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--- lecture_22_slides [2016/04/20 12:18] – [Alternative formula] 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 108: / Line 122: @@
   * This works in $\bR^n$ for any $n$
-==== Example ====
+==== Example, again ====
-Let's redo the previous example using this method.
-We have $\vec{AB}=\c024$ and $\vv=\c41{-5}$, so
-\[\pp=\proj_{\vv}\vec{AB} = \left(\frac{\vec{AB}\cdot \vv}{\|\vv\|^2}\right)\vv = \frac{0(4)+2(1)+4(-5)}{4^2+1^2+5^2}\c41{-5} = -\frac{18}{42}\c41{-5}=-\frac{3}{7}\c41{-5},\]
-so
-\[ \nn=\vec{AB}-\pp=\c024-(-\frac37)\c41{-5}=\c024+\frac37\c41{-5}=\frac17\c{12}{17}{13}\]
-so
-\[ \dist(B,L)=\|\nn\|=\frac17\sqrt{12^2+17^2+13^2} = \frac17\sqrt{602} \approx 3.5051.\]
-===== The distance between skew lines in $\mathbb{R}^3$ =====
-Suppose that $L_1$ and $L_2$ are skew lines in $\rt$: lines which are not parallel and do not cross.
-Let $\vv_1$ be a direction vector along $L_1$, and let $\vv_2$ be a direction vector along $L_2$.
-{{ :sl1.png?nolink&600 |}}
-The shortest distance from $L_1$ and $L_2$ is measured along the direction orthogonal to both $\vv_1$ and $\vv_2$, namely the direction of $\nn=\vv_1\times\vv_2$.
-Let $\Pi$ be the plane with normal vector $\nn$ which contains $L_1$.
-{{ :sl2.png?nolink&600 |}}
-For any point $B$ in $L_2$, we have
-\[\dist(L_1,L_2)=\dist(B,\Pi) = \frac{|\vec{AB}\cdot \nn|}{\|\nn\|}\]
-where $A$ is any point in $\Pi$; for example, we can take $A$ to be any point in $L_2$.
+Find the $\dist(B,L)$ for $B=(1,2,3)$ and $L:\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}$.
-To summarise: for skew lines $L_1$ and $L_2$ with direction vectors $\vv_1$ and $\vv_2$, we have
+  * $\vec{AB}=\c024$ and $\vv=\c41{-5}$
-\[ \dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}\]
+  * $\pp=\proj_{\vv}\vec{AB} = \left(\frac{\vec{AB}\cdot \vv}{\|\vv\|^2}\right)\vv = \frac{-18}{42}\c41{-5}= -\frac{3}{7}\c41{-5}$
-where $\nn=\vv_1\times\vv_2$ and $A$ and $B$ are points with one in $L_1$ and the other in $L_2$.
+  * $\nn=\vec{AB}-\pp=\c024-(-\frac37)\c41{-5}=\frac17\c{12}{17}{13}$
+  * $\dist(B,L)=\|\nn\|=\frac17\sqrt{12^2+17^2+13^2} \approx 3.5051$.
-=== Example ===
-Consider the skew lines
-\[ L_1:\c xyz=\c 101+t_1\c123,\quad t_1\in\mathbb{R}\]
-and
-\[ L_2:\c xyz=\c 321+t_2\c1{-1}1,\quad t_2\in\mathbb{R}.\]
-The direction vectors are $\c123$ and $\c1{-1}1$, so we take $\nn$ to be their cross product:
-\[ \nn=\c123\times\c1{-1}1=\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\1&2&3\\1&-1&1\end{vmatrix}=\c52{-3}\]
-and if $A=(1,0,1)$ and $B=(3,2,1)$ then $A$ and $B$ are points with one in $L_1$ and the other in $L_2$, and $\vec{AB}=\c 220$. Hence
-\[\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}=\frac{2(5)+2(2)+0(-3)}{\sqrt{5^2+2^2+3^2}} = \frac{14}{\sqrt{38}}.\]