MAT 2355 Introduction to Geometry - MAT 2355 Practice Problems. 50 Questions and Worked Solutions.

MAT 2355 Practice Problems.   cos ✓ sin ✓ x0 1. Let ✓ 2 R, set R✓ = and fix v0 = sin ✓ cos ✓ y0 f : R2 ! R2 by f (v) = R✓ v + v0

2 R2 , and define

a) Show that f is an isometry. b) Show that if v0 = 0, then f (u) · f (v) = u · v for every u, v 2 R2 .  2  2 1 0 cos ✓ + sin ✓ 0 Solution: a) First note that R✓t R✓ = = .

2 2 0 1 0 cos ✓ + sin ✓ Now let u, v be (column) vectors in R2 . Then, ||f (u)

f (v)||2 = ||R✓ u + v0 = ||R✓ (u = (R✓ (u = (R✓ (u

(R✓ v + v0 )||2

v)||2

v)) · (R✓ (u v))t R✓ (u

v)) v)

(dot product) (matrix product)

= (u

v)t R✓t R✓ (u

v)

= (u

v)t (u

v)

(matrix product)

= (u

v) · (u

v)

(dot product)

= ||u

v||2 .

Hence, f is an isometry. b) This part follows from part c) if we note that when v0 = 0, then f (0) = 0. c) Recall the theorem proven in class: If f : Rn ! Rn is an isometry and P, Q, R 2 Rn , then (f (Q)

f (P )) · (f (R)

f (P )) = (Q

P ) · (R

P ).

0) · (R

0).

Take P = 0 and f = g in this and we obtain (g(Q)

g(0)) · (g(R)

g(0)) = (Q

Since g(0) = 0, this reduces to

g(Q) · g(R) = Q · R,

2

as required.

2. [Exercise 1.2.1] a) Show that every isometry f : Rn ! Rn is a one-to-one function. (i.e show that if P 6= Q are 2 di↵erent points in Rn , and f is an isometry, then f (P ) 6= f (Q).) b) Assume that an isometry f : Rn ! Rn has an inverse f 1 : Rn ! Rn (i.e. f (f 1 (P )) = P and f 1 (f (Q)) = Q for all P, Q 2 Rn . Show that f 1 must also be an isometry.

Solution: a) If f is an isometry, we show that f (P ) = f (Q) ) P = Q for any P, Q: If f (P ) = f (Q), then d(f (P ), f (Q)) = 0. But f is an isometry, so 0 = d(f (P ), f (Q)) = d(P, Q). This implies that P = Q. b) We assume that f : Rn ! Rn is an isometry and that there is a function g : Rn ! Rn satisfying g(f (P )) = P and f (g(Q)) = Q for all P, Q 2 Rn (Such a function g is called the inverse of f and is denoted f 1 ). We show that g is also an isometry. Let P, Q 2 Rn . Then, d(g(P ), g(Q)) = d(f (g(P )), f (g(Q))) = d(P, Q) Hence, g = f

1

(since f is an isometry)

(since f (g(A)) = A for all A 2 En ).

is an isometry.

3. [Exercise 1.2.2] Fix P 2 E2 and r > 0. Let C = {Q 2 E2 | d(P, Q) = r} be the circle with centre P and radius r. If f : E2 ! E2 is an isometry, prove that f (C) is the circle with centre f (P ) and radius r. (i.e. show that isometries of E2 map circles to circles).

Solution: First we will assume that f has an inverse f 1 which is also an isometry. (This can be shown as follows: We know there is A 2 O(2) and b 2 E2 such that f (x) = Ax + b, and At A = I2 . Then the formula for f 1 is f 1 (u) = At (u b) = At u At b. Thus f has an inverse f 1 which is also an isometry, by 2(b).)

Now, fix P 2 E2 and r > 0. Let C = {Q 2 E2 | d(P, Q) = r} be the circle with centre P and radius r. We show that if f : E2 ! E2 is an isometry, then f (C) is the circle with centre f (P ) and radius r.

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Let C 0 = {Q0 2 E2 | d(f (P ), Q0 ) = r}’and recall that f (C) = {f (Q) | Q 2 C}.

We will show that C 0 ✓ f (C) and f (C) ✓ C 0 , which implies that f (C) = C 0 .

1. C 0 ✓ f (C) Let Q0 2 C 0 . We show that Q0 2 f (C) as follows. Set Q = f 1 (Q0 ), so that f (Q) = Q0 . Then, d(P, Q) = d(f (P ), f (Q))

(since f is an isometry)

= d(f (P ), Q0 ) =r

(since Q0 2 C 0 ).

This show that Q 2 C. But Q0 = f (Q), so Q0 2 f (C). This proves that C 0 ✓ f (C).

2. f (C) ✓ C 0 Let R 2 f (C). We show that R 2 C 0 as follows. Since R 2 f (C), we know that R = f (Q) for some Q 2 C. Then, d(f (P ), R) = d(f (P ), f (Q)) = d(P, Q) =r

(since R = f (Q))

(because f is an isometry)

(since Q 2 C).

This shows that R 2 C 0 , so f (C) ✓ C 0 .

This completes our argument establishing that if f is an isometry and C is the circle with centre P and radius r, then f (C) is the circle with centre f (P ) and radius r. (i.e. isometries map circles to circles).

4. a) Suppose R2 3 (a, b) 6= 0 and c 2 R. Let be a line in R2 . Find a formula for the reflection RL : R2 ! R2 in the line L = {(x, y) 2 R2 | ax + by + c = 0}. b) Check that your formula in (a) really does define an isometry.

2 c) Check that your formula in (a) satisfies RL (P ) = RL (RL (P )) = P for all P 2 R2 .

Solution: a) Suppose (a, b) 6= 0 and c 2 R. Let L = {(x, y) 2 R2 | ax+by +c = 0} be a line in R2 . We find a formula for the reflection R : R2 ! R2 in L as follows.

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First note that the vector n = (a, b) 6= 0 is perpendicular to L, since if v1 = (x1 , y1 ) and v2 = (x2 , y2 ) are 2 di↵erent points on L, then (a, b) · (v1 v2 ) = (ax1 + by1 ) (ax2 + by2 ) = c c = 0. We may then write the equation for L is vector form: L = {v 2 R2 | v · n + c = 0}.

If u = (x, y), let u0 2 L be the point on L which is closest to u. Then, we know that (u0 u) is perpendicular to L, and R(u) = u0 +(u0 u) = 2u0 u = 2(u0 u)+u.

We find an expression for u0 as follows.

Since (u u0 ) is perpendicular to L, we may write u0 = u+sn for some s 2 R. We can solve for s, since we know that u0 2 L, so c = u0 ·n = (u+sn)·n = u·n+s||n||2 .

Thus, c+u·n s= , ||n||2 and so (u0

(c + u · n) n. ||n||2

u) = sn =

Hence, (1)

RL (u) = u

2

(c + u · n) n. ||n||2

n c We could have simplified our lives by setting m = ||n|| , e = ||n|| , and noticing that L = {v 2 R2 | v · n + c = 0} = {v 2 R2 | v · m + e = 0}. Then, ||m|| = 1, and our formula (1) can be written as

(2)

RL (u) = u

2(e + u · m)m.

b) Let’s use the easier formula (2) here: If u, v 2 R2 , then RL (u)

RL (v) = u =u

Now let w = u

v, so that

2(e + u · m)m v

2[(u

(v

v) · m]m

2(e + v · m)m

5

||RL (u)

RL (v)||2 = (RL (u) = (w = ||w||2 = ||w||2 = ||u

Hence, ||RL (u)

RL (v)|| = ||u

RL (v)) · (RL (u)

2(w · m)m) · (w

RL (v))

2(w · m)m)

4(w · m)2 + 4(w · m)2

v||2 .

v||, and so RL is an isometry.

c) Once again we use (2): RL (RL (u)) = RL (u) =u

2(e + RL (u) · m)m

2(e + u · m)m

= u 2em = u,

2(e + (u

2(u · m)m

2em

2(e + u · m)m) · m)m

2(u · m)m + 4e(m · m)m + 4(u · m)m

as required.

