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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 10 "MTHSC 206H" }}{PARA 19 "
" 0 "" {TEXT -1 18 "Test on Chapter 10" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 13 "with(linalg):" }}}{SECT 0 {PARA 0 "" 0 "" {TEXT -1
64 "A dodecahedron is a polyhedron formed from 12 regular pentagons."
}}{PARA 0 "" 0 "" {TEXT -1 144 "A regular pentagon can be traced , in \+
a counter-clockwise direction, by moving a specified distance along th
e first edge, then turning an angle " }{XPPEDIT 18 0 "theta = 2*Pi /5
" "/%&thetaG*(\"\"#\"\"\"%#PiGF&\"\"&!\"\"" }{TEXT -1 65 " , then cont
inuing along the second edge, then turning an angle, " }{XPPEDIT 18 0
"theta" "I&thetaG6\"" }{TEXT -1 30 ", etc. The angle of the turn, " }
{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 248 ", is called the ext
erior angle. Once we have traveled all the way around the pentagon and
made a turn at each vertex, we will be back where we started and we w
ill be facing the direction we started in. In other words we will have
turned a total of " }{XPPEDIT 18 0 "2*Pi" "*&\"\"#\"\"\"%#PiGF$" }
{TEXT -1 28 ", which is why each turn is " }{XPPEDIT 18 0 "theta = 2*P
i /5" "/%&thetaG*(\"\"#\"\"\"%#PiGF&\"\"&!\"\"" }{TEXT -1 22 ". The in
terior angle, " }{XPPEDIT 18 0 "phi" "I$phiG6\"" }{TEXT -1 76 ", is th
e supliment to the exterior angle, so the interior angle is given by \+
" }{XPPEDIT 18 0 "phi = 3*Pi / 5" "/%$phiG*(\"\"$\"\"\"%#PiGF&\"\"&!\"
\"" }{TEXT -1 1 "." }{MPLTEXT 1 0 0 "" }}{SECT 0 {PARA 5 "" 0 ""
{TEXT -1 81 "1. Consider a pentagon lying in the zy-plane. Let the fir
st vertex be the origin " }}{PARA 5 "" 0 "" {XPPEDIT 18 0 "O = [0,0,0]
" "/%\"OG7%\"\"!F%F%" }{TEXT -1 59 ", and let the second vertex lie on
e unit along the y axis, " }{XPPEDIT 18 0 "A = [0, 1, 0]" "/%\"AG7%\"
\"!\"\"\"F%" }{TEXT -1 55 ". Determine the coordinates of the remainin
g vertices, " }{XPPEDIT 18 0 "B" "I\"BG6\"" }{TEXT -1 2 ", " }
{XPPEDIT 18 0 "C" "I\"CG6\"" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "D" "
I\"DG%*protectedG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 14 "vO := [0,0,0];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 14 "vA := [0,1,0];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "t
heta := 2*Pi/5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "vB := vA
+ [0, cos(theta), sin(theta)];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 43 "vC := vB + [0, cos(2*theta), sin(2*theta)];" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 43 "vD := vC + [0, cos(3*theta), sin(3*theta)
];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "evalf(vB); evalf(vC);
evalf(vD);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 240 "Note that the y v
alue for C is half way between O and A as well as half way between B a
nd D. In addition, the z values for B and D are the same, and the z va
lue for C is larger. These are all consistent with the way we set up t
he pentagon.." }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 43 "2. Use the fac
t that the interior angle is " }{XPPEDIT 18 0 "phi = 3*Pi/5" "/%$phiG*
(\"\"$\"\"\"%#PiGF&\"\"&!\"\"" }{TEXT -1 97 " and give a formula for t
he coordinates of D. Do these agree with the coordinates from problem \+
1?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "phi := 3*Pi/5;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "vDa := [0, cos(phi), sin(phi
)];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalf(vDa);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "The two calculations agree." }}}}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 42 " 3. Compute the length of the vec
tor from " }{XPPEDIT 18 0 "O" "I\"OG%*protectedG" }{TEXT -1 4 " to " }
{XPPEDIT 18 0 "D" "I\"DG%*protectedG" }{TEXT -1 1 "." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 11 "norm(vD,2);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 18 "evalf(norm(vD,2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 92 "Note that the length of each side should be one, since we start
ed with a side of length one." }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1
38 " 4. Determine the vector from C to D.." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "vD - vC;" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 86 " \+
5. Determine the \"center\" of the pentagon, i.e. the vector u = (O + \+
A + B + C + D)/5." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "u := (v
O + vA + vB + vC + vD)/5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9
"evalf(u);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Note that the y val
ue is 1/2 which it should be, and the z value is well inside the figur
