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--- lecture_20 [2016/04/12 09:44] – rupert
+++ lecture_20 [2016/04/14 09:50] (current) – rupert
@@ Line 27: / Line 27: @@
 \[ V=Ah=\|\vv\times\ww\|\frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
 or
-\[ V=|\uu\cdot(\vv\times\ww)|,\]
+\[ V=|\uu\cdot(\vv\times\ww)|.\]
-so $V$ is the absolute value of the determinant $\begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{vmatrix}$:
+Now $\uu\cdot(\vv\times \ww)=\det\left(\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)$, so $V$ is the absolute value of this determinant:
-\[ V=\left|\quad \begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{vmatrix}\quad \right|.\]
+\[ V=\left|\quad\det\left( \begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)\quad \right|.\]
 === Example ===
@@ Line 56: / Line 56: @@
 We call a vector with this property a **normal** vector to the plane.
-==== Example ====
+==== Examples ====
-Find a unit normal vector to the plane $x+y-3z=4$.
+. Find a unit normal vector to the plane $x+y-3z=4$.
-=== Solution ===
+Solution: The vector $\nn=\c11{-3}$ is a normal vector to this plane, so $\vv=\frac1{\|\nn\|}\nn=\frac1{\sqrt{11}}\c11{-3}$ is a unit normal vector to this plane. Indeed, $\vv$ is a unit vector and it's in the same direction as the normal vector $\nn$, so $\vv$ is also a normal vector.
-The vector $\nn=\c11{-3}$ is a normal vector to this plane, so $\vv=\frac1{\|\nn\|}\nn=\frac1{\sqrt{11}}\c11{-3}$ is a unit normal vector to this plane. Indeed, $\vv$ is a unit vector and it's in the same direction as the normal vector $\nn$, so $\vv$ is also a normal vector.
- ===== Planes in $\mathbb{R}^3$ =====
+. 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.
-==== Examples ====
-. 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.
 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
@@ Line 74: / Line 69: @@
 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 $C=(4,3,-2)$.
-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 $\vec{AC}$:
-\[ \nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\cp2{-2}131{-2}=\c378\]
-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 ====
-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 //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 ===
-. Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
-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 $2x-4y+5z=10$.
 . Find the equation of the plane parallel to the vectors $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$.
@@ Line 106: / Line 74: @@
 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 $x-z=2$.
-. 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}
-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 $\Pi$.
-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
-\[ 2x+13y-17z=5.\]