5. Show that if f : Rn ! Rn is an isometry and P and Q are points in Rn , then f (P Q) = f (P )f (Q), i.e. that f maps the segment with endpoints P and Q to the segment with endpoints f (P ) and f (Q).

Solution: First note that we know that there is an orthogonal matrix A and vector b 2 Rn such that f (P ) = AP +b for all P 2 Rn . Since P Q = {P +s(Q P ) | s 2 [0, 1]} and f (P )f (Q) = {f (P ) + s(f (Q) f (P )) | s 2 [0, 1]}, we have f (P Q) = {f (P + s(Q

= {AP + s(AQ

P )) | s 2 [0, 1]}

AP ) + b | s 2 [0, 1]}

= {(AP + b) + s((AQ + b) = {f (P ) + s(f (Q) = f (P )f (Q),

(AP + b)) | s 2 [0, 1]}

f (P )) | s 2 [0, 1]}

6

as required.

2

3 0 0 1 6. Let A = 4 1 0 0 5 and consider the map f : R3 ! R3 defined by f (x) = Ax. 0 1 0 a) Show that f is an isometry. b) Is f a reflection ? If so, find the plane of reflection, and if not, give reasons.

Solution: a) Since the columns of A form an orthonormal basis if R3 , A is an orthogonal matrix. Hence, ||f (u)

f (v)||2 = ||Au

Av||2

= ||A(u = A(u

v)||2 v) · A(u

v)

(dot product)

v))t A(u

v)

(matrix product)

= (u

v)t At A(u

v)

= (u

v)t In (u

= (u

v)t (u

v)

(matrix product)

= (u

v) · (u

v)

(dot product)

= (A(u

= ||u

v)

v||2 ,

so f is an isometry. (b) If f were a reflection, it would fix an entire plane–the plane in which the reflection occurs. So we compute F = {x 2 R3 | f (x) = x} = {x 2 R3 | Ax = x}

= {x 2 R3 | (A I)x = 0} 2 3 2 3 1 0 1 1 0 1 But A I = 4 1 1 0 5 ⇠ 40 1 1 5, so F = span{(1, 1, 1)}, which is a 0 1 1 0 0 0 line, not a plane. Hence, f is not a reflection. (In fact, f is a rotation of 2⇡ 3 about an axis parallel to (1, 1, 1). Can you show this?)

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7. Suppose n 6= 0 is a vector in Rn and let H = {v 2 Rn | v · n = 0} be the hyperplane through the origin with normal vector n. If RH : Rn ! Rn is the reflection in H, show that the matrix A of RH is a symmetric matrix. (i.e. satisfies At = A).

Solution: As 0 2 H, we know RH (0) = 0, and so RH (v) = Av, 8v 2 R3 .

Moreover, we also know (maybe you should check this!) RH (RH (v)) = v, 8v 2 R3 ,thus AA = In , i.e. A 1 = A. But we also know that A 2 O(3), so AAt = In too, i.e. A 1 = At . Thus A = A 1 = At .

8. (a) Find 2 lines L0 and L1 so that reflection in L0 followed by reflection in L1 is the isometry f : R2 ! R2 defined by f (x, y) = (x + 1, y 2). Check that your reflections achieve the desired result. (b) Find 2 planes H0 and H1 in R3 so that reflection in H0 followed by reflection in H1 is the isometry f : R3 ! R3 defined by f (x, y, z) = (x + 1, y + 2, z 2).

Check that your reflections achieve the desired result.

Solution: (a) Let L0 {(x, y) | x

5 2y = 0} and  3 L1 4= {(x,  y) | x 2y = 2 }. x 5 Then, RL0 (x, y) = (x, y) + 25 ( x + 2y)(1, 2) = 54 and RL1 (x, y) = 3 y 5 3   5 4 x 1 5 (x, y) + 25 ( 52 x + 2y)(1, 2) = 54 + , so 3 y 2 5 5

RL1 RL0 (x, y) =

3 

5 4 5

4 5 3 5

x = + y



3 5 4 5

1 , 2

4 5 3 5



x + y



1 2

as required. (b) All we need are two planes which share the normal a = (1, 2, 2) and which 3 are exactly ||a|| 2 units apart. So let H0 = {v 2 R | a · v = 0}. A point which a a is exactly ||a|| 2 units from (0, 0, 0) is 2 , and the plane through 2 with normal a is

8 a 9 3 exactly ||a|| 2 units from H0 . So we let H1 = {v 2 R | a · v = a · 2 = 2 }. Then,

RH1 RH0 v = RH1 (v

2

a·v a) ||a||2

a·v =v 2 a+2 ||a||2 a·v =v 2 a+a ||a||2 = v + a,

2

( ||a|| 2 2

a · (v

||a||2

a·v 2 ||a|| 2 a))

a

a·v a·v a + 4 a ||a||2 ||a||2

as required.

9. Let O(n) = {A | A is an n ⇥ n matrix with AAt = In } be the set of n by n orthogonal matrices. Show that a) In 2 O(n)

b) If A, B 2 O(n), then AB 2 O(n) and that c) If A 2 O(n), then A 1 2 O(n).

(i.e. Under the operation of matrix multiplication, O(n) is a group.)

Solution: (a) Since Int In = In In = In , the matrix In is orthogonal. (b) Suppose At A = B t B = In . Then, (AB)t AB = (B t At )AB = B t (At A)B = B t (In )B = B t B = In , so AB is orthogonal. (c) Suppose A 2 O(n). Then At A = In , so A 1 = At . So, (A 1 )t A 1 = (At )t At = AAt = In , so A 1 is orthogonal as well.

10. Show that if A 2 O(n) and b 2 Rn are fixed, then f : Rn ! Rn is defined by f (x) = Ax + b is indeed an isometry.

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Solution: We compute: If u, v, 2 Rn , then ||f (u)

f (v)||2 = ||Au

Av||2

= ||A(u = A(u

v)||2 v) · A(u

v)

(dot product)

v))t A(u

v)

(matrix product)

= (u

v)t At A(u

v)

= (u

v)t In (u

= (u

v)t (u

v)

(matrix product)

= (u

v) · (u

v)

(dot product)

= (A(u

= ||u

v)

v||2 ,

so f is an isometry.

11. a) Show that the composition of two isometries is an isometry. b) Show that any isometry f : Rn ! Rn has an inverse, that is, that there is a function g : Rn ! Rn such that f (g(P )) = P and g(f (P )) = P for all P 2 Rn . c) Show that the inverse g of the isometry f is an isometry.

Solution: (a) This follows directly from the definition of an isometry. (b) We know that there is an orthogonal matrix A and vector b 2 Rn such that f (P ) = AP + b for all P 2 Rn . To find the formula for g : Rn ! Rn such that f (g(P )) = P and g(f (P )) = P for all P 2 Rn , we solve f (g(P )) = P for g(P ). But f (g(P )) = P ) Ag(P ) + b = P ) Ag(P ) = P b ) g(P ) = A 1 P A 1 b = At P At b.

So, g(P ) = At P At b. A short check now shows that this g does indeed satisfy gf (P ) = f g(P ) = P for all P 2 Rn . (c) By a previous problem, we know At 2 O(n), and so by the preceding problem, g is an isometry. (i.e. Under the operation of composition, the set {f : Rn ! Rn | f is an isometry} is a group. It has a (proper ) subgroup isomorphic to O(n).) 

cos ✓ sin ✓ a line L through the origin, and the line of reflection.

12. Show that f : R2 ! R2 defined by f (v) =

sin ✓ v is a reflection in cos ✓

10

Solution: For notational convenience, set ↵ = cos ✓ and A=



↵ ↵

= sin ✓, and

.