e." }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 31 "6. Determine the distance
from " }{XPPEDIT 18 0 "u" "I\"uG6\"" }{TEXT -1 40 " to the line throu
gh the points O and D." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "Project \+
" }{XPPEDIT 18 0 "u" "I\"uG6\"" }{TEXT -1 59 " onto the unit vector fr
om O to D, then take the length of " }{XPPEDIT 18 0 "u" "I\"uG6\"" }
{TEXT -1 29 " minus the projection onto D." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 35 "umd := evalf(u - dotprod(u,vD)*vD);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "norm(umd,2);" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 28 "Note that the distance from " }{XPPEDIT 18 0 "u" "I
\"uG6\"" }{TEXT -1 26 " to the line though O and " }{XPPEDIT 18 0 "D"
"I\"DG%*protectedG" }{TEXT -1 36 " is the same as the z coordinate of \+
" }{XPPEDIT 18 0 "u" "I\"uG6\"" }{TEXT -1 28 " which is the distance f
rom " }{XPPEDIT 18 0 "u" "I\"uG6\"" }{TEXT -1 21 " to the line through
" }{XPPEDIT 18 0 "O" "I\"OG%*protectedG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "A" "I\"AG6\"" }{TEXT -1 1 "." }}}}{SECT 0 {PARA 5 "" 0
"" {TEXT -1 38 "7. Determine the area of the pentagon." }}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 94 "We can proceed in one of two ways. First \+
the pentagon can be viewed as made up of 3 triangles." }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "T1 := norm(crossprod(vA,vB),2)/2;"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "T2 := norm(crossprod(vB,v
C),2)/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "T3 := norm(cros
sprod(vC,vD),2)/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Area[
1] := evalf(T1+T2+T3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "It can \+
also be viewed as 5 congruent triangles centered at " }{XPPEDIT 18 0 "
u" "I\"uG6\"" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 37 "T4 := norm(crossprod(vO-u,vA-u),2)/2;" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 23 "Area[2] := evalf(5*T4);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 45 "The answers agree to within a rounding error." }}}}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 85 "8. At any vertex of a dodecahedro
n, we have three pentagons. Determine a unit vector " }{XPPEDIT 18 0 "
v = [x,y,z]" "/%\"vG7%%\"xG%\"yG%\"zG" }{TEXT -1 30 " so that it makes
an angle of " }{XPPEDIT 18 0 "phi = 3*Pi/5" "/%$phiG*(\"\"$\"\"\"%#Pi
GF&\"\"&!\"\"" }{TEXT -1 66 " with the vector from O to A and also wit
h the vector from O to D." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36
"x:='x': y:='y': z:='z': v:= [x,y,z];" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 37 "y := solve(dotprod(v,vA)=cos(phi),y);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "z := solve(dotprod(v,vD) = cos(phi)
, z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "x := solve(x^2+y^2
+z^2=1,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "x := x[1];" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "v := evalf(v);" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 101 "Note that this vector is consistent with
a dodecahedron whose back face is flat against the yz-plane." }}}}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 59 "9. The vector v and the vector fr
om O to A define a plane, " }{XPPEDIT 18 0 "P[2]" "&%\"PG6#\"\"#" }
{TEXT -1 40 ". Determine the equation for the plane, " }{XPPEDIT 18 0
"P[2]" "&%\"PG6#\"\"#" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 21 "n := crossprod(v,vA);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 26 "nhat := evalm((1/n[1])*n);" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 13 "n := [1,0,2];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1
24 "Since O is in the plane " }{XPPEDIT 18 0 "P[2]" "&%\"PG6#\"\"#" }
{TEXT -1 58 ", we can conclude that nĄ[x,y,z] = 0 defines the equation
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "x:='x':y:='y':z:='z':
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "eqP2 := dotprod(n,[x,y,
z]) = 0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "Note that the yz-plan
e and " }{XPPEDIT 18 0 "P[2]" "&%\"PG6#\"\"#" }{TEXT -1 56 " intersect
along the y-axis. The equation confirms that." }}}}{SECT 0 {PARA 5 "
" 0 "" {TEXT -1 59 "10. Determine the angle between the yz-plane and t
he plane " }{XPPEDIT 18 0 "P[2]" "&%\"PG6#\"\"#" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "theta[2] := arccos(dotprod(n
,[1,0,0])/norm(n,2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "th
eta[2] := evalf(theta[2]);" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 47 "1
1. Determine the distance from D to the plane " }{XPPEDIT 18 0 "P[2]"
