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--- lecture_23 [2016/04/26 10:23] – [The distance between skew lines in $\mathbb{R}^3$] rupert
+++ lecture_23 [2017/05/06 09:59] (current) – rupert
@@ Line 1: / Line 1: @@
 ===== The distance from a point to a line =====
+==== Cross product method ====
+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 $L$.
+Given a point $B$, how can we find $d=\text{dist}(B,L)$, the (shortest) distance from the point $B$ to the line $L$?
-==== Alternative method ====
+{{ :dl.jpg?nolink&600 |}}
+Let $A$ be any point in $L$ and let $\theta$ be the angle between $AB$ and $\vv$. We have
+\[ d=\|\vec{AB}\|\,\sin \theta = \frac{\|\vec{AB}\|\,\|\vv\|\sin \theta}{\|\vv\|} = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}.\]
+So
+\[ \text{dist}(B,L) = \frac{\|\vec{AB}\times \vv\|}{\|\vv\|}\]
+where $A$ is any point in $L$.
+=== Example ===
+To find the distance from the point $B=(1,2,3)$ to the line \[L:\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c xyz=\c10{-1}+t\c41{-5},\quad t\in\mathbb{R}\]
+we can choose $A=(1,0,{-1})$ so that $\vec{AB}=\c024$ and taking $\vv=\c41{-5}$, we obtain
+\[ \vec{AB}\times \vv = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\0&2&4\\4&1&-5\end{vmatrix}=\c{-14}{16}{-8}=2\c{-7}{8}{-4}\]
+so
+\[ \def\dist{\text{dist}}\dist(B,L)=\frac{\|\vec{AB}\times\vv\|}{\|\vv\|}=\frac{2\sqrt{7^2+8^2+4^2}}{\sqrt{4^2+1^2+5^2}} = \frac{2\sqrt{129}}{\sqrt{42}} \approx 3.5051.\]
+==== Dot product method ====
 The method above relies on the cross product, so only works in $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\def\rt{\mathbb{R}^3}\def\rn{\mathbb{R}^n}\def\vv{\vec v}\def\dist{\text{dist}}\rt$. The following alternative method works in $\rn$ for any $n$.
@@ Line 22: / Line 41: @@
 \[ \dist(B,L)=\|\nn\|=\frac17\sqrt{12^2+17^2+13^2} = \frac17\sqrt{602} \approx 3.5051.\]
-===== The distance between skew lines in $\mathbb{R}^3$ =====
+===== The distance between lines in $\mathbb{R}^3$ =====
+==== Skew lines ====
 Suppose that $L_1$ and $L_2$ are skew lines in $\rt$: lines which are not parallel and do not cross.
@@ Line 46: / Line 67: @@
 == Remark ==
-What about the distance between lines which are not skew? This means that either they are non-parallel and they intersect (so that the distance between them will be zero), or the are parallel lines.
+What about the distance between lines which are not skew? This means that either they are non-parallel and they intersect (so that the distance between them will be zero), or they are parallel lines.
-  * The same method and formula work if $L_1$ and $L_2$ are non-parallel lines which intersect, and you get  $\dist(L_1,L_2)=0$ in this case. The reason is that in this case $L_1$ and $L_2$ will lie in one plane, $\Pi$, and $\vec{AB}$ will also be in~$\Pi$, and $\vec n$ will be orthogonal to $\Pi$, so $\vec{AB}\cdot \vec n=0$.
+  * The same method and formula work if $L_1$ and $L_2$ are non-parallel lines which intersect, and you get  $\dist(L_1,L_2)=0$ in this case. The reason is that in this case $L_1$ and $L_2$ will lie in one plane, $\Pi$, and $\vec{AB}$ will also be in $\Pi$, and $\vec n$ will be orthogonal to $\Pi$. So $\frac{\vec{AB}\cdot \vec n}{\|\vec n\|}=\frac{0}{\|\vec n\|}=0=\dist(L_1,L_2)$.
-  * If $L_1$ and $L_2$ are parallel lines (i.e., if the vectors $\vec v_1$ and $\vec v_2$ along the lines are in the same direction), then a different method is required to find $\vec n$ because $\vv_1\times \vv_2=0$ which isn't helpful. However, if you can find a (non-zero) vector $\vec n$ orthogonal to $\vec v_1$ and $\vec v_2$, then the formula $\dist(L_1,L_2)=\frac{\vec AB\cdot \vec n}{\|\vec n\|}$ will give the correct answer.
+  * If $L_1$ and $L_2$ are parallel lines (i.e., if the vectors $\vec v_1$ and $\vec v_2$ along the lines are in the same direction), then $\vv_1\times \vv_2=0$ which isn't helpful, so this method won't work here. In this case, observe that $\dist(L_1,L_2)=\dist(A,L_2)$ where $A$ is any point in $L_1$ (because the lines are parallel), so you can use one of the formulae above for the distance from a point to a line.
 === Example ===
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 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
 \[\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}=\frac{2(5)+2(2)+0(-3)}{\sqrt{5^2+2^2+3^2}} = \frac{14}{\sqrt{38}}.\]
+==== Distance between lines in $\mathbb{R}^3$ in general ====
+The formula $\dist(L_1,L_2)=\frac{|\vec{AB}\cdot \nn|}{\|\nn\|}$ where $\nn=\vec v_1\times \vec v_2$ works for
+  * skew lines (not parallel, not intersecting), as we saw above,
+  * and actually: any non-parallel lines $L_1$, $L_2$. We can see this by noticing that if the lines above intersect, then $L_1$ lies in $\Pi$, so the formula gives $\dist(L_1,L_2)=\dist(B,\Pi)=0$ which is the correct answer.
+What about parallel lines?
+    * The formula can't work because we'd have $\vec v_1=\vec v_2$ so $\vec n=\vec v_1\times \vec v_2=\vec 0$
+    * Instead: observe that when $L_1$ and $L_2$ are parallel, we have $\dist(L_1,L_2)=\dist(A,L_2)$ for any point $A$ in $L_1$
+    * So we can use one of of the point-to-line distance formulae we saw earlier.