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--- lecture_22 [2015/04/16 12:20] – rupert
+++ lecture_22 [2017/04/20 09:02] (current) – rupert
@@ Line 1: / Line 1: @@
- ===== Planes in $\mathbb{R}^3$ =====
-==== Examples ====
+====== The distance to a plane ======
-. Find the equation of the plane with normal vector $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c1{-3}2$ which contains the point $(1,-2,1)$. Then find three other points in this plane.
+===== The distance from a point to a plane =====
-Solution: the equation is $x-3y+2z=d$, and we can find $d$ by subbing in $(x,y,z)=(1,-2,1)$: $1-3(-2)+2(1)=d$, so $d=9$ and the equation of the plane is
+Let $\Pi$ be a plane in $\def\rt{\mathbb{R}^3}\rt$ with equation $ax+by+cz=d$, so that $\def\nn{\vec n}\nn=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c abc$ is a normal vector to $\Pi$. Also let $A$ be any point in $\rt$.
-\[ x-3y+2z=9.\]
-Some other points in this plane are $(9,0,0)$, $(0,1,6)$, $(1,1,\tfrac{11}2)$. (We can find these by inspection).
-. Find the equation of the plane containing the points $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$.
+The shortest path from $A$ to a point in $\Pi$ goes in the same direction as $\nn$. Let $B$ be any point in the plane $\Pi$.
-Solution: $\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in this plane. We want to find a normal vector $\nn$ which is be orthogonal to both of these. The cross product of two vectors is orthogonal to both, so we can take the cross product of $\vec{AB}$ and $\vec{AC}$:
+{{ :dpp.jpg?nolink&600 |}}
-\[ \nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\cp2{-2}131{-2}=\c378\]
-so the equation of the plane is $3x+7y+8z=d$, and we find $d$ by subbing in a point in the plane, say $A=(1,2,0)$, which gives $d=3(1)+7(2)+8(0)=17$. So the equation is
-\[ 3x+7y+8z=17.\]
-==== Orthogonal planes and parallel planes ====
+From the diagram, we see that the shortest distance from $A$ to $\Pi$ is given by
+\[ \text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|\]
+where
+\[ \pp=\text{proj}_{\nn}{\vec{AB}}.\]
+Using the formula for $\text{proj}_{\vec w}\vec v$ and the fact that $\|c\vec v\|=|c|\,\|\vec v\|$ where $c$ is a scalar and $\vec v$ is a vector, we obtain the formula
+\[ \text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}.\]
-Let $\Pi_1$ be a plane with normal vector $\nn_1$, and let $\Pi_2$ be a plane with normal vector $\nn_2$.
+==== Example ====
-  - $\Pi_1$ and $\Pi_2$ are //orthogonal// or //perpendicular// planes if they meet at right angles. The following conditions are equivalent:
+To find the distance from  $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$, choose any point $B$ in $\Pi$; for example, let $B=(2,1,0)$. Then $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$, so
-    - $\Pi_1$ and $\Pi_2$ are orthogonal planes;
+\[ \def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7.\]
-    - $\nn_1\cdot\nn_2=0$;
-    - $\nn_1$ is a vector in $\Pi_2$;
-    - $\nn_2$ is a vector in $\Pi_1$.
-  - $\Pi_1$ and $\Pi_2$ are //parallel// planes if they have the same normal vectors.
-=== Examples ===
+==== Remark: the distance from the origin to a plane ====
-. Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
+If we write $\def\rt{\mathbb{R}^3}\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}0=(0,0,0)$ for the origin in $\rt$ and apply the formula above to the plane $\Pi:ax+by+cz=d$ with $B=(d/a,0,0)$ (assuming that $a\ne 0$) then we obtain
+\[ \dist(0,\Pi)=\frac{|d|}{\|\nn\|}\]
+where $\nn$ is the normal vector $\nn=\c abc$.
-Solution: The plane $x-y+z=5$ has normal vector $\c1{-1}1$, so this is a vector in $\Pi$. Moreover, $\vec{AB}=\c3{-5}4$ is also a vector in $\Pi$, so it has normal vector
+So as $d$ varies (with the normal vector $\nn$ fixed), we obtain parallel planes at different distances to the origin $0$; the larger $d$ is, the further the plane is from $0$.
-\[ \n=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}.\]
-So the equation of $\Pi$ is $x-y-2z=d$ and subbing in $A=(1,3,-3)$ gives $d=1-3-2(-3)=4$, so the equation of $\Pi$ is \[ x-y-2z=4.\]
+===== The distance between parallel planes =====
+If $\Pi_1$ and $\Pi_2$ are parallel planes, then the shortest distance between them is given by
+\[ \dist(\Pi_1,\Pi_2)=\dist(A,\Pi_2)\]
+for any point $A$ is $\Pi_1$. The reason is that for parallel planes, changing $A$ to a different point in $\Pi_1$ does not change $\dist(A,\Pi_2)$.
+Of course, if the planes $\Pi_1$ and $\Pi_2$ are not parallel, then they intersect (in many points: in a whole line). So for non-parallel planes we always have $\dist(\Pi_1,\Pi_2)=0$.
+==== 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
+\[ \frac{|\tfrac{11}2-4|}{\|\nn\|} = \frac{|\tfrac 32|}{\sqrt{1^2+3^2+(-5)^2}} = \frac3{2\sqrt{35}}.\]
+====== Lines in $\mathbb{R}^3$ ======
+A line $L$ in $\rt$ has an equation of the form
+\[ L: \c xyz=\c abc+t\c def, \quad t\in \mathbb R\]
+where $a,b,c,d,e,f$ are fixed numbers.
+The variable $t$ is called a "free parameter": it's "free" because it can take any value, and it's a "parameter" because this is just another name for a variable.
+The equation above is called a **parametric equation** for the line $L$, because of the free parameter $t$.
+What do $a,b,c,d,e,f$ mean?
+    * Set $t=0$: **$A=(a,b,c)$ is a point in $L$**
+    * Set $t=1$: $B=(a+d,b+e,c+f)$ is another
+    * So **$\vec {AB}=\c def$ is a direction along $L$**.
+==== Example ====
+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}$).
-.