Commit 458a30d0 authored by Jim Hefferon's avatar Jim Hefferon

done edits on gr2

parent 14455503
This diff is collapsed.
......@@ -1773,10 +1773,10 @@
\subsection{Subsection One.II.1: Vectors in Space}
\begin{ans}{One.II.1.1}
\begin{exparts*}
\partsitem \( \colvec{2 \\ 1} \)
\partsitem \( \colvec{-1 \\ 2} \)
\partsitem \( \colvec{4 \\ 0 \\ -3} \)
\partsitem \( \colvec{0 \\ 0 \\ 0} \)
\partsitem \( \colvec[r]{2 \\ 1} \)
\partsitem \( \colvec[r]{-1 \\ 2} \)
\partsitem \( \colvec[r]{4 \\ 0 \\ -3} \)
\partsitem \( \colvec[r]{0 \\ 0 \\ 0} \)
\end{exparts*}
\end{ans}
......@@ -1784,13 +1784,13 @@
\begin{exparts}
\partsitem No, their canonical positions are different.
\begin{equation*}
\colvec{1 \\ -1}
\colvec[r]{1 \\ -1}
\qquad
\colvec{0 \\ 3}
\colvec[r]{0 \\ 3}
\end{equation*}
\partsitem Yes, their canonical positions are the same.
\begin{equation*}
\colvec{1 \\ -1 \\ 3}
\colvec[r]{1 \\ -1 \\ 3}
\end{equation*}
\end{exparts}
......@@ -1798,8 +1798,8 @@
\begin{ans}{One.II.1.3}
That line is this set.
\begin{equation*}
\set{\colvec{-2 \\ 1 \\ 1 \\ 0}
+\colvec{7 \\ 9 \\ -2 \\ 4}t \suchthat t\in\Re }
\set{\colvec[r]{-2 \\ 1 \\ 1 \\ 0}
+\colvec[r]{7 \\ 9 \\ -2 \\ 4}t \suchthat t\in\Re }
\end{equation*}
Note that this system
\begin{equation*}
......@@ -1818,19 +1818,19 @@
\begin{exparts}
\partsitem Note that
\begin{equation*}
\colvec{2 \\ 2 \\ 2 \\ 0}
-\colvec{1 \\ 1 \\ 5 \\ -1}
=\colvec{1 \\ 1 \\ -3 \\ 1}
\colvec[r]{2 \\ 2 \\ 2 \\ 0}
-\colvec[r]{1 \\ 1 \\ 5 \\ -1}
=\colvec[r]{1 \\ 1 \\ -3 \\ 1}
\qquad
\colvec{3 \\ 1 \\ 0 \\ 4}
-\colvec{1 \\ 1 \\ 5 \\ -1}
=\colvec{2 \\ 0 \\ -5 \\ 5}
\colvec[r]{3 \\ 1 \\ 0 \\ 4}
-\colvec[r]{1 \\ 1 \\ 5 \\ -1}
=\colvec[r]{2 \\ 0 \\ -5 \\ 5}
\end{equation*}
and so the plane is this set.
\begin{equation*}
\set{\colvec{1 \\ 1 \\ 5 \\ -1}
+\colvec{1 \\ 1 \\ -3 \\ 1}t
+\colvec{2 \\ 0 \\ -5 \\ 5}s
\set{\colvec[r]{1 \\ 1 \\ 5 \\ -1}
+\colvec[r]{1 \\ 1 \\ -3 \\ 1}t
+\colvec[r]{2 \\ 0 \\ -5 \\ 5}s
\suchthat t,s\in\Re}
\end{equation*}
\partsitem No; this system
......@@ -1849,20 +1849,20 @@
\begin{ans}{One.II.1.5}
The vector
\begin{equation*}
\colvec{2 \\ 0 \\ 3}
\colvec[r]{2 \\ 0 \\ 3}
\end{equation*}
is not in the line.