We find the set L of fixed points of f : R2 ! R2 defined by f (v) = Av. A short computation (using ↵2 + 2 = 1) shows that, L = {v | Av = v} = {v | (A

I)v = 0}

is found to be the  line L = {(x, y) | (↵ 1)x + y = 0} unless ↵ = 1 and = 0, in 1 0 which case A = , and L is the x axis.

0 1 If ↵ = 1, A is the matrix of the reflection in the x axis. So assume ↵ 6= 1, and compute the formula for the reflection in L: RL (x, y) = (x, y) = (x, y)

((↵

1)x + y) (↵ 2(1 ↵)

1, )

((1

↵)x

(1 + ↵)y)

2

= (↵x + y, x

y, x

(using ↵2 +

2

= 1)

↵y)

= f (x, y), so that f = RL , as required.

13. Let L1 and L2 be two lines in R2 with direction vectors {v1 , v2 }, which meet at a single point P . Show that RL2 RL1 is the rotation about P , in the positive sense determined the orientation {v1 , v2 }, through the angle which is twice the (acute) angle between L1 and L2 .

Solution: Let L1 and L2 be 2 lines in R2 which meet at a point P . Since each reflection reverses the orientation, their product will preserve it, and moreover, RL2 RL1 (P ) = P . Hence, RL2 RL1 is an orientation preserving isometry of R2 which fixes at least one point, and is thus a rotation about P (it cannot be a non-trivial translation, since this would have no fixed points.)

To find the angle ✓ of rotation, let a1 and a2 be (unit) normal vectors for L1 and L2 and let ' be the (acute) angle between a1 and a2 (which is the same as

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the angle between L1 and L2 ), so that a1 · a2 = cos ' 0. Since P 2 L1 \ L2 , we may write L1 = {v | a1 · v = a1 · P } and L2 = {v | a2 · v = a2 · P }. Note that RLi (v) = v + 2ai · (P v)ai for i = 1, 2.

It suffices to compute the cosine of the angle ✓ (✓ 2 [0, ⇡])between the segment P (P + a1 ) and the segment P (RL2 RL1 (P + a1 )) : cos ✓ = (RL2 RL1 (P + a1 ) = (RL2 (P = (P

a1 )

P ) · ((P + a1 )

P ) · a1

a1 + 2(a1 · a2 )a2

= ( a1 + 2(a1 · a2 )a2 ) · a1 =

P)

P ) · a1

1 + 2(a1 · a2 )2

= 2 cos2 ' = cos 2'.

1

Since ' is acute, cos ✓ = cos 2' implies that ✓ = 2', as required. If cos ' = a1 .a2 = 0, the rotation is by ⇡ and there are no orientation issues. So, suppose cos ' = a1 .a2 > 0 Then, det [ (P + a1

P ) (RL2 RL1 (P + a1 )

P ) ] = det [ a1 = det [ a1

a1 + 2(a1 · a2 )a2 ]

a2 ]

= 2(a1 · a2 ) det [ a1 which has the same sign as det [ a1 rotation by ✓.

a2 ] ,

a2 ], showing that {a1 , a2 } is the sense of the

14. (Bonus) Prove or find a counterexample: if any isometry f : R3 ! R3 satisfies f (0) = 0, then f (u) ⇥ f (v) = f (u ⇥ v) or f (u) ⇥ f (v) = f (u ⇥ v)for all u, v 2 R3 , where “⇥” denotes the cross product in R3 . If it is true, what property of f determines the choice of sign?

Suppose f (u) = Au. Then A 2 O(3), so {Ae1 , Ae2 , Ae3 } is an orthonormal basis of R3P . Hence, by the expansion theorem from first year linear algebra, namely, that 3 w = i=1 (w · Aei )Aei , for any w 2 R3 , we have

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3 X Au ⇥ Av = (Au ⇥ Av · Aei )Aei

= = =

i=1 3 X i=1 3 X i=1 3 X

det[Au Av Aei ]Aei det(A[u v ei ])Aei det A det[u v ei ]Aei

i=1

3 X = det A (u ⇥ v · ei )Aei i=1

Moreover, using the expansion theorem again (and properties of multiplication by a matrix) 3 ⇣X ⌘ A(u ⇥ v) = A (u ⇥ v · ei )ei i=1

3 X = (u ⇥ v · ei )Aei i=1

Hence, Au ⇥ Av = (det A).A(u ⇥ v) = That is f (u) ⇥ f (v) =

⇢

⇢

A(u ⇥ v), if det A = 1, A(u ⇥ v), if det A = 1.

f (u ⇥ v), if det A = 1, f (u ⇥ v), if det A = 1.

15. (a) Let P 6= Q be distinct points in Rn . Show that H = {v 2 Rn | ||v ||v Q||} is a hyperplane and that reflection in H interchanges P and Q.

P || =

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(b) Suppose a 2 Rn , b 2 R with ||a|| = 1 and let H = {v 2 Rn | v · a = b} be a hyperplane in Rn . Show that if P and Q are any points in Rn such that P Q is parallel to a , and P · a b = (Q · a b) > 0, then H = {v 2 Rn | ||v P || = ||v Q||}. (What does “P · a b = (Q · a b)” mean geometrically?)

Solution: As we have seen, ||v

P || = ||v

Q|| () v · (2(P

Q)) = ||P ||2

||Q||2 ,

so H is certainly a hyperplane. Moreover,

RH (v) = v + 2 =v+2

||Q||2 v · (2(P Q)) 2(P 4||P Q||2 P ||2 ||v Q||2 2(P Q) 4||P Q||2

||P ||2 ||v

Q)

so that RH (P ) = P + 2 =P

||P

(P

P ||2 ||P Q||2 2(P 4||P Q||2 Q) = Q.

Q)

Since RH (RH (v)) = v, RH (Q) = RH (RH (P )) = P , as required. (b) Let’s assume that P Q is parallel to a, and that ||a|| = 1. If P Q = a, then Q = P a, so P · a b = (Q · a b) implies that 2P · a = 2b.

2 2 Moreover, v 2 H () v · (2(P Q)) = 2 b, while ||Q|| = ||P || 2 a · P + 2.

2 Hence, ||P ||2 ||Q||2 = 2 a · P = 2 b = v · (2(P Q)). Thus, v 2 H () 2 2 v · (2(P Q)) = ||P || ||Q|| () ||v P ||2 = ||v Q||2 , just as in part (a).

“P · a b = (Q · a b)” means that P and Q are on opposite sides of H. 16. Let T1 be the triangle whose vertices are (1, 1),p(1, 4) and (4, 4) and T2 the p p 3 2 3 2 triangle whose vertices are (0, 0), ( 3 2, 0) and ( 2 , 2 ). Find an isometry f of R2 such that f (T1 ) = T2 .

Solution: First, we move A = (1, 1) to D = (0, 0) by reflection RL (x, y) = (1 y, 1 x) in the line L = {(x, y) | x + y = 1}.

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Then, C 0 := RL (C) = RL (4, 4) = ( 3, 3), which we wish to move to E = p ( 3 2, 0), while keeping the origin D fixed. (Draw yourself a picture to see why.)

This we can do by reflection in the line K that is equidistant from C 0 and E, since D is also equidistant from C 0 and E. So we know that RK (C 0 ) = RK (E) and RK (D) = D. But a short calculation shows that F and B 0 := RL (B) = ( 3, 0) are also equidistant from C 0 and E, so RK (B 0 ) = F .

