599 lines
16 KiB
TeX
Executable File
599 lines
16 KiB
TeX
Executable File
% use [nosolutions] flag to hide solutions.
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% use [solutions] flag to show solutions.
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\documentclass[solutions]{../../resources/ormc_handout}
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\usepackage{../../resources/macros}
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\uptitlel{Intermediate 2}
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\uptitler{\smallurl{}}
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\title{Vectors 2}
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\subtitle{
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Prepared by Mark on \today \\
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Based on a handout by Oleg Gleizer
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}
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\begin{document}
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\maketitle
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\section{Review}
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\definition{}
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A \textit{vector} is a directed line segment. In the vector below, $A$ is its initial point and $B$ is its terminal point.
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\begin{center}
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\begin{tikzpicture}
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\begin{normalsize}
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\draw [->, line width = 2pt] (0,0) -- (4,-1);
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\coordinate [label = left: {$A$}] (a) at (0,0);
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\coordinate [label = right: {$B$}] (b) at (4,-1);
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\coordinate [label = above: {$v$}] (v) at (2,-1);
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\end{normalsize}
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\end{tikzpicture}
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\end{center}
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\definition{Equivalence}<vec_eq>
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We say two vectors are equal if the quadrilateral they form is a parallelogram. In other words, two vectors are equal if they have the same length and direction.
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\begin{center}
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\begin{tikzpicture}
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\begin{normalsize}
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\draw [->, line width = 2pt] (0,0) -- (4, -1);
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\coordinate [label = left: {$A$}] (a) at (0, 0);
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\coordinate [label = right: {$B$}] (b) at (4, -1);
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\coordinate [label = above: {$v$}] (v) at (2, -1);
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\draw [->, line width = 2pt] (-1,-1) -- (3,-2);
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\coordinate [label = left: {$C$}] (c) at (-1,-1);
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\coordinate [label = right: {$D$}] (d) at (3,-2);
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\coordinate [label = below: {$w$}] (w) at (1,-1.6);
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\draw (-1,-1) -- (0,0);
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\draw (3,-2) -- (4,-1);
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\end{normalsize}
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\end{tikzpicture}
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\end{center}
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\definition{Addition}
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To add two vectors $v$ and $w$, we move $w$ so that the initial point of $w$ coincides with the terminal point of $v$. The vector originating at the initial point of $v$ and terminating at the terminal point of $w$ is the sum $v + w$. \\
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\begin{center}
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\begin{tikzpicture}
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\begin{normalsize}
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\draw [->, line width = 2pt] (0,0) -- (4,-1);
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\coordinate [label = above: {$v$}] (v) at (2,-.9);
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\draw [->, line width = 2pt] (4,-1) -- (2,-2);
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\coordinate [label = right: {$w$}] (w) at (3, -1.7);
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\draw [->, line width = 2pt] (0,0) -- (2,-2);
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\coordinate [label = left: {$v + w$}] (p) at (1,-1.2);
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\end{normalsize}
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\end{tikzpicture}
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\end{center}
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Note that $v+w = w+v$. If we create a parallelogram with sides $w$ and $v$ (\ref{vec_eq}), the sums create the same diagonal:
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\begin{center}
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\begin{tikzpicture}
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\begin{normalsize}
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\draw [->, line width = 1pt] (0,0) -- (4,-1);
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\coordinate [label = above: {$v$}] (v) at (2,-.9);
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\draw [->, line width = 1pt] (4,-1) -- (2,-2);
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\coordinate [label = right: {$w$}] (w) at (3, -1.7);
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\draw [<-, line width = 1pt] (2,-2) -- (-2,-1);
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\coordinate [label = above: {$v$}] (v) at (0.2,-2);
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\draw [<-, line width = 1pt] (-2,-1) -- (0, 0);
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\coordinate [label = right: {$w$}] (w) at (-1.5, -0.3);
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\draw [->, line width = 2pt] (0,0) -- (2,-2);
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\coordinate [label = left: {$v + w$}] (p) at (1,-1.2);
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\end{normalsize}
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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\definition{The Zero Vector}
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A vector that has coinciding initial and terminal points is called the {\it zero vector} and is denoted as $\overrightarrow{0}$. According to the above definition of the vector addition,
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\begin{equation*}
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v + \overrightarrow{0} =
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\overrightarrow{0} + v =
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v
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\end{equation*}
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for any vector $v$.
