Commit 7eaa46cf authored by Jim Hefferon's avatar Jim Hefferon

final edits of projplane

parent 292c30a0
This diff is collapsed.
This diff is collapsed.
......@@ -24092,7 +24092,7 @@ ans = 0.017398
\begin{ans}{1}
From the dot product
\begin{equation*}
0=\colvec{1 \\ 0 \\ 0}\dotprod\rowvec{L_1 &L_2 &L_3}
0=\colvec[r]{1 \\ 0 \\ 0}\dotprod\rowvec{L_1 &L_2 &L_3}
=L_1
\end{equation*}
we get that the equation is $L_1=0$.
......@@ -24102,15 +24102,15 @@ ans = 0.017398
\begin{exparts}
\partsitem This determinant
\begin{equation*}
0=\begin{vmatrix}
0=\begin{vmat}
1 &4 &x \\
2 &5 &y \\
3 &6 &z
\end{vmatrix}
\end{vmat}
=-3x+6y-3z
\end{equation*}
shows that the line is $L=\rowvec{-3 &6 &-3}$.
\partsitem $\colvec{-3 \\ 6 \\ -3}$
\partsitem $\colvec[r]{-3 \\ 6 \\ -3}$
\end{exparts}
\end{ans}
......@@ -24121,13 +24121,13 @@ ans = 0.017398
\qquad
v=\colvec{v_1 \\ v_2 \\ v_3}
\end{equation*}
can be found from this determinant equation.
comes from this determinant equation.
\begin{equation*}
0=\begin{vmatrix}
0=\begin{vmat}
u_1 &v_1 &x \\
u_2 &v_2 &y \\
u_3 &v_3 &z
\end{vmatrix}
\end{vmat}
=(u_2v_3-u_3v_2)\cdot x
+ (u_3v_1-u_1v_3)\cdot y
+ (u_1v_2-u_2v_1)\cdot z
......@@ -24158,8 +24158,8 @@ ans = 0.017398
to the line from the sun through the pinhole \Dash shows the circle
of the sun projecting to an image that is an ellipse.
(Another example is that in many pictures in this
Topic, the circle that is the sphere's equator is drawn as an ellipse,
that is, is seen by a viewer of the drawing as an ellipse.)
Topic, we've shown the circle that is the sphere's equator as an ellipse,
that is, a viewer of the drawing sees a circle as an ellipse.)
The solar eclipse picture also shows the converse.
If we picture the projection as going from left to right
......@@ -24200,19 +24200,19 @@ ans = 0.017398
basis $B$ with respect to which the points have these
homogeneous coordinate vectors.
\begin{equation*}
\rep{\vec{t}_0}{B}=\colvec{1 \\ 0 \\ 0}
\rep{\vec{t}_0}{B}=\colvec[r]{1 \\ 0 \\ 0}
\quad
\rep{\vec{t}_1}{B}=\colvec{0 \\ 1 \\ 0}
\rep{\vec{t}_1}{B}=\colvec[r]{0 \\ 1 \\ 0}
\quad
\rep{\vec{t}_2}{B}=\colvec{0 \\ 0 \\ 1}
\rep{\vec{t}_2}{B}=\colvec[r]{0 \\ 0 \\ 1}
\quad
\rep{\vec{v}_0}{B}=\colvec{1 \\ 1 \\ 1}
\rep{\vec{v}_0}{B}=\colvec[r]{1 \\ 1 \\ 1}
\end{equation*}
\partsitem
First, any $U_0$ on $T_0V_0$
\begin{equation*}
\rep{\vec{u}_0}{B}=a\colvec{1 \\ 0 \\ 0}
+b\colvec{1 \\ 1 \\ 1}
\rep{\vec{u}_0}{B}=a\colvec[r]{1 \\ 0 \\ 0}
+b\colvec[r]{1 \\ 1 \\ 1}
=\colvec{a+b \\ b \\ b}
\end{equation*}
has homogeneous coordinate vectors of this form
......@@ -24224,8 +24224,8 @@ ans = 0.017398
a homogeneous coordinate vector of this form for some $u_0\in\Re$).
Similarly, $U_2$ is on $T_1V_0$
\begin{equation*}
\rep{\vec{u}_2}{B}=c\colvec{0 \\ 1 \\ 0}
+d\colvec{1 \\ 1 \\ 1}
\rep{\vec{u}_2}{B}=c\colvec[r]{0 \\ 1 \\ 0}
+d\colvec[r]{1 \\ 1 \\ 1}
=\colvec{d \\ c+d \\ d}
\end{equation*}
and so has this homogeneous coordinate vector.
......@@ -24234,8 +24234,8 @@ ans = 0.017398
\end{equation*}
Also similarly, $U_1$ is incident on $T_2V_0$
\begin{equation*}
\rep{\vec{u}_1}{B}=e\colvec{0 \\ 0 \\ 1}
+f\colvec{1 \\ 1 \\ 1}
\rep{\vec{u}_1}{B}=e\colvec[r]{0 \\ 0 \\ 1}
+f\colvec[r]{1 \\ 1 \\ 1}
=\colvec{f \\ f \\ e+f}
\end{equation*}
and has this homogeneous coordinate vector.
