PR v Geometri

8/13/2019 PR v Geometri

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PR V MA3041 GEOMETRI

KEVIN MANDIRA LIMANTA(10110106)

1. Prove that lines P+ [v] and Q + [w] intersect if and only if

Q P, v w= 0

Answer:

Let P+ [v] and Q + [w] intersect at R, then we have

R= P+ v=Q + w

so that Q P =v w . Hence, we have

Q P, v w= v w,v w

=v, v w w, v w

= 0

as desired.

Conversely, suppose that QP, v w= 0, then we can represent QPas linear combinationofv and w, namely

Q P =v + w

Hence we have that P+v =Q w. But the left side is on line P+ [v], while the right side ison line Q + [w]. Since both sides are equal, it must be the case that those two lines intersect.

2. Let A and B be two distinct points of S2. Show that

{X S2 |d(X, A) =d(X, B)}

is a line, and find an expression for its pole.

Answer:

To show that l = {X S2 | d(X, A) = d(X, B)} is a line, we must show that there existsvector, which is its pole, such that X, = 0 for all Xl. But since

d(X, A) = cos1X, A= cos1X, B= d(X, B),

we have that

X, A= X, B

or X, A B= 0.

Hence, we have shown the existence of, namelyA B, and l is a line {X S2 | X, A B= 0}.

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3. Prove Theorem 11. In particular, show that

(a) The poles ofmare P

| P|, where is a pole of l.

(b) The point of intersection are

P P, 1 P, 2 .

(c) The distances from P to l are cos1

1 P, 2

Answer:

Choosem=

X S2 |

X,

OP

|OP |

= 0

. To prove the uniqueness, suppose that there exists

such that , = 0 and the line that has as its pole also passes through P. Then we have and P is on the line, and the line connecting any two distinct points must be unique.

(a) Since the line m that is perpendicular to l is unique, then we must have

= OP

|OP |

(b) Vector that is related is the projection ofOPtowards, that isP P, . We must have theunit vector, so the point of intersection is

P P,

|P P, |=

P P,

1 P, 2

(c) The distance from P to l is the distance between Pand the point of intersection Q, that is

d(P, Q) = cos1 P, Q

= cos1

P,

P P, 1 P, 2

= cos1

P,

P P, 1 P, 2

= cos1 P, P P,

1 P, 2= cos1

P, P P, P, 1 P, 2

= cos1 1 P,

1 P, 2

= cos1

1 P, 2

4. Verify that Theorem 15 and Theorem 16 can be proved with the same calculations as were used inthe Euclidean case.

Answer:

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If the matrix corresponds to , , and are

=

ref 0

0 1

, =

ref 0

0 1

, and =

ref 0

0 1

then we can choose so that it corresponds to matrix

ref ( + ) 00 1

.

Next, suppose that line l passes through point Pthat corresponds with matrix

l =

ref(L) 0

0 1

then choose line m and m that correspond with matrix

m=

ref(M) 0

0 1

and m =

ref(M) 0

0 1

such that 2L= M+ M.

We have

lm=

ref(L) 0

0 1

ref(M) 0

0 1

=

rot 2(L M) 0

0 1

=

rot 2(M L) 0

0 1

=

ref(M

) 00 1

ref(L) 0

0 1

= ml

5. Let P and Q be distinct non-antipodal points. Under what circumstances will the group generatedby{HP, HQ} be finite? (Note: A half-turn on S

2 is again a product of reflections in two perpen-dicular lines.)