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\definition{Inverse Vectors}
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A vector $w$ such that $w + v = \overrightarrow{0}$ is called the \textit{inverse} of $v$ and is denoted as $-v$. The vector $-v$ lies either on the same straight line as $v$ or on a parallel one, has the same length as $v$, but points in the opposite direction: \\
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\begin{center}
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\begin{tikzpicture}
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\begin{normalsize}
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\draw [->, line width = 2pt] (0,0) -- (4, -1);
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\coordinate [label = above: {$v$}] (v) at (2, -1);
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\draw [<-, line width = 2pt] (-1,-1) -- (3,-2);
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\coordinate [label = below: {$-v$}] (w) at (1,-1.6);
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\end{normalsize}
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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\section{Vectors}
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\problem{}
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For the given vector $\overrightarrow{v}$
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and point $A$, construct the vector
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$\overrightarrow{w} = \overrightarrow{v}$
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having $A$ as its initial point
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on the graph paper below.
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Use the grid instead of a compass and ruler. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-8,-8) grid (1,1);
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\draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1);
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\filldraw (-6,-4) circle (2pt);
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\coordinate [label = left:{$A$}] (a) at (-6,-4);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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For the given vector $\overrightarrow{v}$
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and point $A$, construct the vector
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$\overrightarrow{w} = -\overrightarrow{v}$
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having $A$ as its initial point
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-12,-7) grid (1,1);
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\draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1);
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\filldraw (-6,-4) circle (2pt);
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\coordinate [label = below:{$A$}] (a) at (-6,-4.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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For the given vector $\overrightarrow{v}$
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and point $A$, construct the vector
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$\overrightarrow{w} = 1.5 \overrightarrow{v}$
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having $A$ as its initial point
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-9,-9) grid (1,1);
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\draw [line width = 1.5pt, ->] (-6,-1) -- (-2,-3);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.1);
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\filldraw (-7,-4) circle (2pt);
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\coordinate [label = left:{$A$}] (a) at (-7,-4.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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For the given vector $\overrightarrow{v}$
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and point $A$, construct the vector
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$\overrightarrow{w} = -2 \overrightarrow{v}$
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having $A$ as its initial point
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-16,-8) grid (1,1);
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\draw [line width = 1.5pt, ->] (-5,-2) -- (-1,-4);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-3.1);
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\filldraw (-6,-5) circle (2pt);
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\coordinate [label = below:{$A$}] (a) at (-6,-5.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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For the given vector $\overrightarrow{v}$
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and point $A$, construct the vector
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$\overrightarrow{w} = -\frac13 \overrightarrow{v}$
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having $A$ as its initial point
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-11,-6) grid (1,1);
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\draw [line width = 1.5pt, ->] (-7,-1) -- (-1,-4);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.4);
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\filldraw (-6,-4) circle (2pt);
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\coordinate [label = below:{$A$}] (a) at (-6,-4.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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For the given vectors $\overrightarrow{v}$
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and $\overrightarrow{w}$,
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construct the vector
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$\overrightarrow{w} + \overrightarrow{v}$
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-12,-10) grid (1,1);
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\draw [line width = 1.5pt, ->] (-5,0) -- (0,-4);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.4,-2.1);
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\draw [line width = 1.5pt, ->] (-10,-1) -- (-9,-5);
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\coordinate [label = left:{$\overrightarrow{w}$}] (w) at (-9.5,-3.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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For the given vectors $\overrightarrow{v}$
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and $\overrightarrow{w}$,
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construct the vector
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$\overrightarrow{w} - \overrightarrow{v}$
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-12,-6) grid (1,1);
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\draw [line width = 1.5pt, ->] (-11,0) -- (-6,-4);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-8.4,-2.1);
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\draw [line width = 1.5pt, ->] (-1,-1) -- (0,-5);
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\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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For the given vectors $\overrightarrow{v}$
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and $\overrightarrow{w}$,
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construct the vector
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$2\overrightarrow{v} - 3\overrightarrow{w}$
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-13,-10) grid (1,1);
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\draw [line width = 1.5pt, ->] (-12,-1) -- (-7,-5);
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\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-9.6,-3.1);
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\draw [line width = 1.5pt, ->] (-1,-2) -- (0,-5);
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\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.6);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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For the given vectors $\overrightarrow{v}$
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and $\overrightarrow{w}$,
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construct the vector
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$1.75\overrightarrow{v} - \frac23 \overrightarrow{w}$
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on the graph paper below.