Thus, if we set f = RK RL , then f (A) = RK RL (A) = RK (D) = d, f (B) = RK RL (B) = RK (B 0 ) = F and f (C) = RK RL (C) = RK (C 0 ) = E. Since f is a product of 2 reflections in lines that do intersect, f is a rotation in this case. (How do we know L and K p intersect? We know that 0 2 K and C 0 +E C 0 +E 3 2 K, so the vector 2 ) = 2 (1 + 2, 1) is actually a direction vector for 2 K, and this is clearly not parallel to (1, 1), a direction vector for L.) 17. If H and K are perpendicular planes in R3 , show that RH (K) = K, where RH denotes the reflection in H. Solution: Suppose aH , aK are (unit) normals to H and K respectively and choose h0 2 H so that H = {v 2 R3 | v ·aH = h0 ·aH } and K = {v 2 R3 | v ·aK = bK }. Let H 0 = {v 2 R3 | v · aH = 0}. Then, 0 2 H 0 , so RH 0 (v) · RH 0 (w) = v · w for all v, w 2 R3 .

Moreover, RH (v) = v + 2(bH aH · v)aH = RH 0 (v) + 2bH aH . In particlular, since aH · aK = 0, RH 0 (aK ) = aK . Hence, if k 2 K, RH (k) · aK = (RH 0 (k)

2bH aH ) · aK

= RH 0 (k) · aK

= RH 0 (k) · RH 0 (aK ) = k · aK = bK , so RH (k) 2 K. Thus RH (K) ✓ K, and one can see that equality holds by applying RH to both sides of this set equation.

18. Show that the reflection RH in a hyperplane H ⇢ Rn reverses orientation.

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Solution: It suffices to show this for a hyperplane through the origin, so let 0 6= a 2 Rn be a normal vector for H so that H = {v 2 Rn | a · v = 0}. Note that H is a subspace of Rn of dimension n 1, and that if {v1 , . . . , vn 1 } is a basis for H, then {a, v1 , . . . , vn 1 } is a basis for Rn . Note that RH (a) = a and that RH fixes each of v1 , . . . , vn 1 . Hence we may compute: det [ RH a RH v1 . . . RH vn 1 ] = det [ a v1 . . . vn 1 ] =

det [ a v1 . . . vn 1 ] ,

and so the reflection RH reverses orientation.

19. Consider the plane H = {v 2 R3 | v · a = 0} which contains the origin and which is perpendicular to the unit vector a 2 R3 . Suppose ✓ 2 [0, 2⇡], and let ⇢ : R3 ! R3 be the isometry which is rotation by ✓ about the line through the origin with direction a. As usual, let RH denote the reflection in H. Show that ⇢RH = RH ⇢.

Solution: We show that ⇢RH (P ) = RH ⇢(P ) for all P 2 S2 . First note that if v · a = 0, then RH (v) = v. In addition, both ⇢ and ⇢ fix 0 and so are they are linear and preserve dot products. Note that if w · a = 0, then ⇢(w) · a = ⇢(w) · ⇢(a) = w · a = 0, so ⇢(w) · a = 0 as well. That is, w 2 H ) ⇢(w) 2 H. Thus w 2 H ) RH (w) = w and RH ⇢(w) = ⇢(w). Note also that (1)

RH (a) =

a

and (2)

⇢(a) = a.

Now let v 2 R3 be any point. Then, we may write v = (v · a)a + (v

(v · a)a),

16

where v 0 = (v (3)

(v · a)a) satisfies v 0 · a = 0, so

v 0 2 H, ⇢(v 0 ) 2 H and so RH (v 0 ) = v 0 , and RH ⇢(v 0 ) = ⇢(v 0 ).

Hence, we can write v = sa + v 0 where s = v · a. Then, ⇢RH (v) = ⇢RH (sa + v 0 ) = ⇢(RH (sa) + RH (v 0 )) (because RH is linear) = ⇢( sa + v 0 ) (using equations 0, 1 and 3) =

sa + ⇢(v 0 ) (because ⇢ is linear, and using equation 2)

On the other hand, RH ⇢(v) = RH ⇢(sa + v 0 ) = RH (⇢(sa) + ⇢(v 0 )) (because ⇢ is linear) = RH (sa + ⇢(v 0 )) (using equation 2) = RH (sa) + RH ⇢(v 0 ) (because RH is linear) =

sa + ⇢(v 0 ) (using equations 1 and 3)

= ⇢RH (v), as required.

20. Suppose 3 distinct points A, B, C lie on a line L, with B between A and C, and let D be a point not on L. (a) Express the fact that “B is between A and C” algebraically. (b) Using (a) or otherwise, show that if B is between A and C and f is an isometry, then f (B) is between f (A) and f (C). (c) Show that \ABC = ⇡ and that \DBA + \DBC = ⇡. (Hint: use a suitable isometry, and (b), to simplify your life.)

Solution: We know that angles and lines are preserved by isometries. So is “between-ness”:if f is an isometry and A, B, C lie on a line L, with B between

17

A and C, then f (A), f (B), f (C) lie on a line f (L), with f (B) between f (A) and f (C). Proof: B lies between A and C on the line determined by A and C i↵ (B A) · (B C) < 0. Since isometries preserve dot products of this form (any translation part cancels) (f (B) f (A)) · (f (B) f (C)) = (B A) · (B C) < 0, so f (B) lies between f (A) and f (C) on the line determined by f (A) and f (C).

Thus, we may (by applying an isometry if necessary) assume that L is the x axis, A = ( x0 , 0), B = (0, 0), C = (x1 , 0) with x0 , x1 > 0, and D = (x, y) with y > 0.

Then, a short computation shows that cos(\DBA) = p 2x 2 , sin(\DBA) = x +y y y x p 2 2 , cos(\DBC) = p 2 2 , and sin(\DBC) = p 2 2 , so x +y x +y x +y

cos(\DBA + \DBC) = cos(\DBA) cos(\DBC) x2 = 2 x + y2 = 1.

sin(\DBA) sin(\DBC)

y2 x2 + y 2

Thus, \DBA + \DBC = ⇡.

21. Suppose A, B, C, D are points in the plane with A 6= D, and let L be the line L through A and D. Suppose that B and C are on the same side of L. (a) Express the fact that “B and C are on the same side of the line through A and D” algebraically (Hint: use orientations in the plane). (b) Using (a) or otherwise, show that if B and C are on the same side of the line through A and D and f is an isometry, then f (B) and f (C) are on the same side of the line through f (A) and f (D) (c) If \CAD = \BAD, show that A, B and C are collinear. (Hint: use a suitable isometry, and (b), to simplify your life.)

Solution: An isometry certainly preserves angles. It also preserves “on the same side as-ness”, i.e. if f is an isometry and A, B, C, D are points in the plane with B and C on the same side of the line through A and D, then f (A), f (B), f (C), f (D) are points in the plane with f (B) and f (C) on the same side of the line through

18

f (A) and f (D). Proof: A, B, C, D are points in the plane with B and C on the same side of the line through A and D i↵ (C A)·(D E) and (B A)·(D E) have the same sign, where E is a point chosen so that (D E) is a normal for the line through A and D. Since f preserves angles, f (D) f (E) will be a normal for the line through f (A) and f (D), and since (C A)·(D E) = (f (C) f (A))·(f (D) f (E)), and (B A) · (D E) = (f (B) f (A)) · (f (D) f (E)), f (B) and f (C) will on the same side of this line if B and C on the same side of the line through A and D. Thus, using an isometry if necessary, we may assume that A = (0, 0), D = (x0 , 0) (with x0 > 0) , B = (x1 , y1 ) and C = (x2 , y2 ) with y1 , y2 0. We may further assume that neither of B or C is A, since in those cases, the conclusion is obvious.

Hence, x21 + y12 > 0 and x22 + y22 > 0. Then, the line through A and B is the line L = {(x, y) 2 R2 | (x, y) · ( y1 , x1 ) = 0} Then, \CAD = \BAD implies that

and that

x x p 1 p 2 = cos(\BAD) = cos(\CAD) , x21 + y12 x22 + y22

y y p 1 p 2 = sin(\BAD) = sin(\CAD) .

