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% https://git.betalupi.com/Mark/latex-packages
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
% Last built with version 1.0.6
\documentclass[solutions]{ormc_handout}
\begin{document}
\maketitle
<Intermediate 2>
<ORMC Summer Sessions>
{Vectors 2}
{
Prepared by Mark on \today \\
Based on a handout by Oleg Gleizer
}
\section{Review}
\definition{}
A \textit{vector} is a directed line segment. In the vector below, $A$ is its initial point and $B$ is its terminal point.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-1);
\coordinate [label = left: {$A$}] (a) at (0,0);
\coordinate [label = right: {$B$}] (b) at (4,-1);
\coordinate [label = above: {$v$}] (v) at (2,-1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\definition{Equivalence}<vec_eq>
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.
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4, -1);
\coordinate [label = left: {$A$}] (a) at (0, 0);
\coordinate [label = right: {$B$}] (b) at (4, -1);
\coordinate [label = above: {$v$}] (v) at (2, -1);
\draw [->, line width = 2pt] (-1,-1) -- (3,-2);
\coordinate [label = left: {$C$}] (c) at (-1,-1);
\coordinate [label = right: {$D$}] (d) at (3,-2);
\coordinate [label = below: {$w$}] (w) at (1,-1.6);
\draw (-1,-1) -- (0,0);
\draw (3,-2) -- (4,-1);
\end{normalsize}
\end{tikzpicture}
\end{center}
\definition{Addition}
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 $v + w$. \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4,-1);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [->, line width = 2pt] (4,-1) -- (2,-2);
\coordinate [label = right: {$w$}] (w) at (3, -1.7);
\draw [->, line width = 2pt] (0,0) -- (2,-2);
\coordinate [label = left: {$v + w$}] (p) at (1,-1.2);
\end{normalsize}
\end{tikzpicture}
\end{center}
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:
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 1pt] (0,0) -- (4,-1);
\coordinate [label = above: {$v$}] (v) at (2,-.9);
\draw [->, line width = 1pt] (4,-1) -- (2,-2);
\coordinate [label = right: {$w$}] (w) at (3, -1.7);
\draw [<-, line width = 1pt] (2,-2) -- (-2,-1);
\coordinate [label = above: {$v$}] (v) at (0.2,-2);
\draw [<-, line width = 1pt] (-2,-1) -- (0, 0);
\coordinate [label = right: {$w$}] (w) at (-1.5, -0.3);
\draw [->, line width = 2pt] (0,0) -- (2,-2);
\coordinate [label = left: {$v + w$}] (p) at (1,-1.2);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\definition{The Zero Vector}
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,
\begin{equation*}
v + \overrightarrow{0} =
\overrightarrow{0} + v =
v
\end{equation*}
for any vector $v$.
\definition{Inverse Vectors}
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: \\
\begin{center}
\begin{tikzpicture}
\begin{normalsize}
\draw [->, line width = 2pt] (0,0) -- (4, -1);
\coordinate [label = above: {$v$}] (v) at (2, -1);
\draw [<-, line width = 2pt] (-1,-1) -- (3,-2);
\coordinate [label = below: {$-v$}] (w) at (1,-1.6);
\end{normalsize}
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\section{Vectors}
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below.