Because
\begin{equation*}
\colvec{2 \\ 0 \\ 3}
-\colvec{-1 \\ 0 \\ -4}
=\colvec{3 \\ 0 \\ 7}
\colvec[r]{2 \\ 0 \\ 3}
-\colvec[r]{-1 \\ 0 \\ -4}
=\colvec[r]{3 \\ 0 \\ 7}
\end{equation*}
that plane can be described in this way.
\begin{equation*}
\set{\colvec{-1 \\ 0 \\ -4}
+m\colvec{1 \\ 1 \\ 2}
+n\colvec{3 \\ 0 \\ 7}
\set{\colvec[r]{-1 \\ 0 \\ -4}
+m\colvec[r]{1 \\ 1 \\ 2}
+n\colvec[r]{3 \\ 0 \\ 7}
\suchthat m,n\in\Re}
\end{equation*}
......@@ -1899,12 +1899,12 @@
gives \( k=-(1/9)+(8/9)m \), so \( s=-(1/3)+(2/3)m \) and \( t=1+2m \).
The intersection is this.
\begin{equation*}
\set{\colvec{1 \\ 1 \\ 0}+
\colvec{0 \\ 3 \\ 0}(-\frac{1}{9}+\frac{8}{9}m)+
\colvec{2 \\ 0 \\ 4}m
\set{\colvec[r]{1 \\ 1 \\ 0}+
\colvec[r]{0 \\ 3 \\ 0}(-\frac{1}{9}+\frac{8}{9}m)+
\colvec[r]{2 \\ 0 \\ 4}m
\suchthat m\in\Re}
=\set{\colvec{1 \\ 2/3 \\ 0}
+\colvec{2 \\ 8/3 \\ 4}m
=\set{\colvec[r]{1 \\ 2/3 \\ 0}
+\colvec[r]{2 \\ 8/3 \\ 4}m
\suchthat m\in\Re}
\end{equation*}
......@@ -1921,7 +1921,7 @@
\end{equation*}
gives \( s=6 \) and \( t=8 \), so this is the solution set.
\begin{equation*}
\set{\colvec{1 \\ 9 \\ 10} }
\set{\colvec[r]{1 \\ 9 \\ 10} }
\end{equation*}
\partsitem This system
\begin{equation*}
......@@ -1934,7 +1934,7 @@
gives \( t=-2 \), \( w=-1 \), and \( s=2 \) so their intersection
is this point.
\begin{equation*}
\colvec{0 \\ -2 \\ 3}
\colvec[r]{0 \\ -2 \\ 3}
\end{equation*}
\end{exparts}
......@@ -1947,14 +1947,14 @@
\end{center}
is not the result of doubling
\begin{equation*}
\colvec{2 \\ 0 \\ 0}
+\colvec{-0.5 \\ 1 \\ 0}\cdot 1
\colvec[r]{2 \\ 0 \\ 0}
+\colvec[r]{-0.5 \\ 1 \\ 0}\cdot 1
\end{equation*}
instead it is
\begin{equation*}
\colvec{2 \\ 0 \\ 0}
+\colvec{-0.5 \\ 1 \\ 0}\cdot 2
=\colvec{1 \\ 2 \\ 0}
\colvec[r]{2 \\ 0 \\ 0}
+\colvec[r]{-0.5 \\ 1 \\ 0}\cdot 2
=\colvec[r]{1 \\ 2 \\ 0}
\end{equation*}
which has a parameter twice as large.
\partsitem The vector
......@@ -1963,18 +1963,18 @@
\end{center}
is not the result of adding
\begin{equation*}
(\colvec{2 \\ 0 \\ 0}
+\colvec{-0.5 \\ 1 \\ 0}\cdot 1)
(\colvec[r]{2 \\ 0 \\ 0}
+\colvec[r]{-0.5 \\ 1 \\ 0}\cdot 1)
+
(\colvec{2 \\ 0 \\ 0}
+\colvec{-0.5 \\ 0 \\ 1}\cdot 1)
(\colvec[r]{2 \\ 0 \\ 0}
+\colvec[r]{-0.5 \\ 0 \\ 1}\cdot 1)
\end{equation*}
instead it is
\begin{equation*}
\colvec{2 \\ 0 \\ 0}
+\colvec{-0.5 \\ 1 \\ 0}\cdot 1
+\colvec{-0.5 \\ 0 \\ 1}\cdot 1
=\colvec{1 \\ 1 \\ 1}
\colvec[r]{2 \\ 0 \\ 0}
+\colvec[r]{-0.5 \\ 1 \\ 0}\cdot 1
+\colvec[r]{-0.5 \\ 0 \\ 1}\cdot 1
=\colvec[r]{1 \\ 1 \\ 1}
\end{equation*}
which adds the parameters.