2 2 2 2 x1 + y1 x2 + y2 p 2 2 x +y Thus, if we let k = p 22 22 , we have x2 = kx1 and y2 = ky1 . Then, x1 +y1

(x2 , y2 ) · ( y1 , x1 ) = k(x1 , y1 ) · ( y1 , x1 ) = k.0 = 0, so that C = (x2 , y2 ) 2 L, as required.

22. (Bonus) Suppose A is a real, n ⇥ n orthogonal matrix. a) Show that if is a (possibly) complex eigenvalue of A, then | | = 1, where | | denotes the modulus of the complex number . b) Show that if is an eigenvalue of A then ¯ (where the bar denotes complex conjugation) is also an eigenvalue of A. c) Show that eigenvectors of A corresponding to di↵erent eigenvalues are orthogonal. d) Suppose v 2 Cn is a (possibly complex) eigenvector of A with eigenvalue .

Show that v̄ is an eigenvector with eigenvalue ¯ .

19

e) Show that if v is a eigenvector of A with eigenvalue = ei✓ 2 / R, and we v+v̄ v v̄ define v1 = 2 , and v2 = 2i , show that {v1 , v2 } is an orthogonal subset of Rn , with ||v1 || = ||v2 ||, and finally that Av1 = (cos ✓)v1 + (sin ✓)v2 , (*)

Av2 =

(sin ✓)v1 + (cos ✓)v2 .

Solution: To denote the fact that the entries of a vector V may be complex, we will write v 2 C n . If the entries are actually real, we will say that v is a real vector, or write v 2 Rn .

First note that if for a (possibly complex) matrix B, we define B ⇤ := B̄ t , then (B ⇤ )⇤ = B, (µB)⇤ = µ̄B ⇤ if µ 2 C and (AB)⇤ = B ⇤ A⇤ . If v 2 Cn , then v ⇤ v 0 and v ⇤ v = 0 () v = 0. Moreover, if A is a real, orthogonal matrix, and v, w 2 Cn , then (Av)⇤ (Aw) = v ⇤ A⇤ Aw = v ⇤ At Aw = v ⇤ w. (a) Suppose that v 6= 0 is an eigenvector of A with (possibly) complex eigenvalue . Then, 0 6= v ⇤ v = (Av)⇤ Av = ( v)⇤ ( v) = ¯ v ⇤ v, so that | |2 = ¯ = 1. (Note that this means that the eigenvalue is of the form = ei✓ , for some ✓ 2 [0, 2⇡).) (b) The eigenvalues of A are the roots of the polynomial cA (x) = det(A xI).

Since the entries of A are real, this polynomial has real coefficients (even though some of its roots may be complex). Thus, if is an eigenvalue, then, cA ( ) = 0 = 0̄ = cA ( ) = cA ( ¯ ), the latter because the coefficients of cA are real. Hence, ¯ is a root of cA (x), and so is an eigenvalue of A as well. (c) Two vectors v, w 2 Cn are orthogonal if v ⇤ w = 0. So suppose v, w 6= 0, and that Av = v and Aw = µw, so that A 1 w = µ1 w = µ̄w. Then, since A is orthogonal, (Av)⇤ w = ¯ v ⇤ w, and as well, (Av)⇤ w = v ⇤ A⇤ w = v ⇤ At w = v ⇤ A 1 w = µ̄v ⇤ w. Thus, either = µ or v ⇤ w = 0, i.e. eigenvectors of A corresponding to di↵erent eigenvalues are orthogonal. (d) If Av = v, taking the complex conjugate of both sides yields Av̄ = ¯ v̄, since A is real. Moreover, since 0 6= v = (v̄), v̄not = 0, and so v̄ is en eigenvector with eigenvalue ¯ . (e) A vector w 2 Cn actually lies in Rn i↵ w = w̄, and a quick computation shows that this is the case for v1 and v2 . The fact that v1 · v2 = 0 and ||v1 || = ||v2 || hold follows v · v̄ = 0 and ||v̄|| = ||v||. Finally, that the equations in (*) are satisfied is a straightforward calculation.

20

23. (Bonus) Suppose A is a real, 3 ⇥ 3 orthogonal matrix with det A = 1. Show that there is an orthonormal basis {v1 , v2 , v3 } of R3 and a real number ✓ 2 [0, 2⇡) such that Av1 = (cos ✓)v1 + (sin ✓)v2 , Av2 = (**) That is, if P = [ v1

(sin ✓)v1 + (cos ✓)v2 ,

Av3 = v3 . v2

v3 ], then 2

cos ✓ 1 4 P AP = sin ✓ 0

sin ✓ cos ✓ 0

3 0 05 1

Solution: See me for a solution.

24. (Bonus) Suppose A is a real, 3 ⇥ 3 orthogonal matrix with det A = 1. Show that there is an orthonormal basis {v1 , v2 , v3 } of R3 and a real number ✓ 2 [0, 2⇡) such that either Av1 = (cos ✓)v1 + (sin ✓)v2 ,

(i)

(that is, if P = [ v1

v2

Av1 = (cos ✓)v1 + (sin ✓)v2 ,

(ii)

(that is, if P = [ v1

Av2 = (sin ✓)v1 + (cos ✓)v2 , Av3 = v3 , 2 cos ✓ sin ✓ 1 v3 ], then P AP = 4 sin ✓ cos ✓ 0 0 Av2 = (sin ✓)v1 Av3 = v3 .

v2

(cos ✓)v2 ,

v3 ], then 2

cos ✓ 1 P AP = 4 sin ✓ 0

sin ✓ cos ✓ 0

3 0 0 5 .)

1

3 0 0 5) , or 1

21

Solution: See me for a solution.

25. (Bonus) Suppose A is a real, 4 ⇥ 4 orthogonal matrix with det A = 1. Show that there is an orthonormal basis {v1 , v2 , w1 , w2 } of R4 and two real numbers ✓, ' 2 [0, 2⇡) so that Av1 = (cos ✓)v1 + (sin ✓)v2 , Av2 =

(sin ✓)v1 + (cos ✓)v2 ,

Aw1 = (cos ')w1 + (sin ')w2 , (*)

Aw2 =

That is, if P = [ v1

v2

w1

(sin ')w1 + (cos ')w2 .

w2 ], then 2

cos ✓ 6 sin ✓ P 1 AP = 4 0 0

sin ✓ cos ✓ 0 0

0 0 cos ' sin '

3 0 0 7 5 sin ' cos '

Solution: Consider the eigenvalues of A. If 62 R is an eigenvalue, then so is ¯ , by the previous exercise. Thus the only possibilities are: Case 1: All the eigenvalues are real In this case, we work entirely in R4 . Since | | = 1 and the determinant is the product of the eigenvalues, the eigenvalues (with multiplicities) are {1, 1, 1, 1} or {1, 1, 1, 1}. subcase 1(i): {1, 1, 1, 1} Let E1 be the eigenspace corresponding to

= 1,

and set W = E1? , so that R4 = W E1 . We show that W = 0 as follows. Note that for all v 2 E1 , Av = v, so that A 1 v = v as well. Then, since A 1 is also orthogonal, if w 2 W , and v 2 E1 , we have (Aw) · v = (A 1 (Aw)) · (A 1 v) = w · v = 0, so Aw 2 W . Suppose that W 6= 0. If {w1 , · · · , wm , v1 , . . . , vk } is a basis of R4 such that B = {w1 , · · · , wm } is an orthonormal basis for W and {v1 , . . . , vk } is an orthonormal basis  0 for E1 , then the matrix of the linear A 0 transformation u 7! Au w.r.t. B is . Moreover, one can check that 0 Ik A0 is again an orthogonal (m ⇥ m) matrix, and that det(A xI4 ) = det(A0 xIm ) det(Ik xIk ) = det(A0 xIm )(1 x)k . Hence, det(A0 xIm ) = (1 x)m ,

22

and so 1 is an eigenvalue of A0 . In particular, there is an eigenvector in W of A0 , and hence of A, with eigenvalue 1 , contradicting W = E1? . Hence, W =0, so E1 = R4 , and thus A = I4 . We may then choose {v1 , v2 , w1 , w2 } to be any orthonormal basis of R4 and take ✓ = ' = 0 and we see that the equations (*) are satisfied. subcase 1(ii): {1, 1, 1, 1} Let E1 and E1 be the eigenspaces corresponding

to = 1, 1 and set W = (E1 E 1 )? , so that R4 = W E1 E 1 . Again, we show that W = 0. Note that if v 2 E1 or v 2 E 1 , Av = ±v, so that A 1 v = ±v as well. Then, since A 1 is also orthogonal, if w 2 W , and v 2 E±1 , we have (Aw) · v = (A 1 (Aw)) · (A 1 v) = ±w · v = 0, so Aw 2 W .

