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--- lecture_20_slides [2016/04/13 15:19] – rupert
+++ lecture_20_slides [2016/04/13 15:43] (current) – rupert
@@ Line 108: / Line 108: @@
 ==== Example 3 ====
+What's the equation of the plane parallel to  $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$?
+  * A normal vector is $\nn=\c111\times\c1{-1}1=\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\cp1111{-1}1=\c{2}0{-2}$
+  * So the equation is $2x+0y-2z=2(3)-2(1)=4$,
+  * or $2x-2z=4$
+  * or $x-z=2$.
+==== Example 4 ====
 Find the equation of the plane $\Pi$ containing $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$.
@@ Line 115: / Line 124: @@
   * $\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in $\Pi$
   * Need $\nn$, orthogonal to both. Use cross product!
-  * $\nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\cp2{-2}131{-2}=\c378$
+  * $\nn=\vec{AB}\times\vec{AC}=\cp2{-2}131{-2}=\c378$
-  * Equation is $3x+7y+8z=d$; find $d$ by subbing in a point in $\Pi$
+  * Equation is $3x+7y+8z=d$; find $d=17$ by subbing in $A=(1,2,0)$
-  * e.g. $A=(1,2,0)$ is in $\Pi$, so $d=3(1)+7(2)+8(0)=17$.
   * Answer: $ 3x+7y+8z=17$.
-==== Orthogonal planes and parallel planes ====
+==== Parallel planes ====
 Let $\Pi_1$ be a plane with normal vector $\nn_1$, and let $\Pi_2$ be a plane with normal vector $\nn_2$.
-  - $\Pi_1$ and $\Pi_2$ are //orthogonal// or //perpendicular// planes if they meet at right angles. The following conditions are equivalent:
+  * $\Pi_1$ and $\Pi_2$ are //parallel// planes if $\nn_1$ and $\nn_2$ are are in the same direction
-    - $\Pi_1$ and $\Pi_2$ are orthogonal planes;
+    * If $\Pi_1$ has equation $ax+by+cz=d_1$ then any parallel plane $\Pi_2$  has an equation which may be written with the same left hand side: $ax+by+cz=d_2$.
-    - $\nn_1\cdot\nn_2=0$;
+    * i.e., we can assume that $\nn_1=\nn_2=\c abc$.
-    - $\nn_1$ 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. In other words, if $\Pi_1$ has equation $ax+by+cz=d_1$ then any parallel plane $\Pi_2$ has an equation with the same left hand side: $ax+by+cz=d_2$.
-=== Examples ===
+==== Example ====
-. Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
+The plane parallel to $2x-4y+5z=8$ passing through $(1,2,3)$ is
+  * $2x-4y+5z=2(1)-4(2)+5(3) = 10$
+  * i.e., $2x-4y+5z=10$.
-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
-\[ \nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}.\]
-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 plane parallel to $2x-4y+5z=8$ passing through $(1,2,3)$ is $2x-4y+5z=2(1)-4(2)+5(3) = 10$, or $2x-4y+5z=10$.
+==== Orthogonal planes ====
-. Find the equation of the plane parallel to the vectors $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$.
+Let~$\Pi_1$ be a plane with normal vector $\nn_1$ and let $\Pi_2$ be a plane with normal vector $\nn_2$.
-Solution: a normal vector is $\nn=\c111\times\c1{-1}1=\cp1111{-1}1=\c{2}0{-2}$, so the equation is $2x+0y-2z=2(3)-2(1)=4$, or $2x-2z=4$, or $x-z=2$.
+$\Pi_1$ and $\Pi_2$ are //orthogonal// or //perpendicular// planes if they meet at right angles. The following conditions are equivalent:
+  - $\Pi_1$ and $\Pi_2$ are orthogonal planes;
+  - $\nn_1\cdot\nn_2=0$;
+  - $\nn_1$ is a vector in $\Pi_2$;
+  - $\nn_2$ is a vector in $\Pi_1$.
-. Find the equation of the plane $\Pi$ which contains the line of intersection of the planes
+==== Example 1 ====
-\[ \Pi_1: x-y+2z=1\quad\text{and}\quad \Pi_2: 3x+2y-z=4,\]
-and is perpendicular to the plane $\Pi_3:2x+y+z=3$.