Answer:

Since for any Pwe haveHPHPX=HPX+ 2HPX, PP

= (X+ 2X, PP) + 2X+ 2X, PP, PP

=X2X, PP2X, PP+ 4X, PP, P

=X4X, PP+ 4X, PP

=X

we can conclude that H2P =I, and hence

HP, HQ= {HmPH

nQ|m, n Z}

={I, HP, HQ, HPHQ}

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But every element in above set is an inverse of itself except for HPHQ. It must be the case thenthat HPHQ is the inverse of itself, that is if they commute. Hence we must have HPHQ= HQHP.Direct calculation shows that

HPHQ(X) =X2X, QQ 2X, PP+ 4X, QP, QP

while

HQHP(X) =X2X, QQ 2X, PP+ 4X, PP, QQ

So, we must have

X, QP, QP =X, PP, QQ

Left side is a multiple of P, while right side is a multiple of Q. So they are equal if and only ifP, Q= 0, that is when vectors OP and OQ are orthogonal.

6. If an isometry of S

2

leaves Pfixed and takes Q to Q, show that P, Q= 0.

Answer:

SupposeTis the isometry, then we have T P =P and T Q= Q. We then have

d(P, Q) =d(T P , T Q) =d(P, Q)

so that

cos1P, Q= cos1P, Q

or P, Q= P, Q= P, Q. This gives P, Q= 0, as desired.

7. Find all isometries of S2 satisfying 2 =I, but =I. Such an isometry is said to be an involu-tion. If and are involutions, is an involution?

Answer:

Reflection, rotation about , half-turn, and glide reflection, which is reflection followed by transla-tion about . Product of involutions andneed not be involution. Take andsuch that theydont commute, that is, = . There exists such a pair, take = l and = m where l andmare not perpendicular.

8. Let Pbe a point of S2, and l a line of S2. Show that

(lHP)2 =I

if and only ifPis a pole ofl or P l.

Answer:

() We divide this into two cases. Let Pbe a pole ofl. For any x S2, we have

HPx= x + 2x, P and l =x 2x, P

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We have three cases:

(a) Ify, P= 0, then P is pole ofl

(b) IfP, = 0, then P is in l

(c) If lP= 0, then lP= 0, which contradicts the fact that lP= 1 because lP S2

.In every cases, P is pole of l or P l.

9. Let l be a line of S2 with pole P. Show that {I, HP, l, E} is a group, and give its multiplicationtable. (E is the antipodal map)

Answer:

Since associative property holds and there is I as identity element, it suffices to show that ev-ery element has an inverse. Clearly Iis the inverse of itself, as well as HPand l because they areall involutions. Note that we have EE= lHPlHP = llHPHP =I, so that E is an inverse of

itself also.

The multiplication table is given as follows:

I HP l EI I HP l E

HP HP I E ll l E I HPE E l HP I

10. Let F ={P,Q,R} be a figure consisting of three mutually perpendicular points. Find S(F).

Answer:

Because P, Q, and R are three mutually perpendicular points, then there are lines in S2 whichpoles are P, Q, and R. Then the group S(F) consists of 24 rotations, 9 reflections, and 15 glidereflections, including the antipodal map.

11. Under what circumstances will a reflection and a half-turn commute?

Answer:

Let lx= x 2x, QQ with Q pole ofl, and HPx= x + 2x, PP. We have

lHPx= x + 2x, PP+ 2x, QQ 4x, PP, QQ

and

HPlx= x + 2x, QQ + 2x, PP4x, QP, QP

= lHPx + 4x, PP, QQ 4x, QP, QP

Hence, in order to l and HPx be commute, we have x, PP, QQ= x, QP, QP, or

P, Q (x, PQ x, QP) = 0

Hence there are three circumstances: P, Q= 0, P =Q, and P =Q. But, we need to considerone more cases, namely P, Q = 0, P = Q. If that were to happen, we get P+Q = 0 and6


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P, Q = 1. So, for x= P+ Q

P+ Q, we have

x, QP x, PQ=

P+ Q

P+ Q, Q

P

P+ Q

P+ Q, P

Q

= 1P+ Q(P+ Q, PQ P+ Q, QP)

= 1

P+ Q(P, QQ + Q P, QP P)

= 1

P+ Q(P, Q+ 1) (PQ)= 0

which cannot happen.

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