Supposing that dim W > 0, we then argue as in the previous subcase to show that A has an eigenvector in W with either 1 or 1 as eigenvalue , which is a contradiction. Hence, R4 = E1 E 1 , and since det A = 1, dim E1 = 2 = dim E 1 . Finally, take {v1 , v2 } to be any orthonormal basis of E1 , let {w1 , w2 } be any orthonormal basis of E 1 , and let ✓ = 0 and ' = ⇡.

Since E1 · E 1 = 0, by 7(c), {v1 , v2 , w1 , w2 } is an orthonormal basis of R4 , and the equations (*) are satisfied.

Case 2: Exactly 2 eigenvalues are real In this case we have 2 complex eigenvalues { , ¯ } and 2 real eigenvalues {±1, ±1}, so we will initially work in C4 . The condition det A = 1 means we again have only 2 subcases: subcase 2(i): {1, 1, , ¯ } Let E1 , E , and E ¯ be the eigenspaces correspond-

ing to the distinct eigenvalues. Set W = (E1 E E ¯ )? , so that R4 = W E1 E E ¯ . We can show that W = 0 exactly as before. Hence, 4 C = E1 E E ¯ . By 7, complex conjugation is an isomorphism from E to E ¯ , so dim E = dim E ¯ = 1, and dim E1 = 2 (complex dimensions).

Note that since A is real, the Gaussian algorithm applied to Av = v, followed

by Gram-Schmidt, yields an orthonormal basis {v1 , v2 } of E1 consisting of real vectors v1 , v2 2 R4 . w̄ w w̄ If {w} is a basis for E , then {w̄} is a basis for E ¯ , and {w1 = w+ 2 , w2 = 2i } is an orthogonal (because w · w̄ = 0) basis for a 2-dimensional subspace H of R4 , as both w1 , w2 2 R4 . The subspace H is orthogonal to E1 because H⇢E E ¯ , which is orthogonal to E1 , by 7.

Lastly, if we write = ei , a quick computation shows that Aw1 = (cos )w1 +

23

(sin )w2 and Aw2 = (sin )w1 + (cos )w2 . Another short calculation shows that ||w1 || = ||w2 ||, so we may normalize w1 and w2 to obtain an orthonormal basis {v1 , v2 , w1 , w2 } of R4 for which the equations (*) above are satisfied with ✓ = 0 and ' = . subcase 2(ii): { 1, 1, , ¯ } This subcase is handled in exactly the same way as subcase 2(i), except that here, ✓ = ⇡.

Case 3: No eigenvalues are real In this case, the eigenvalues (with multiplicities) are { , ¯ , µ, µ̄}. This case is dealt with just as for E E ¯ in subcase 2(i): Either = µ so that dim E = 2, in which case we take ✓ = ' so that = ei✓ , or 6= µ, when we take ✓, ' so that = ei✓ and µ = ei' .

26. Show that any two line segments AB and CD with non-zero lengths in R3 are similar.

Solution: Let g be a similarity with constant k > 0, and let f : Rn ! Rn be defined by f (x) = g(x) k . Then, as we saw in class, f is an isometry. So, if P, Q and n R2R ,

(g(Q) g(P )) · (g(R) g(P )) ||g(Q) g(P )|| ||g(R) g(P )|| ((kf (Q) kf (P )) · (kf (R) kf (P )) = (because g = kf ) ||kf (Q) kf (P )|| ||kf (R) kf (P )|| ((f (Q) f (P )) · (f (R) f (P )) = (because the factors of k2 cance ||f (Q) f (P )|| ||f (R) f (P )|| (Q P ) · (R P ) = (because f is an isometry) ||Q P || ||R P || = cos(\QP R).

cos(\g(Q)g(P )g(R)) =

Since both angles lie in [0, ⇡], this shows that \g(Q)g(P )g(R) = \QP R, as required.

Solution: First translate A to C with the isometry f1 . Then, choose a rotation f2 about C = f1 (A) so that f2 (f1 (B)) is on the line L through C and D, and is on the same side (inside L) of C as D. Then, f2 (f1 (A)) = C and f2 (f1 (B)) =

24

C + k(D C) for some k 2 R. Because f2 (f1 (B)) is on the same side (inside L) of C as D, and A 6= B and C 6= D we know that k > 0. Now consider the function g1 : Rn ! Rn defined by g1 (x) = k1 (x C) + C. Then, g1 is a similarity with g1 (f2 (f1 (A)) = g1 (C) = C and g1 (f2 (f1 (B))) = g1 (C + k(D C)) = D. Thus, g = g1 f2 f1 is a similarity (being a composition of similarities) with g(A) = C and g(B) = D. We know from class that there is E 2 O(n), b 2 Rn and l > 0 such that g(x) = lEx + b. (In fact, one can easily show that l = k1 , where k is the constant of the previous paragraph.)

If P 2 AB, then P = A + s(B A) for some s 2 [0, 1], so g(P ) = lE(A + s(B

A)) + b

= lEA + s(lEB

lEA) + b

= g(A) + s(g(B)

g(A))

= C + s(D

C) 2 CD.

Since AB = {A + s(B A) | s 2 [0, 1]} and CD = {C + s(D this shows that g(AB) = CD, as required.

C) | s 2 [0, 1]},

Solution: If either is the identity rotation, the first case holds, since the identity may be taken to be a rotation about any axis.

Each non-trivial rotation R of S2 determines and is determined by a pair of antipodal points P, P of S2 , and for points v 2 S2 , R(v) = v i↵ v 2 {P, P }.

So let P, P and Q, Q be determined by r and r0 respectively. Then rr0 = r0 r implies that r0 r(P ) = r0 (P ) = rr0 (P ), so r0 (P ) is fixed by r. Hence, by previous remarks, r0 (P ) 2 {P, P }.

If r0 (P ) = P , then P = Q or Q, so r and r0 will have the same axis.

If r0 (P ) = P , then P · Q = r0 (P ) · r0 (Q) = P · Q, so P · Q = 0 and hence the axes are perpendicular. Then, r0 (P ) = P also shows that r0 is a rotation by ⇡ about an axis perpendicular to that of r. One shows similarly that r(Q) = Q, and so r is also a rotation through ⇡ about an axis perpendicular to that of r0 .

(The latter case is possible, as is illustrated by r and r0 which 2 are rotations 3 by 1 0 0 ⇡ about the z- and x- axes respectively. Their matrices are 4 0 1 0 5 and 0 0 1

25

2

3 0 0 5 respectively. Since they are diagonal matrices, they commute.)

1 2 3 0 0 1 27. Let g : S2 ! S2 be defined by g(v) = Av where A = 4 1 0 0 5.

0 1 0 1 40 0

0 1 0

a) Show that g is an orientation reversing isometry of S2 . b) Find a great circle C, a point a 2 S2 and ✓ 2 [0, 2⇡) such that g is rotation by ✓ about a followed by reflection in C. c) Check your answer in (b) by an explicit computation.