Use the grid instead of a compass and ruler. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-8,-8) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1);
\filldraw (-6,-4) circle (2pt);
\coordinate [label = left:{$A$}] (a) at (-6,-4);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = -\overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-7) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,-1) -- (-1,-3);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-2.1);
\filldraw (-6,-4) circle (2pt);
\coordinate [label = below:{$A$}] (a) at (-6,-4.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = 1.5 \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-9,-9) grid (1,1);
\draw [line width = 1.5pt, ->] (-6,-1) -- (-2,-3);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.1);
\filldraw (-7,-4) circle (2pt);
\coordinate [label = left:{$A$}] (a) at (-7,-4.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = -2 \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-16,-8) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,-2) -- (-1,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.8,-3.1);
\filldraw (-6,-5) circle (2pt);
\coordinate [label = below:{$A$}] (a) at (-6,-5.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vector $\overrightarrow{v}$
and point $A$, construct the vector
$\overrightarrow{w} = -\frac13 \overrightarrow{v}$
having $A$ as its initial point
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-11,-6) grid (1,1);
\draw [line width = 1.5pt, ->] (-7,-1) -- (-1,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-3.8,-2.4);
\filldraw (-6,-4) circle (2pt);
\coordinate [label = below:{$A$}] (a) at (-6,-4.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$\overrightarrow{w} + \overrightarrow{v}$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-10) grid (1,1);
\draw [line width = 1.5pt, ->] (-5,0) -- (0,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.4,-2.1);
\draw [line width = 1.5pt, ->] (-10,-1) -- (-9,-5);
\coordinate [label = left:{$\overrightarrow{w}$}] (w) at (-9.5,-3.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$\overrightarrow{w} - \overrightarrow{v}$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-6) grid (1,1);
\draw [line width = 1.5pt, ->] (-11,0) -- (-6,-4);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-8.4,-2.1);
\draw [line width = 1.5pt, ->] (-1,-1) -- (0,-5);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$2\overrightarrow{v} - 3\overrightarrow{w}$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-13,-10) grid (1,1);
\draw [line width = 1.5pt, ->] (-12,-1) -- (-7,-5);
\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-9.6,-3.1);
\draw [line width = 1.5pt, ->] (-1,-2) -- (0,-5);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-.5,-3.6);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$1.75\overrightarrow{v} - \frac23 \overrightarrow{w}$
on the graph paper below.
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-19,-9) grid (1,1);
\draw [line width = 1.5pt, ->] (-18,-1) -- (-10,-5);
\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-14.4,-3.1);
\draw [line width = 1.5pt, ->] (-3,0) -- (0,-6);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-1.5,-3);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
For the given vectors $\overrightarrow{v}$
and $\overrightarrow{w}$,
construct the vector
$\overrightarrow{v} + \overrightarrow{w}$
originating at the same point
as the vector $\overrightarrow{v}$
and the vector
$\overrightarrow{w} + \overrightarrow{v}$
originating at the same point
as the vector $\overrightarrow{w}$.
Is $\overrightarrow{v} + \overrightarrow{w} =
\overrightarrow{w} + \overrightarrow{v}$?
Why or why not?
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-17,-7) grid (1,1);
\draw [line width = 1.5pt, ->] (-16,-1) -- (-11,-5);
\coordinate [label = left:{$\overrightarrow{v}$}] (v) at (-13.6,-3.1);
\draw [line width = 1.5pt, ->] (-8,-6) -- (-5,-1);
\coordinate [label = right:{$\overrightarrow{w}$}] (w) at (-6.5,-4.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\section{Pythagoras' Theorem}
\problem{}
Formulate and prove the Pythagoras' theorem.
\vfill
\pagebreak
\problem{}
Use the Pythagors' theorem to find $x$
for the following right triangles. \\
\noindent a.~~
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) -- (3,0) -- (0,3) -- (0,0);
\coordinate [label = left:{$1$}] (a) at (0,1.5);
\coordinate [label = below:{$1$}] (b) at (1.5,-.1);
\coordinate [label = right:{$x$}] (x) at (1.6,1.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
\bigskip
\noindent b.~~
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) -- (4,0) -- (0,3) -- (0,0);
\coordinate [label = left:{$3$}] (a) at (0,1.5);
\coordinate [label = below:{$x$}] (x) at (2,-.1);
\coordinate [label = right:{$5$}] (b) at (2.1,1.6);
\end{tikzpicture}
\end{normalsize}
\end{center}
\bigskip
\noindent c.~~
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,0) -- (4.5,0) -- (0,3) -- (0,0);
\coordinate [label = left:{$x$}] (x) at (0,1.5);
\coordinate [label = below:{$\sqrt{2}$}] (a) at (2,-.1);
\coordinate [label = right:{$\sqrt3$}] (b) at (2.2,1.7);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
Construct a segment of length $\sqrt{2} a$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-7,-7) grid (1,1);
\draw [line width = 1.5pt] (-5,-1) -- (-1,-1);
\coordinate [label = above:{$a$}] (a) at (-3,-.9);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
Construct a segment of length $\sqrt{5} a$
on the graph paper below. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-12,-6) grid (1,1);
\draw [line width = 1.5pt] (-5,-1) -- (-1,-1);
\coordinate [label = above:{$a$}] (a) at (-3,-.9);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
The side length of a grid square below
is one unit. Find the length
the vector $\overrightarrow{v}$. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-5,-3) grid (1,1);
\draw [line width = 1.5pt, ->] (-4,0) -- (0,-2);
\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-2.7,-2.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\pagebreak
\problem{}
The side length of a grid square below
is three units. Find the length
the vector $\overrightarrow{w}$. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture} [scale=.7]
\draw[step=1cm, gray, very thin] (-6,-3) grid (1,1);
\draw [line width = 1.5pt, <-] (-5,0) -- (0,-2);
\coordinate [label = above:{$\overrightarrow{w}$}] (w) at (-2.5,-2);
\end{tikzpicture}
\end{normalsize}
\end{center}
\medskip
\section{Rationals}
A number is called {\it rational}
if it can be represented as a ratio $p/q$
of an integer $p$ and a positive integer $q$
such that $p$ and $q$ have no common
factors. Otherwise, that number is called {\it irrational} \\
\problem{}
Decide whether the following numbers
are rational or irrational.
