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Notebook[{
Cell[CellGroupData[{
Cell["Calculus: The Language of Change", "Title"],
Cell[TextData[{
StyleBox["Exercise Set 15.6\n",
FontVariations->{"Underline"->True}],
"Angles and Projections"
}], "Subtitle"],
Cell["solutions by Randall Childs and David Theobald", "Subsubtitle"],
Cell[TextData[{
"\tSee the text for explicit instructions on using a ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" notebook to solve Exercises 2 and 3."
}], "Text"],
Cell[CellGroupData[{
Cell["Exercise 1", "Subsection"],
Cell[TextData[StyleBox["Your Own Condition for Perpendicularity",
FontWeight->"Bold"]], "Text"],
Cell["\<\
Give the simplest possible algebraic condition you can use to test \
whether two vectors are perpendicular. \
\>", "Text"],
Cell[TextData[{
"HINT: What is Cos[",
Cell[BoxData[
FormBox[
StyleBox[\(\[Pi]\/2\),
FontSize->16], TraditionalForm]]],
"]? "
}], "Text"],
Cell[TextData[{
"For example, are ",
Cell[BoxData[
FormBox[
RowBox[{"(", GridBox[{
{"1"},
{\(-3\)},
{"2"}
}], ")"}], TraditionalForm]]],
" and ",
Cell[BoxData[
FormBox[
RowBox[{"(", GridBox[{
{"2"},
{"2"},
{"1"}
}], ")"}], TraditionalForm]]],
" perpendicular? "
}], "Text"],
Cell[TextData[{
"Are ",
Cell[BoxData[
FormBox[
RowBox[{"(", GridBox[{
{"1"},
{\(-3\)},
{"2"}
}], ")"}], TraditionalForm]]],
" and ",
Cell[BoxData[
FormBox[
RowBox[{"(", GridBox[{
{"1"},
{"2"},
{"1"}
}], ")"}], TraditionalForm]]],
" ? "
}], "Text"],
Cell["\<\
What are the angles between them? Does the denominator \
matter?\
\>", "Text"],
Cell["\<\
YOUR CONDITION FOR PERPENDICULARITY: A is perpendicular to B if \
and only if ????\
\>", "Text"],
Cell["\<\
Two vectors are perpendicular if and only if they form a right \
angle. In other words, the angle between them is\
\>", "Text"],
Cell[BoxData[
\(TextForm\`\(\ \[Pi]\/2\ \)\)], "Text",
TextAlignment->Center,
FontSize->16],
Cell[TextData[{
"We know that Cos[",
Cell[BoxData[
FormBox[
StyleBox[\(\[Pi]\/2\),
FontSize->16], TraditionalForm]]],
"] = 0."
}], "Text"],
Cell["\<\
So we can restate the perpendicularity condition as: two vectors \
are perpendicular if and only if the cosine of the angle between them is \
zero.\
\>", "Text"],
Cell["\<\
The dot (or inner) product gives us a formula for the cosine of the \
angle between two vectors.\
\>", "Text"],
Cell[TextData[{
StyleBox["Cos[\[Theta]] = ",
FontSize->14],
Cell[BoxData[
FormBox[
StyleBox[
FractionBox[
RowBox[{"\[LeftAngleBracket]",
StyleBox["X\[Bullet]Y",
FontWeight->"Bold"], "\[RightAngleBracket]"}],
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StyleBox["X",
FontWeight->"Bold"], "||",
StyleBox["Y",
FontWeight->"Bold"]}], "|"}]],
FontSize->18], TraditionalForm]],
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}], "Text",
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Cell[TextData[{
"where \[Theta] is the angle between the vectors ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
"."
}], "Text"],
Cell[TextData[{
"So Cos[\[Theta]] is zero if and only if \[LeftAngleBracket]",
StyleBox["X\[Bullet]Y",
FontWeight->"Bold"],
"\[RightAngleBracket] is zero. The value of the denominator, |",
StyleBox["X",
FontWeight->"Bold"],
"||",
StyleBox["Y",
FontWeight->"Bold"],
"|, does not affect this condition. If the cosine of the included angle is \
zero, then the product of that with the denominator is still zero. The only \
way the denominator could cause trouble is if it were zero. But that would \
require either ",
StyleBox["X",
FontWeight->"Bold"],
" or ",
StyleBox["Y",
FontWeight->"Bold"],
" to be the zero vector, and the idea of an angle between any vector and \
the zero vector makes no geometric sense. In order for two vectors to have \
an included angle, neither may be the zero vector. Therefore, the \
denominator is nonzero, and our condition is well-defined."