-Solution: To find the line of intersection of $\Pi_1$ and $\Pi_2$, we must solve the system of linear equations \begin{gather}x-y+2z=1\\3x+2y-z=4.\end{gather}
+Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
-We can solve this linear system in the usual way, by applying EROs to the matrix $\begin{bmatrix}1&-1&2&1\\3&2&-1&4\end{bmatrix}$:
-\begin{align*}
-\def\go#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}}
-\def\ar#1{\\[6pt]\xrightarrow{#1}&}
-&\go{1&-1&2&1}{3&2&-1&4}
-\ar{R2\to R2-3R1}
-\go{1&-1&2&1}{0&5&-7&1}
-\ar{R1\to 5R1+R2}
-\go{5&0&3&6}{0&5&-7&1}
-\ar{R1\to\tfrac15R1,\,R2\to\tfrac15R2}
-\go{1&0&3/5&6/5}{0&1&-7/5&1/5}
-\end{align*}
-So the line $L$ of intersection is given by
-\[ L: \c xyz=\c{\tfrac65}{\tfrac15}0+t\c{-\tfrac35}{\tfrac75}1,\quad t\in\mathbb{R}.\]
-So $\c{-\tfrac35}{\tfrac75}1$ is a direction vector along $L$, and also $5\c{-\tfrac35}{\tfrac75}1=\c{-3}75$ is a vector along $L$. So $\c{-3}75$ is a vector in the plane $\Pi$. Moreover, taking $t=2$ gives the point $(0,3,2)$ in the line $L$, so this is a point in $\Pi$.
-Since $\Pi$ is perpendicular to $\Pi_3$, which has normal vector $\nn_3=\c211$, the vector $\c211$ is in $\Pi$.
+  * $x-y+z=5$ has normal $\c1{-1}1$; this is in $\Pi$.
+  * $\vec{AB}=\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\c3{-5}4$ is also a vector in $\Pi$
+  * Normal for $\Pi$:  $\def\nn{\vec n}\nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}$.
+  * Sub in $A$ (or $B$): get $x-y-2z=4$.
+==== Example 2 ====
+Find the equation of the plane $\Pi$ which contains the line of intersection of the planes
+\[ \Pi_1: x-y+2z=1\quad\text{and}\quad \Pi_2: 3x+2y-z=4,\]
+and is perpendicular to the plane $\Pi_3:2x+y+z=3$.
-So a normal vector for $\Pi$ is
+  * First find the line of intersection of $\Pi_1$ and $\Pi_2$
-\[ \nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}\]
+  * Solve $x-y+2z=1$, $3x+2y-z=4$
-hence $\Pi$ has equation $-2x-13y+17z=d$, and subbing in the point $(0,3,2)$ gives $d=0-13(3)+17(2)=-39+34=-5$, so $\Pi$ has equation $-2x-13y+17z=-5$, or
+  * $\left[\begin{smallmatrix}1&-1&2&1\\3&2&-1&4\end{smallmatrix}\right]\to_{EROs}\left[\begin{smallmatrix}1&0&3/5&6/5\\0&1&-7/5&1/5\end{smallmatrix}\right]$
-\[ 2x+13y-17z=5.\]
+  * Line $L$ of intersection is $\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$.
+==== ====
+  * $\Pi$ contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$, orthogonal to $\Pi_3: 2x+y+z=3$
+  * $5\c{-3/5}{7/5}1=\c{-3}75$ is a vector along $L$, so in $\Pi$
+  * $\Pi_3$ has  normal vector $\nn_3=\c211$, which is in $\Pi$.
+  * Normal vector for $\Pi$: $\nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}$
+  * $\Pi$ has equation $-2x-13y+17z=d$
+==== ====
+  * $\Pi$ has equation $-2x-13y+17z=d$ and contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$
+  * Take $t=2$: $(0,3,2)$ in $L$, so in $\Pi$
+  * Sub in: $d=0-13(3)+17(2)=-39+34=-5$
+  * Answer: $-2x-13y+17z=-5$, or $2x+13y-17z=5$.