Solution: a) The columns of A are clearly an orthonormal basis of R3 , so g is an isometry. A short calculation shows that det A = 1, so g reverses orientation. b) We first solve Av = v and obtain v = s(1, 1, 1) for s 2 R. So, we can take p a = 33 (1, 1, 1) and so C = {v 2 S2 | v · a = 0}. To find the angle ✓, we take p w = 22 (1, 1, 0) 2 C, and compute cos ✓ = w ·g(w) = 12 (1, 1, 0)·(0, 1, 1) = 12 .

1 Thus, ✓ = ⇡3 or 5⇡ 3 . To see which, we compute w⇥g(w) = 2 (1, 1, 0)⇥(0, 1, 1) = 1 5⇡ 2 ( 1, 1, 1), which is anti-parallel to a, so ✓ = 3 . p 2(v·a) 1 1 3 c) We know that RC (v) = v 3 a and that R✓,a (v) = 2 (v · a)a + 2 v 2 (a ⇥ v).

Hence,(where by Q. 5 we reverse the order for convenience of calculation ), R✓,a RC (v) = R✓,a (v

2(v · a)a)

p 1 1 3 = ((v 2(v · a)a) · a)a + (v 2(v · a)a) (a ⇥ (v 2 2 2 p 3 1 3 = (v · a)a + v (a ⇥ v) 2 2 2 1 1 1 = (v · (1, 1, 1))(1, 1, 1) + v ((1, 1, 1) ⇥ v) 2 2 2 = ( z, x, y) 2 32 3 0 0 1 x =4 1 0 0 54y 5, 0 1 0 z

2(v · a)a))

26

as required.

28. Suppose ⇢ and ⇢0 are rotations of S2 such that ⇢⇢0 = ⇢0 ⇢. Show that either i) ⇢ and ⇢0 are rotations about the same point, by any 2 angles, or, ii) ⇢ and ⇢0 are rotations, both by ⇡, about “perpendicular points”.

Solution: If either is the identity rotation, the first case holds, since the identity may be taken to be a rotation about any axis.

Each non-trivial rotation R of S2 determines a pair of antipodal points P, P of S2 , and for points v 2 S2 , R(v) = v i↵ v 2 {P, P }.

Clearly, if r and r0 have the same axis, they commute, since a computation shows that R',a R✓,a (v) = R'+✓,a (v) = R✓,a (v)R',a .

So let P, P and Q, Q be determined by r and r0 respectively. Then rr0 = r0 r implies that r0 r(P ) = r0 (P ) = rr0 (P ), so r0 (P ) is fixed by r. Hence, by previous remarks, r0 (P ) 2 {P, P }.

If r0 (P ) = P , then P = Q or Q, so r and r0 will have the same axis.

If r0 (P ) = P , then P · Q = r0 (P ) · r0 (Q) = P · Q, so P · Q = 0 and hence the axes are perpendicular. Then, r0 (P ) = P also shows that r0 is a rotation by ⇡ about an axis perpendicular to that of r. One shows similarly that r(Q) = Q, and so r is also a rotation through ⇡ about an axis perpendicular to that of r0 .

(The latter case is possible, as is illustrated by r and r0 which 2 are rotations 3 by 1 0 0 ⇡ about the z- and x- axes respectively. Their matrices are 4 0 1 0 5 and 0 0 1 2 3 1 0 0 40 1 0 5 respectively. Since they are diagonal matrices, they commute.)

0 0 1 29. Show that any two line segments AB and CD with non-zero lengths in R3 are similar.

D|| Solution: Let k = ||C ||A B|| . Consider the similarity defined by f (x) = kx, and define A0 = f (A), B 0 = f (B). Then, ||A0 B 0 || = k||A B|| = ||C D||. Now, by a previsou exercise, choose an isometry g satisfying g(A0 ) = C, g(B 0 ) = D. To finish, note h = gf is a similarity (being a composition of similarities) with h(A) = C and h(B) = D.

27

We know from class that there is E 2 O(n), b 2 Rn and l > 0 such that h(x) = lEx + b.

If P 2 AB, then P = A + s(B A) for some s 2 [0, 1], so h(P ) = lE(A + s(B

A)) + b

= lEA + s(lEB

lEA) + b

= h(A) + s(h(B)

h(A))

= C + s(D

C) 2 CD.

Since AB = {A + s(B A) | s 2 [0, 1]} and CD = {C + s(D this shows that h(AB) = CD, as required.

Recall from class that if of is defined as

C) | s 2 [0, 1]},

: [0, 1] ! R3 is a smooth curve, then the length l( ) l( ) :=

Z 1 0

|| 0(t)||dt.

Recall as well that if P, Q 2 S2 , the spherical distance d(P, Q) from P to Q is the length of the shortest (smooth) path on S2 which starts at P and ends at Q. 30. (Bonus) Suppose that f : R3 ! R3 is a Euclidean isometry, and that [0, 1] ! R3 is a smooth curve. i) Show that l( ) = l(f

:

).

Now suppose in addition that f (0) = 0. ii) Show that f (S2 ) = S2 . iii) Conclude that if P, Q 2 S2 , then d(P, Q) = d(f (P ), f (Q)). (You may not assume in this question that d(P, Q) = arccos(P · Q). This will be established in the next question, which will rely on this question!)

Solution: See me for a solution.

31. Suppose P, Q 2 S2 . Recall from class that if that if either P or Q is N , then d(P, Q) = arccos(P · Q). Now suppose neither P nor Q is N , and let H be the plane equidistant from P and N .

28

i) Show that the reflection RH in H fixes the origin, that and so conclude that H \ S2 is a great circle. ii) Show that d(P, Q) = d(N, RH Q). iii) Conclude that d(P, Q) = arccos(P · Q).

Solution: See me for a solution.

32. Given any 2 points P, Q 2 S2 with P 6= Q show that there is a great circle containing P and Q. Show that this great circle is unique unless P 2 {Q, Q}, and that in these cases there are infinitely many great circles containing P and Q. Solution: If P 6= Q and P 2 {Q, Q}, then P = Q. In this case, let V = {a 2 R | a · Q = 0} be the subspace perpendicular to Q. Because dim V = 2, there are infinitely many distinct lines L in V through the origin. If a1 , a2 2 V are non-zero direction vectors for 2 such distinct lines, the planes H1 = {w 2 R3 | w · a1 = 0} and H2 = {u 2 R3 | u · a2 = 0} through the origin perpendicular to them will also be distinct. Note that because a1 , a2 2 V , we have Q · a1 = 0 = Q · a2 and so P = Q and Q belong to both H1 and H2 . Hence, P and Q belong to infinitely many distinct great circles.

If P 62 {Q, Q} then {P, Q} ⇢ S2 is linearly independent: suppose aP + bQ = 0.

Taking dot products with P then Q yields a + bP · Q = 0 and aP · Q + b = 0(*).

Thus, a a(P · Q)2 = 0. If a 6= 0, this implies P · Q = ±1. As ||P || = ||Q|| = 1, this is only possible if P 2 {Q, Q}, a contradiction. Hence, a = 0, so by (*), b = 0.

Thus, there is a unique plane K through the origin containing P and Q, namely K = span{P, Q}. This shows that if P 62 {Q, Q}, then there is a unique great circle, K \ S2 , containing P and Q. 3

33. A pair of points of the form {P, P } is called antipodal.

a) Show that any two distinct great circles intersect in a pair of antipodal points.

b) Show that antipodal points remain antipodal after an isometry of S2 .