\end{exparts}
......@@ -2067,16 +2067,16 @@
\end{ans}
\begin{ans}{One.II.2.12}
We express each displacement as a vector (rounded to one
decimal place because that's the accuracy of the problem's statement)
We express each displacement as a vector, rounded to one
decimal place because that's the accuracy of the problem's statement,
and add to find the total displacement
(ignoring the curvature of the earth).
\begin{equation*}
\colvec{0.0 \\ 1.2}
+\colvec{3.8 \\ -4.8}
+\colvec{4.0 \\ 0.1}
+\colvec{3.3 \\ 5.6}
=\colvec{11.1 \\ 2.1}
\colvec[r]{0.0 \\ 1.2}
+\colvec[r]{3.8 \\ -4.8}
+\colvec[r]{4.0 \\ 0.1}
+\colvec[r]{3.3 \\ 5.6}
=\colvec[r]{11.1 \\ 2.1}
\end{equation*}
The distance is \( \sqrt{11.1^2+2.1^2}\approx 11.3 \).
......@@ -2092,7 +2092,7 @@
\end{equation*}
can also be described with parameters in this way.
\begin{equation*}
\set{\colvec{-3 \\ 1 \\ 0}y+\colvec{1 \\ 0 \\ 1}z
\set{\colvec[r]{-3 \\ 1 \\ 0}y+\colvec[r]{1 \\ 0 \\ 1}z
\suchthat y,z\in\Re}
\end{equation*}
......@@ -2225,11 +2225,11 @@
No.
These give an example.
\begin{equation*}
\vec{u}=\colvec{1 \\ 0}
\vec{u}=\colvec[r]{1 \\ 0}
\quad
\vec{v}=\colvec{1 \\ 0}
\vec{v}=\colvec[r]{1 \\ 0}
\quad
\vec{w}=\colvec{1 \\ 1}
\vec{w}=\colvec[r]{1 \\ 1}
\end{equation*}
\end{ans}
......@@ -2296,9 +2296,9 @@
To see that `less than' can happen, in \( \Re^2 \) take
\begin{equation*}
\vec{u}=\colvec{1 \\ 0}
\vec{u}=\colvec[r]{1 \\ 0}
\qquad
\vec{v}=\colvec{0 \\ 1}
\vec{v}=\colvec[r]{0 \\ 1}
\end{equation*}
and note that \( \vec{u}\dotprod\vec{v}=0 \).
For `equal to', note that \( \vec{u}\dotprod\vec{u}=1 \).
......@@ -2560,8 +2560,8 @@
\begin{equation*}
\theta
=
\arccos(\,\frac{\colvec{7 \\ 12}\dotprod\colvec{10 \\ 12}}{
\norm{\colvec{7 \\ 12}\,}\cdot\norm{\colvec{10 \\ 12}\,} }\,)
\arccos(\,\frac{\colvec[r]{7 \\ 12}\dotprod\colvec[r]{10 \\ 12}}{
\norm{\colvec[r]{7 \\ 12}\,}\cdot\norm{\colvec[r]{10 \\ 12}\,} }\,)
=\arccos(\frac{214}{\sqrt{244}\sqrt{193}})
\approx \text{$0.17$~rad}
\end{equation*}
......@@ -2587,8 +2587,8 @@
\begin{equation*}
\theta
=
\arccos(\,\frac{\colvec{7 \\ 12}\dotprod\colvec{m \\ d}}{
\norm{\colvec{7 \\ 12}\,}\cdot\norm{\colvec{m \\ d}\,} }\,)
\arccos(\,\frac{\colvec[r]{7 \\ 12}\dotprod\colvec{m \\ d}}{
\norm{\colvec[r]{7 \\ 12}\,}\cdot\norm{\colvec{m \\ d}\,} }\,)
\end{equation*}
Of course, we cannot take $m$ or $d$ negative and so we cannot
get a vector orthogonal to the given one.
......@@ -665,8 +665,8 @@ and we will see how to describe the
solution set.
\smallskip
\noindent\textbf{Note}\hspace*{.4em}
\medskip
\noindent\textbf{Note}\hspace*{.2em}
\textit{For all exercises,
you must justify your answer.
For instance, if a question asks whether a system has a solution then you
......
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