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--- lecture_20_slides [2016/04/13 15:28] – [Example 1] rupert
+++ lecture_20_slides [2016/04/13 15:43] (current) – rupert
@@ Line 145: / Line 145: @@
 ==== Orthogonal 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 orthogonal planes;
+  - $\nn_1\cdot\nn_2=0$;
+  - $\nn_1$ is a vector in $\Pi_2$;
+  - $\nn_2$ is a vector in $\Pi_1$.
-  - $\Pi_1$ and $\Pi_2$ are //orthogonal// or //perpendicular// planes if they meet at right angles. The following conditions are equivalent:
+==== Example 1 ====
-    - $\Pi_1$ 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 $\Pi_1$.
-  - $\Pi_1$ and $\Pi_2$ are
-=== Examples ===
+Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
-. Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
+  * $x-y+z=5$ 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 $x-y-2z=4$.
-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
+==== Example 2 ====
-\[ \nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}.\]
-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 \[ x-y-2z=4.\]
-. Find the equation of the plane $\Pi$ which contains the line of intersection of the planes
+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 $\Pi_3:2x+y+z=3$.
-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}
+  * First find the line of intersection of $\Pi_1$ and $\Pi_2$
-We can solve this linear system in the usual way, by applying EROs to the matrix $\begin{bmatrix}1&-1&2&1\\3&2&-1&4\end{bmatrix}$:
+  * Solve $x-y+2z=1$, $3x+2y-z=4$
-\begin{align*}
+  * $\left[\begin{smallmatrix}1&-1&2&1\\3&2&-1&4\end{smallmatrix}\right]\to_{EROs}\left[\begin{smallmatrix}1&0&3/5&6/5\\0&1&-7/5&1/5\end{smallmatrix}\right]$
-\def\go#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}}
+  * Line $L$ of intersection is $\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$.
-\def\ar#1{\\[6pt]\xrightarrow{#1}&}
+==== ====
-&\go{1&-1&2&1}{3&2&-1&4}
+  * $\Pi$ contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$, orthogonal to $\Pi_3: 2x+y+z=3$
-\ar{R2\to R2-3R1}
+  * $5\c{-3/5}{7/5}1=\c{-3}75$ is a vector along $L$, so in $\Pi$
-\go{1&-1&2&1}{0&5&-7&1}
+  * $\Pi_3$ has  normal vector $\nn_3=\c211$, which is in $\Pi$.
-\ar{R1\to 5R1+R2}
+  * Normal vector for $\Pi$: $\nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}$
-\go{5&0&3&6}{0&5&-7&1}
+  * $\Pi$ has equation $-2x-13y+17z=d$
-\ar{R1\to\tfrac15R1,\,R2\to\tfrac15R2}
+==== ====
-\go{1&0&3/5&6/5}{0&1&-7/5&1/5}
+  * $\Pi$ has equation $-2x-13y+17z=d$ and contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$
-\end{align*}
+  * Take $t=2$: $(0,3,2)$ in $L$, so in $\Pi$
-So the line $L$ of intersection is given by
+  * Sub in: $d=0-13(3)+17(2)=-39+34=-5$
-\[ L: \c xyz=\c{\tfrac65}{\tfrac15}0+t\c{-\tfrac35}{\tfrac75}1,\quad t\in\mathbb{R}.\]
+  * Answer: $-2x-13y+17z=-5$, or $2x+13y-17z=5$.
-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 $\Pi$.
-So a normal vector for $\Pi$ is
-\[ \nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}\]
-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
-\[ 2x+13y-17z=5.\]