", 1.1\ P]]},
DisplayFunction \[Rule] $DisplayFunction]; \)\)}], "Input"],
Cell["\<\
One particular X = Q on this line (besides P = {1,2}) is the vector X = Q \
= {-2,0}. Note that Q is NOT perpendicular to the Red V, but that the \
Brown (Q-P) is perpendicular to V. Take a look:\
\>", "Text"],
Cell[BoxData[{
\(\(Q = {\(-2\), 0}; \)\),
\(\(line2 =
View2D[{graph, Arrow[V, Tail \[Rule] P, Red],
Graphics[Text["\ ", .6\ P]], Arrow[Q, Blue],
Graphics[Text["\ ", .6\ P]]},
DisplayFunction \[Rule] $DisplayFunction]; \)\)}], "Input"],
Cell["\<\
One particular X on this line besides P = {2,1,-3} is the vector X = \
{-2,0,-1}. Note that X is NOT perpendicular to the Red V, but that the \
Black (X-P) is perpendicular to V. Take a look:\
\>", "Text"],
Cell[BoxData[{
\(\(X = {\(-2\), 0, \(-1\)}; \)\),
\(\(plane2 =
View3D[{plane2, Arrow[X, Blue], Arrow[X - P, Tail \[Rule] P]},
DisplayFunction \[Rule] $DisplayFunction]; \)\)}], "Input"],
Cell[BoxData[
\(\(Show[plane1, plane2]; \)\)], "Input"],
Cell[CellGroupData[{
Cell["Geometric Summary of a(x-x1) + b(y-y1) + c(z-z1) = 0", "Subsubsection"],
Cell["The implicit equation", "Text"],
Cell["\ta (x - x1) + b (y - y1)+ c(z-z1) = 0", "Text"],
Cell["\<\
for given values of the constants a, b, c, x1, y1, z1, is the plane through \
the point\
\>", "Text"],
Cell["\tP = {x1,y1,z1} perpendicular to the vector V = {a,b,c}.", "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["Practice Exercise", "Subsubsection"],
Cell["\<\
1) Sketch the plane through the origin given by 3x + 2y + 4z = 0 and show \
that each vector X lying in it is perpendicular to the vector V = {3,2,4}. \
Do this by hand and then by computer.\
\>", "Text"],
Cell["\<\
2) Sketch the plane in 3-D through the point {1,-2,-1} perpendicular to \
{3,2,4}. Do this by hand and then by computer.\
\>", "Text"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["\<\
Geometry of k in a x + b y + c z = k: Distance to the origin\
\>", "Subsection"],
Cell["The equation", "Text"],
Cell["\ta x + b y + c z = k", "Text"],
Cell["\<\
when k is not zero requires a little more effort to plot. For example,\
\>", "Text"],
Cell["\t2 x - 3 y + 4 z = -8", "Text"],
Cell["\<\
is perpendicular to the vector V = {2,-3,4}, but we need to find some point \
on the lint to plot it. We can set x = y = 0 and solve\
\>", "Text"],
Cell["\t4 z = 8 => z = -2", "Text"],
Cell["\<\
showing us that {0,0,-2} lies on the plnae. Now we draw V (in Red) with \
its tail at P = {0,0,2} and sketch the perpendicular line:\
\>", "Text"],
Cell[BoxData[{
\(\(Clear[P, V, X, x, y, z]; \)\),
\(\(V = {2, \(-3\), 4}; \)\),
\(\(P = {0, 0, \(-2\)}; \)\),
\(\(X = {x, y, z}; \)\),
\(eq = V . \((X - P)\) == 0\),
\(Solve[eq, z]\)}], "Input"],
Cell[BoxData[{
\(\(explicitPlane =
Plot3D[1\/4\ \((\(-8\) - 2\ x + 3\ y)\), {x, \(-5\), 5}, {y, \(-5\),
5}, AxesLabel \[Rule] {"\", .6\ Q]],
Arrow[{\(-3\), \(-2\)}, Tail \[Rule] {1, 2}, Brown],
Graphics[Text["\