt{\bR^3}\nn$
-  * Take a point $A$
-  * $\dist(A,\Pi)=\frac{|\vec{AB}\cdot \nn|}{\|n\|}$ where $B$ is any point in $\Pi$
 ==== The distance from the origin to a plane ====
@@ Line 73: / Line 87: @@
 ==== Example ====
-Find the parametric equation of the line in $\rt$ which passes through $A=(2,1,-3)$ and $B=(4,-1,5)$.
+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}$).
+===== The distance from a point to a line =====
+==== ====
+How can we find $d=\text{dist}(B,L)$, the distance from the point $B$ to a line $L$?
+  * Let $A$ be a point on $L$
+  * Let $\def\vv{\vec v}\vv$ be a direction vector along $L$.
+  * {{ :dl.jpg?nolink&600 |}}
+  * $\text{dist}(B,L) =\|\vec{AB}\|\,\sin \theta = \frac{\|\vec{AB}\|\,\|\vv\|\sin \theta}{\|\vv\|} = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}$.
+==== Example ====
+Find the distance from the point $B=(1,2,3)$ to the line \[L:\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}.\]
+  * Choose $A=\c10{-1}$, in $L$, and  $\vv=\c41{-5}$, along $L$
+  * $\vec{AB}=\c024$, so $ \vec{AB}\times \vv = \cp02441{-5}=\c{-14}{16}{-8}=2\c{-7}{8}{-4}$
+  * So $\def\dist{\text{dist}}\dist(B,L)=\frac{\|\vec{AB}\times\vv\|}{\|\vv\|}=\frac{2\sqrt{7^2+8^2+4^2}}{\sqrt{4^2+1^2+5^2}} = \frac{2\sqrt{129}}{\sqrt{42}} \approx 3.5051.$
+==== Alternative formula ====
+  * $\dist(B,L)=\tfrac{\|\vec{AB}\times \vv\|}{\|\vv\|}$ only works in $\rt$
+    * because the cross product is only defined in $\rt$
+  * {{ :dl2.jpg?nolink&400 |}}
+  * $\def\pp{\vec p}\dist(B,L)=\|\nn\|=\|\vec{AB}-\pp\|$
+    * where $\pp=\def\proj{\text{proj}}\proj_{\vv}\vec{AB}$.
+  * This works in $\bR^n$ for any $n$
+==== Example, again ====
+Find the $\dist(B,L)$ for $B=(1,2,3)$ and $L:\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}$.
+  * $\vec{AB}=\c024$ and $\vv=\c41{-5}$
+  * $\pp=\proj_{\vv}\vec{AB} = \left(\frac{\vec{AB}\cdot \vv}{\|\vv\|^2}\right)\vv = \frac{-18}{42}\c41{-5}= -\frac{3}{7}\c41{-5}$
+  * $\nn=\vec{AB}-\pp=\c024-(-\frac37)\c41{-5}=\frac17\c{12}{17}{13}$
+  * $\dist(B,L)=\|\nn\|=\frac17\sqrt{12^2+17^2+13^2} \approx 3.5051$.