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--- lecture_21_slides [2016/04/14 11:45] – [The distance between parallel planes] rupert
+++ lecture_21_slides [2017/04/18 09:18] (current) – rupert
@@ Line 75: / Line 75: @@
   * Sub in: $d=0-13(3)+17(2)=-39+34=-5$
   * Answer: $-2x-13y+17z=-5$, or $2x+13y-17z=5$.
-====== The distance to a plane ======
-===== Distance from $A$ to $\Pi$ =====
-  * $A$: 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 $\def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7$.
-==== The distance from the origin to a plane ====
-  * We write $0=(0,0,0)$ for the origin in $\rt$
-  * Distance from $0$ to a plane $\Pi:ax+by+cz=d$ ?
-  * Take $B=(d/a,0,0)$ (assuming that $a\ne 0$)
-  * We get $\dist(0,\Pi)=\frac{|d|}{\|\nn\|}$ where $\nn$ is the normal vector $\nn=\c abc$.
-  * In particular, if $\nn$ is a unit vector, then $\dist(0,\Pi)=|d|$.
-  * As $d$ varies (with $\nn$ fixed), we obtain parallel planes at different distances to the origin $0$
-  * The larger $d$ is, the further the plane is from $0$.
-===== The distance between planes =====
-  * Let $\Pi_1$ and $\Pi_2$ be two planes. What is the (shortest) distance between them?
-  * If they're not parallel, then they intersect! So $\dist(\Pi_1,\Pi_2)=0$.
-  * If they're parallel, then $\dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2)$ for any point $A$ in $\Pi_1$.
-    * Why?
-    * Since the planes are parallel, $\dist(A,\Pi_2)$ doesn't change if $A$ changes in $\Pi_1$
-    * So this is also the shortest distance, for any choice of $A$ in $\Pi_1$
-==== Example ====
-The distance between the planes $3x+4y-2z=5$ and $3x+4y-3z=1$ is $0$, since the normal vectors $\c34{-2}$ and $\c34{-3}$ are not scalar multiples of one another, so they are in different directions, so the planes are not parallel.
-==== Example ====
-The planes $\Pi_1:3x+4y-2z=5$ and $\Pi_2:3x+4y-2z=1$ have the same normal vector $\c34{-2}$, so they are parallel. Their distance is given by $\dist(A,\Pi_2)$ where $A$ is any point in $\Pi_1$, and to find this we also need a point $B$ in $\Pi_2$.
-We can choose $A=(1,0,-1)\in \Pi_1$ and $B=(1,0,1)\in \Pi_2$. (Of course, there are lots of different possible choices here, but they should all give the same answer!) Then $\vec {AB}=\c002$ and
-\[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2) = \frac{|\nn\cdot \vec{AB}|}{\|n\|}=\frac{|0+0+(-2)2|}{\sqrt{3^2+4^2+(-2)^2}} = \frac4{\sqrt{29}}.\]
-==== Exercise: a formula for the distance between parallel planes ====
-Show that the distance between the parallel planes $\Pi_1:ax+by+cz=d_1$ and $\Pi_2:ax+by+cz=d_2$ is \[\dist(\Pi_1,\Pi_2)=\frac{|d_2-d_1|}{\|\nn\|},\] where $\nn=\c abc$.
-==== Example ====
-To find the distance between $x+3y-5z=4$ and $2x+6y-10z=11$ we can rewrite the second equation as $x+3y-5z=11/2$ to see that this is a parallel plane to the first, with common normal vector $\nn=\c13{-5}$. By the formula in the exercise the distance between these planes is
-\[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]