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-19,-9) grid (1,1);
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\draw [line width = 1.5pt, ->] (-18,-1) -- (-10,-5);
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\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-14.4,-3.1);
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\draw [line width = 1.5pt, ->] (-3,0) -- (0,-6);
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\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-1.5,-3);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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For the given vectors $\overrightarrow{v}$
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and $\overrightarrow{w}$,
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construct the vector
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$\overrightarrow{v} + \overrightarrow{w}$
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originating at the same point
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as the vector $\overrightarrow{v}$
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and the vector
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$\overrightarrow{w} + \overrightarrow{v}$
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originating at the same point
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as the vector $\overrightarrow{w}$.
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Is $\overrightarrow{v} + \overrightarrow{w} =
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\overrightarrow{w} + \overrightarrow{v}$?
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Why or why not?
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-17,-7) grid (1,1);
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\draw [line width = 1.5pt, ->] (-16,-1) -- (-11,-5);
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\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-13.6,-3.1);
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\draw [line width = 1.5pt, ->] (-8,-6) -- (-5,-1);
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\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-6.5,-4.5);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\pagebreak
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\section{Pythagoras' Theorem}
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\problem{}
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Formulate and prove the Pythagoras' theorem.
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\vfill
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\pagebreak
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\problem{}
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Use the Pythagorean theorem to find $x$
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for the following right triangles. \\
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\noindent a.~~
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture}
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\draw (0,0) -- (3,0) -- (0,3) -- (0,0);
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\coordinate [label = left:{$1$}] (a) at (0,1.5);
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\coordinate [label = below:{$1$}] (b) at (1.5,-.1);
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\coordinate [label = right:{$x$}] (x) at (1.6,1.5);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\bigskip
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\noindent b.~~
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture}
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\draw (0,0) -- (4,0) -- (0,3) -- (0,0);
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\coordinate [label = left:{$3$}] (a) at (0,1.5);
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\coordinate [label = below:{$x$}] (x) at (2,-.1);
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\coordinate [label = right:{$5$}] (b) at (2.1,1.6);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\bigskip
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\noindent c.~~
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture}
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\draw (0,0) -- (4.5,0) -- (0,3) -- (0,0);
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\coordinate [label = left:{$x$}] (x) at (0,1.5);
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\coordinate [label = below:{$\sqrt{2}$}] (a) at (2,-.1);
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\coordinate [label = right:{$\sqrt3$}] (b) at (2.2,1.7);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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Construct a segment of length $\sqrt{2} a$
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-7,-7) grid (1,1);
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\draw [line width = 1.5pt] (-5,-1) -- (-1,-1);
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\coordinate [label = above:{$a$}] (a) at (-3,-.9);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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Construct a segment of length $\sqrt{5} a$
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on the graph paper below. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-12,-6) grid (1,1);
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\draw [line width = 1.5pt] (-5,-1) -- (-1,-1);
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\coordinate [label = above:{$a$}] (a) at (-3,-.9);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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The side length of a grid square below
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is one unit. Find the length
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the vector $\overrightarrow{v}$. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-5,-3) grid (1,1);
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\draw [line width = 1.5pt, ->] (-4,0) -- (0,-2);
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\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.7,-2.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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The side length of a grid square below
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is three units. Find the length
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the vector $\overrightarrow{w}$. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture} [scale=.7]
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\draw[step=1cm, gray, very thin] (-6,-3) grid (1,1);
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\draw [line width = 1.5pt, <-] (-5,0) -- (0,-2);
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\coordinate [label = above:{$\overrightarrow{w}$}] (w) at (-2.5,-2);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\medskip
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\section{Rationals}
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A number is called {\it rational}
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if it can be represented as a ratio $p/q$
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of an integer $p$ and a positive integer $q$
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such that $p$ and $q$ have no common
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factors. Otherwise, that number is called {\it irrational} \\
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\problem{}
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Decide whether the following numbers
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are rational or irrational.