......@@ -24245,8 +24245,8 @@ ans = 0.017398
\partsitem
Because $V_1$ is $T_0U_2\,\intersection\,U_0T_2$ we have this.
\begin{equation*}
g\colvec{1 \\ 0 \\ 0}+h\colvec{1 \\ u_2 \\ 1}
=i\colvec{u_0 \\ 1 \\ 1}+j\colvec{0 \\ 0 \\ 1}
g\colvec[r]{1 \\ 0 \\ 0}+h\colvec{1 \\ u_2 \\ 1}
=i\colvec{u_0 \\ 1 \\ 1}+j\colvec[r]{0 \\ 0 \\ 1}
\qquad\Longrightarrow\qquad
\begin{aligned}
g+h &= iu_0 \\
......@@ -24267,8 +24267,8 @@ ans = 0.017398
Since $V_2$ is the intersection
$T_0U_1\,\intersection\,T_1U_0$
\begin{equation*}
k\colvec{1 \\ 0 \\ 0}+l\colvec{1 \\ 1 \\ u_1}
=m\colvec{0 \\ 1 \\ 0}+n\colvec{u_0 \\ 1 \\ 1}
k\colvec[r]{1 \\ 0 \\ 0}+l\colvec{1 \\ 1 \\ u_1}
=m\colvec[r]{0 \\ 1 \\ 0}+n\colvec{u_0 \\ 1 \\ 1}
\qquad\Longrightarrow\qquad
\begin{aligned}
k+l &= nu_0 \\
......@@ -24289,7 +24289,7 @@ ans = 0.017398
Because $V_1$ is on the $T_1U_1$ line its
homogeneous coordinate vector has the form
\begin{equation*}
p\colvec{0 \\ 1 \\ 0}+q\colvec{1 \\ 1 \\ u_1}
p\colvec[r]{0 \\ 1 \\ 0}+q\colvec{1 \\ 1 \\ u_1}
=\colvec{q \\ p+q \\ qu_1}
\tag*{($*$)}\end{equation*}
but a previous part of this question established that $V_1$'s
......@@ -24313,7 +24313,7 @@ ans = 0.017398
Now, the $T_2U_2$ line consists of the points whose homogeneous
coordinates have this form.
\begin{equation*}
r\colvec{0 \\ 0 \\ 1}+s\colvec{1 \\ u_2 \\ 1}
r\colvec[r]{0 \\ 0 \\ 1}+s\colvec{1 \\ u_2 \\ 1}
=\colvec{s \\ su_2 \\ r+s}
\end{equation*}
Taking $s=1$ and $r=u_1u_2-1$ shows that the
......@@ -1721,9 +1721,9 @@ beginfig(24); % a walk around the projective plane
pickup pencircle scaled line_width_dark;
drawarrow subpath (t0,t2) of meridian;
drawarrow subpath (t1,t3) of proj_line;
pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian;
drawarrow subpath (t11,t13) of proj_line;
% pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian withcolor blue;
drawarrow subpath (t11,t13) of proj_line withcolor blue;
endgroup;
%
pickup pencircle scaled line_width_light;
......@@ -1772,9 +1772,9 @@ beginfig(25); % a walk around the projective plane
pickup pencircle scaled line_width_dark;
drawarrow subpath (t0,t2) of meridian;
drawarrow subpath (t1,t3) of proj_line;
pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian;
drawarrow subpath (t11,t13) of proj_line;
% pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian withcolor blue;
drawarrow subpath (t11,t13) of proj_line withcolor blue;
endgroup;
%
pickup pencircle scaled line_width_light;
......@@ -1822,9 +1822,9 @@ beginfig(26); % a walk around the projective plane; back fig thru south pole
pickup pencircle scaled line_width_dark;
drawarrow subpath (t0,t2) of meridian;
drawarrow subpath (t1,t3) of proj_line;
pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian;
drawarrow subpath (t11,t13) of proj_line;
% pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian withcolor blue;
drawarrow subpath (t11,t13) of proj_line withcolor blue;
endgroup;
%
pickup pencircle scaled line_width_light;
......@@ -1872,9 +1872,9 @@ beginfig(27); % a walk around the projective plane; back fig near equator
pickup pencircle scaled line_width_dark;
drawarrow subpath (t0,t2) of meridian;
drawarrow subpath (t1,t3) of proj_line;
pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian;
drawarrow subpath (t11,t13) of proj_line;
% pickup pencircle scaled line_width_light;
drawarrow subpath (t10,t12) of meridian withcolor blue;
drawarrow subpath (t11,t13) of proj_line withcolor blue;
endgroup;
%
pickup pencircle scaled line_width_light;
......
This diff is collapsed.
This diff is collapsed.
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment