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--- lecture_24 [2015/04/23 11:31] – rupert
+++ lecture_24 [2016/01/22 19:06] (current) – rupert
@@ Line 1: / Line 1: @@
+~~REVEAL~~
 ===== The distance from a point to a line =====
@@ Line 13: / Line 14: @@
 where $A$ is any point in $L$.
-==== Example ====
+=== Example ===
 To find the distance from the point $B=(1,2,3)$ to the line \[L:\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}\]
@@ Line 20: / Line 21: @@
 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 method ====
+The method above relies on the cross product, so only works in $\rt$. The following alternative method works in $\rn$ for any $n$.
+{{ :dl2.jpg?nolink&600 |}}
+Observe that $\dist(B,L)$ is the length of the vector $\def\nn{\vec n}\nn=\vec{AB}-\def\pp{\vec p}\pp$ where $\pp=\def\proj{\text{proj}}\proj_{\vv}\vec{AB}$.
+=== Example ===
+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 $L_2$.
+To summarise: for skew lines $L_1$ and $L_2$ with direction vectors $\vv_1$ and $\vv_2$, we have
+\[ \dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}\]
+where $\nn=\vv_1\times\vv_2$ and $A$ and $B$ are points with one in $L_1$ and the other in $L_2$.
+=== 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
+\[\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}=\frac{2(5)+2(2)+0(-3)}{\sqrt{5^2+2^2+3^2}} = \frac{14}{\sqrt{38}}.\]