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In each case, give a reason. \\
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\begin{enumerate}[itemsep=2mm]
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\item $\displaystyle{\frac{375}{376}}$
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\item $10$
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\item $0.5$
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\item $-5$
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\item $1.2345$
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\item $0.111111111...$
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\item $\sqrt{2}$
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\item $\sqrt[3]{10}$
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\end{enumerate}
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\vfill
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\pagebreak
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\problem{}
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Find $\left\lfloor \sqrt[3]{10} \right\rfloor$
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and $\left\lceil \sqrt[3]{10} \right\rceil$.
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\vfill
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\problem{}
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Simplify $\sqrt{8}$.
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\vfill
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\pagebreak
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\problem{}
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A man is crossing a river in a boat.
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The speed of the boat is three meters per second.
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The speed of the water in the river
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is one meter per second.
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In what direction should the man steer the boat,
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if he wants the vessel to move
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perpendicular to the banks?
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Construct the velocity vector. \\
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture}
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\draw (0,10) -- (10,10);
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\draw (0,0) -- (10,0);
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\draw (4.5,1) -- (5.5,1) --(5.5,3) -- (5,3.5) -- (4.5,3) --(4.5,1) ;
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\filldraw (5,2.1) circle (2pt);
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\draw [line width = 2pt, ->] (5,2.1) -- (7,2.1);
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\coordinate [label = below: {$1\frac{m}{s}$}] (w) at (6.3,1.9);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\bigskip
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The width of the river
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is $10\sqrt{2}$ meters.
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How long would it take the man
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to cross the river?
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\vfill
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\pagebreak
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\problem{}
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You need to slide a heavy box over
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the floor from point $A$ to point $B$.
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The box is about twice as tall as you are.
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Which way is easier, to push or to pull?
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Why? \vspace{60pt}
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture}
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\filldraw [gray!90!blue] (1,0) -- (1,2) --
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(4,2) -- (4,0) -- (1,0);
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\draw (0,0) -- (10,0);
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\filldraw (2.5,0) circle (1.5pt);
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\coordinate [label = below: {$A$}] (a) at (2.5,-.1);
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\filldraw (7.5,0) circle (1.5pt);
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\coordinate [label = below: {$B$}] (b) at (7.5,-.1);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\problem{}
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The dot on the picture below
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represents a spaceship.
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There are three forces acting on the ship.
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$\overrightarrow{T}$ is the thrust
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of the ship's engine.
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$\overrightarrow{P}$ is the gravitational pull
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of the neighboring planet.
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$\overrightarrow{S}$ is the gravitational pull
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of the planet's home star.
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You are the captain.
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Use a compass and a ruler to figure out
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where the resulting force would steer the ship.
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\vspace{100pt}
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\begin{center}
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\begin{normalsize}
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\begin{tikzpicture}
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\filldraw (0,0) circle (2pt);
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\draw [line width = 1pt, ->] (0,0) -- (225:6);
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\coordinate [label = below: {$\overrightarrow{T}$}]
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(t) at (230:3);
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\draw [line width = 1pt, ->] (0,0) -- (4,0);
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\coordinate [label = above: {$\overrightarrow{P}$}]
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(p) at (2,.1);
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\draw [line width = 1pt, ->] (0,0) -- (150:3);
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\coordinate [label = right: {$\overrightarrow{S}$}]
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(s) at (140:1.5);
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\end{tikzpicture}
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\end{normalsize}
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\end{center}
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\vfill
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\end{document} |