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--- lecture_20_slides [2016/04/13 15:35] – [Example 1] rupert
+++ lecture_20_slides [2016/04/13 15:43] (current) – rupert
@@ Line 159: / Line 159: @@
   * $x-y+z=5$ 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 $\Pi$
-  * So $\Pi$  has normal vector $\def\nn{\vec n}\nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}$.
+  * Normal for $\Pi$:  $\def\nn{\vec n}\nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}$.
-  * Equation of $\Pi$ is $x-y-2z=d$, and $A=(1,3,-3)$ in $\Pi$, so $d=1-3-2(-3)=4$
+  * Sub in $A$ (or $B$): get $x-y-2z=4$.
-  * Answer: $x-y-2z=4$.
 ==== Example 2 ====
@@ Line 170: / Line 169: @@
   * First find the line of intersection of $\Pi_1$ and $\Pi_2$
-  * Solve \begin{gather}x-y+2z=1\\3x+2y-z=4.\end{gather}
+  * Solve $x-y+2z=1$, $3x+2y-z=4$
   * $\left[\begin{smallmatrix}1&-1&2&1\\3&2&-1&4\end{smallmatrix}\right]\to_{EROs}\left[\begin{smallmatrix}1&0&3/5&6/5\\0&1&-7/5&1/5\end{smallmatrix}\right]$
-  * Line $L$ of intersection is $\c xyz=\c{\tfrac65}{\tfrac15}0+t\c{-\tfrac35}{\tfrac75}1$, $t\in\mathbb{R}$.
+  * Line $L$ of intersection is $\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$.
-  * $\c{-\tfrac35}{\tfrac75}1$ is a direction vector along $L$
-  * Another one is $5\c{-\tfrac35}{\tfrac75}1=\c{-3}75$
-  * So $\c{-3}75$ is a vector in the plane $\Pi$.
-  * Take $t=2$ gives $(0,3,2)$ in $L$, so this is in $\Pi$.
 ==== ====
+  * $\Pi$ contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$, orthogonal to $\Pi_3: 2x+y+z=3$
-Since $\Pi$ is perpendicular to $\Pi_3$, which has normal vector $\nn_3=\c211$, the vector $\c211$ is in $\Pi$.
+  * $5\c{-3/5}{7/5}1=\c{-3}75$ is a vector along $L$, so in $\Pi$
+  * $\Pi_3$ has  normal vector $\nn_3=\c211$, which is in $\Pi$.
-So a normal vector for $\Pi$ is
+  * Normal vector for $\Pi$: $\nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}$
-\[ \nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}\]
+  * $\Pi$ has equation $-2x-13y+17z=d$
-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
+==== ====
-\[ 2x+13y-17z=5.\]
+  * $\Pi$ 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 $2x+13y-17z=5$.