}], "Text"],
Cell[CellGroupData[{
Cell["Condition for Perpendicularity", "Subsubsection"],
Cell[TextData[{
StyleBox["X",
FontWeight->"Bold",
FontSlant->"Plain"],
" is perpendicular to ",
StyleBox["Y",
FontWeight->"Bold",
FontSlant->"Plain"],
" if and only if their dot product is zero."
}], "Text",
TextAlignment->Center,
FontSize->18,
FontSlant->"Italic"],
Cell[TextData[{
"\tNeither of the two pairs of vectors given by the text is a pair of \
perpendicular vectors. The dot product of the first pair is -2, and the dot \
product of the second pair is -3. Use the notebook ",
StyleBox["AnglPerp",
FontWeight->"Bold"],
" to compute the angles between these pairs of vectors."
}], "Text"],
Cell["\<\
\tUsing the condition for perpendicularity, convince yourself \
that\
\>", "Text"],
Cell[BoxData[
FormBox[
RowBox[{
RowBox[{"(", "\[NegativeThinSpace]", GridBox[{
{GridBox[{
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{GridBox[{
{"4"},
{"4"}
}]},
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}], "\[NegativeThinSpace]", ")"}]}], TextForm]], "Text",
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Cell["\<\
are perpendicular vectors. Enter the following cell to see a graph \
of these vectors. The graph uses a \"perfect\" vantage point which allows \
these vectors to actually appear perpendicular.\
\>", "Text"],
Cell[BoxData[
\(X = {2, \(-3\), 2}; \nY = {4, 4, 2}; \n
xvect\ = \ Graphics3D[\ Line[{{0, 0, 0}, \ X}]]; \n
vecty\ = \ Graphics3D[\ Line[{{0, 0, 0}, \ Y}]]; \n
Show[{xvect, vecty}, \ Axes\ -> \ True, \
PlotRange\ -> \ {{\(-5\), 5}, {\(-5\), 5}, {\(-5\), 5}}, \
ViewPoint\ -> \ {\(-7\), 2, 10}]; \)], "Input"]
}, Closed]],
Cell[CellGroupData[{
Cell["Why does this cell work?", "Subsubsection"],
Cell[TextData[{
"\tThe first two lines give the names ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
" to the given vectors. The next two take sets of three-dimensional \
graphical information involving ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
" and assign a name to each set. The fifth one tells ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" to actually draw or \"Show\" those sets of graphical information with a \
set of guidelines telling the computer how to draw the graph."
}], "Text"],
Cell[TextData[{
"\tThe result is that ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" draws a box showing all points in three-dimensional space for which -5\
\[LessEqual]x1\[LessEqual]+5, -5\[LessEqual]x2\[LessEqual]+5, and -5\
\[LessEqual]x3\[LessEqual]+5. Then it draws line segments representing the \
vectors ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
". The specific numbers chosen in the ViewPoint command allow ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
" to appear perpendicular in our graph (which we must remember is only a \
two-dimensional ",
StyleBox["model",
FontSlant->"Italic"],
" of three-dimensional space)."
}], "Text"],
Cell[TextData[{
"\tTo see that this is true, paste a copy of the above cell somewhere else \
in this notebook. Think of the numbers in the ViewPoint command as a \
position vector, and replace it with a vector which is not parallel to \
(-7,2,10). Exercise 3 from Exercise Set 15.4 gives an easy algebraic \
condition to determine whether two position vectors are parallel. If you \
enter the cell with this new ViewPoint, then ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
" will no longer appear to be perpendicular."
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["\<\
Don't look at this until you've learned about cross products!\
\>",
"Subsubsection"],
Cell[TextData[{
"\tEnter the cell below. It does what it looks like it does: compute the \
cross (or outer) product of ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
". Its output is a familiar vector. Is this a coincidence? HINT: The \
solutions author asked ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" to perform this calculation ",
StyleBox["before",
FontSlant->"Italic"],
" choosing a ViewPoint for the graph of ",
StyleBox["X",
FontWeight->"Bold"],
" and ",
StyleBox["Y",
FontWeight->"Bold"],
"."
}], "Text"],
Cell[BoxData[
\(X\[Cross]Y\)], "Input"]
}, Closed]]
}, Closed]]
}, Open ]]
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StyleDefinitions -> "CalcTLCStyle.nb"
]
(***********************************************************************
Cached data follows. If you edit this Notebook file directly, not using
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End of Mathematica Notebook file.
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