"&%\"PG6#\"\"#" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 31 "(1/norm(n,2))*dotprod(n,vD-vO);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "d[2] := evalf(\");" }}}}{SECT 0 {PARA 5 "" 0 ""
{TEXT -1 44 "12. Determine the distance from (1/2)A to u." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "u;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 19 "norm(u-(1/2)*vA,2);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "d[3] := evalf(\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 56 "Surprise! This is just the value of the z coordinate of " }
{XPPEDIT 18 0 "u" "I\"uG6\"" }{TEXT -1 1 "." }}}}{SECT 0 {PARA 5 "" 0
"" {TEXT -1 37 "13. Determine a point w in the plane " }{XPPEDIT 18 0
"P[2]" "&%\"PG6#\"\"#" }{TEXT -1 109 " which is the same distance from
A/2 as u, and whose y coordinate is 1/2, and whose z coordinate is ne
gative." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "x:='x':y:='y':z:=
'z':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "y := 1/2;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "dotprod((u-(1/2)*vA),(u-(1/2
)*vA));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "K[1] := evalf(\"
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "x := solve(dotprod(n,
[x,y,z])=0,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "z := solv
e(x^2+(y-(1/2))^2+z^2=K[1],z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 10 "z := z[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "w := \+
[x,y,z];" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 92 "14. Determine the e
quation of the line which passes thru w and is perpendicular to the pl
an " }{XPPEDIT 18 0 "P[2]" "&%\"PG6#\"\"#" }{TEXT -1 1 "." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "x:='x':y:='y':z:='z': s:='s':t:='t'
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "w;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 21 "v[1] := t -> n*t + w;" }}}}{SECT 0 {PARA 5 "
" 0 "" {TEXT -1 97 "15. Determine the point of intersection between th
e line from the preceding problem and the line " }{XPPEDIT 18 0 "p(t) \+
= [1,0,0]*t + u" "/-%\"pG6#%\"tG,&*&7%\"\"\"\"\"!F+F*F&F*F*%\"uGF*" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "v[2] := t -
> [1,0,0]*t + u;;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "v[1](s)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "v[2](t);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq1 := s + w[1] = t;" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq3 := 2*s + w[3] = u[3];" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "s := solve(eq3,s);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "t := solve(eq1,t);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "v[1](s);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 8 "v[2](t);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "C[0] := evalf(\");" }}}}{SECT 0 {PARA 0 "" 0 ""
{TEXT -1 194 "Note we have determined a point of intersection between \+
two lines, each of which is perpendicular to its respective pentagon a
nd passes thru the center of that pentagon. The intersection point, "
}{XPPEDIT 18 0 "C[0]" "&%\"CG6#\"\"!" }{TEXT -1 160 ", should be the c
enter of the dodecahedron. We can do a couple of calculations which wi
ll establish the consistency of this result. First, a sphere centered \+
at " }{XPPEDIT 18 0 "C[0]" "&%\"CG6#\"\"!" }{TEXT -1 19 " and just tou
ching " }{XPPEDIT 18 0 "u" "I\"uG6\"" }{TEXT -1 159 " would be complet
ely inside the dodecahedron and its surface area should be lower bound
on the surface area for the dodecahedron. Second, a sphere centered a
t " }{XPPEDIT 18 0 "C[0]" "&%\"CG6#\"\"!" }{TEXT -1 19 " and just touc
hing " }{XPPEDIT 18 0 "O" "I\"OG%*protectedG" }{TEXT -1 223 " would co
mpletely contain the dodecahedron, so its surface area would be an upp
er bound for the surface area of the dodecahedron. Third, the surface \+
area of the dodecahedron should be 12 times the area of a single penta
gon." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "R[1] := evalf(norm(u
-C[0],2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "SA[1] :=evalf
(4*Pi*R[1]^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "R[2] := e
valf(norm(C[0],2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "SA[2
] := evalf(4*Pi*R[2]^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20
"SA[3] := 12*Area[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "The resu
lts from the surface area calculations confirm that the point " }
{XPPEDIT 18 0 "C[0]" "&%\"CG6#\"\"!" }{TEXT -1 51 " is consistent with
the center of the dodecahedron." }}}{PARA 4 "" 0 "" {TEXT -1 0 "" }}}
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