In each case, give a reason. \\
\begin{enumerate}[itemsep=2mm]
\item $\displaystyle{\frac{375}{376}}$
\item $10$
\item $0.5$
\item $-5$
\item $1.2345$
\item $0.111111111...$
\item $\sqrt{2}$
\item $\sqrt[3]{10}$
\end{enumerate}
\vfill
\pagebreak
\problem{}
Find $\left\lfloor \sqrt[3]{10} \right\rfloor$
and $\left\lceil \sqrt[3]{10} \right\rceil$.
\vfill
\problem{}
Simplify $\sqrt{8}$.
\vfill
\pagebreak
\problem{}
A man is crossing a river in a boat.
The speed of the boat is three meters per second.
The speed of the water in the river
is one meter per second.
In what direction should the man steer the boat,
if he wants the vessel to move
perpendicular to the banks?
Construct the velocity vector. \\
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\draw (0,10) -- (10,10);
\draw (0,0) -- (10,0);
\draw (4.5,1) -- (5.5,1) --(5.5,3) -- (5,3.5) -- (4.5,3) --(4.5,1) ;
\filldraw (5,2.1) circle (2pt);
\draw [line width = 2pt, ->] (5,2.1) -- (7,2.1);
\coordinate [label = below: {$1\frac{m}{s}$}] (w) at (6.3,1.9);
\end{tikzpicture}
\end{normalsize}
\end{center}
\bigskip
The width of the river
is $10\sqrt{2}$ meters.
How long would it take the man
to cross the river?
\vfill
\pagebreak
\problem{}
You need to slide a heavy box over
the floor from point $A$ to point $B$.
The box is about twice as tall as you are.
Which way is easier, to push or to pull?
Why? \vspace{60pt}
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\filldraw [gray!90!blue] (1,0) -- (1,2) --
(4,2) -- (4,0) -- (1,0);
\draw (0,0) -- (10,0);
\filldraw (2.5,0) circle (1.5pt);
\coordinate [label = below: {$A$}] (a) at (2.5,-.1);
\filldraw (7.5,0) circle (1.5pt);
\coordinate [label = below: {$B$}] (b) at (7.5,-.1);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\problem{}
The dot on the picture below
represents a spaceship.
There are three forces acting on the ship.
$\overrightarrow{T}$ is the thrust
of the ship's engine.
$\overrightarrow{P}$ is the gravitational pull
of the neighbouring planet.
$\overrightarrow{S}$ is the gravitational pull
of the planet's home star.
You are the captain.
Use a compass and a ruler to figure out
where the resulting force would steer the ship.
\vspace{100pt}
\begin{center}
\begin{normalsize}
\begin{tikzpicture}
\filldraw (0,0) circle (2pt);
\draw [line width = 1pt, ->] (0,0) -- (225:6);
\coordinate [label = below: {$\overrightarrow{T}$}]
(t) at (230:3);
\draw [line width = 1pt, ->] (0,0) -- (4,0);
\coordinate [label = above: {$\overrightarrow{P}$}]
(p) at (2,.1);
\draw [line width = 1pt, ->] (0,0) -- (150:3);
\coordinate [label = right: {$\overrightarrow{S}$}]
(s) at (140:1.5);
\end{tikzpicture}
\end{normalsize}
\end{center}
\vfill
\end{document}