Solution: (a)Let C1 = H1 \ S2 and C2 = H1 2 \ S2 be two distinct great circles, where H1 and H2 are distinct planes through the origin. Then, C1 \ C2 = H1 \ H2 \ S2 . But H1 \ H2 is a line through the origin (parallel to the cross product

29

of their normals). So, for some direction vector a 6= 0 for this line, C1 \ C2 = H1 \ H2 \ S2 = {sa | 2 R} \ S2

= {sa | 2 R, ||sa|| = 1}

= {sa | 2 R, |s| ||a|| = 1} 1 = {sa | 2 R, |s| = } ||a|| 1 = {sa | 2 R, s = ± } ||a|| a a ={ , }, ||a|| ||a|| which is a pair of antipodal points, as required. (b) Let f : S2 ! S2 be an isometry of S2 . Then, we know that f (v) = Av for some orthogonal matrix A. If {P, P } is a pair of antipodal points, then f ({P, P }) = {f (P ), f ( P )} = {AP, A( P )} = {AP, (AP )}, which is clearly an antipodal pair.

2

0 2 2 34. Let g : S ! S be defined by g(v) = Av where A = 4 1 0 a) Show that g is an orientation reversing isometry of S2 .

0 0 1

3 1 0 5.

0

b) Find a great circle C, a point a 2 S2 and ✓ 2 [0, 2⇡) such that g is rotation by ✓ about a followed by reflection in C. c) Check your answer in (b) by an explicit computation.

Solution: a) The columns of A are clearly an orthonormal basis of R3 , so g is an isometry. A short calculation shows that det A = 1, so g reverses orientation. b) We first solve Av = v and obtain v = s(1, 1, 1) for s 2 R. So, we can take p a = 33 (1, 1, 1) and so C = {v 2 S2 | v · a = 0}. To find the angle ✓, we take p w = 22 (1, 1, 0) 2 C, and compute cos ✓ = w ·g(w) = 12 (1, 1, 0)·(0, 1, 1) = 12 .

30 1 Thus, ✓ = ⇡3 or 5⇡ 3 . To see which, we compute w⇥g(w) = 2 (1, 1, 0)⇥(0, 1, 1) = 1 5⇡ 2 ( 1, 1, 1), which is anti-parallel to a, so ✓ = 3 . p 2(v·a) 1 1 3 c) We know that RC (v) = v 3 a and that R✓,a (v) = 2 (v · a)a + 2 v 2 (a ⇥ v).

Hence,(where by Q. 5 we reverse the order for convenience of calculation ),

R✓,a RC (v) = R✓,a (v

as required.

2(v · a)a)

p 1 1 3 = ((v 2(v · a)a) · a)a + (v 2(v · a)a) (a ⇥ (v 2 2 2 p 3 1 3 = (v · a)a + v (a ⇥ v) 2 2 2 1 1 1 = (v · (1, 1, 1))(1, 1, 1) + v ((1, 1, 1) ⇥ v) 2 2 2 = ( z, x, y) 2 32 3 0 0 1 x =4 1 0 0 54y 5, 0 1 0 z

2(v · a)a))

35. Let a 2 S2 , 0 6= b 2 R and H = {v 2 R3 | a · v = b}. Show that there is a non-zero point h0 2 H such that H = {v 2 R3 | h0 · v = ||h0 ||2 }. Show that h0 is the closest point on H to the origin.

Solution: Let. h0 = ba. Thus, h0 · v = b(a · v) and h0 · h0 = b2 , so H = {v 2 R3 | h0 · v = ||h0 ||2 } is clear. Now, if v 2 H, then ||v||2 = ||v h0 +h0 ||2 = ||v h0 ||2 2(v h0 )·h0 +||h0 ||2 = ||v h0 ||2 +||h0 ||2

||h0 ||2 ,

so h0 is the closest point on H to the origin.

(Note that if b = 0, H = {v 2 R3 | ||v

a|| = ||v + a||}.)

36. A subset C ⇢ R3 is a circle if there is a plane K, a point k 2 K and 0 < r 2 R such that C = K \ {v 2 R3 | ||v k|| = r}. a) Show that the intersection of any plane K with S2 is either empty, consists of a single point, or is a circle in R3 . (Hint: use the previous question.)

31

b) Show that if C = K \ {v 2 R3 | ||v k|| = r} is a circle, then for any normal a to K, C = K \ {v 2 R3 | ||v (k + a)||2 = r2 + ||a||2 }.

Solution: If K passes through the origin, K \ S2 = {v 2 K | ||v 0|| = 1}, so we take k = 0 and r = 1 to see that K \ S2 is a circle. Now suppose K does not pass through the origin, and that K \ S2 6= ;.

By Q3., there is k0 2 K such that K = {v 2 R3 | k0 · v = ||k0 ||2 }. Suppose v 2 S2 \ K. Then, ||v

k0 ||2 = 1

2k0 .v + ||k0 ||2 = 1

||k0 ||2 .

If ||k0 || = 1, then v = k0 , so S2 \ K = {k0 } consists of a single point. Otherwise, r2 = 1 ||k0 ||2 > 0 and K \ S2 is the circle K \ {v 2 R3 | ||v k0 || = r}. (b) If v, k 2 K and a is normal to K, then ||v k a||2 = ||v k||2 2(v k) · a + ||a||2 . Hence, ||v k a||2 = r2 + ||a||2 () ||v k||2 = r2 , i.e .

37. Let K be a plane, k 2 K and r > 0 be given, and suppose the circle C = K \ {v 2 R3 | ||v k|| = r} lies in S2 . Show that C = K \ S2 .

Solution: Since C ⇢ K and C ⇢ S2 , then C ⇢ K \ S2 is clear. It remains to show that K \ S2 ⇢ C. By problem 34, we can find k0 6= 0 such that either k0 2 K and K = {v 2 R3 | k0 · v = ||k0 ||2 } or K = {v 2 R3 | k0 · v = 0} (and k0 2 / K). In the first case, we know that k0 2 K is the closest point on K to 0.

Case 1: k0 2 K. We shall first show that k = k0 . Assume not and denote ||k0 k|| = ⇢ > 0, and define r v1 = k + (k0 ⇢

r k) = k0 + (1 + )(k ⇢

k0 )

r )(k ⇢

k0 )

and v2 = k

r (k0 ⇢

k) = k0 + (1

(Draw yourslf a picture: v1 and v2 are the points of intersection of the line through k and k0 with the circle C.)

32

Since k, k0 2 K, and K is a plane, vi 2 K, for i = 1, 2. Moreover, the first expressions for v1 and v2 show that ||vi k|| = r, for i = 1, 2, so that vi 2 K \ {v 2 R3 | ||v k|| = r} ⇢ S2 , by the hypotheses of the problem. Hence, ||v1 ||2 = ||v2 ||2 since both are equal to 1.

Now, recalling that k 2 K () k · k0 = ||k0 ||2 () k0 · (k k0 ) = 0, the second expressions for the vi (and the Pythagorean theorem) show that r ||v1 ||2 = ||k0 ||2 + (1 + )2 ⇢2 ⇢ and ||v2 ||2 = ||k0 ||2 + (1

r 2 2 ) ⇢ ⇢

But ||v1 ||2 = ||v2 ||2 () (1 + ⇢r )2 ⇢2 = (1 ⇢r )2 ⇢2 i↵ ⇢ = 0 or (1 + ⇢r )2 = (1 ⇢r )2 () ⇢r = 0, which is impossible, since r > 0. Hence, ⇢ = 0, and so k = k0 .

Now that we know k = k0 , we next need to show that 1 = r2 +||k0 ||2 . So suppose v 2 C. Then 1 = ||v||2 = ||v k0 +k0 ||2 = ||v k0 ||2 +||k0 ||2 (since (v k0 )· k0 = 0), so 1 = r2 + ||k0 ||2 .

Now suppose that v 2 K \ S2 . We need to show that v 2 C. Since v 2 K, it su↵uces to show that ||v k0 ||2 = r2 . But, arguing as before, ||v k0 ||2 = ||v||2 2v · k0 + ||k0 ||2 = 1 + ||k0 ||2 = r2 , as required.

Case 2: k0 2 / K. Here we show that k = 0 and r = 1. This is handled in the same way as case 1.

Assume k 6= 0 and denote ||k|| = ⇢ > 0, and define r v1 = (1 + )k ⇢ and v2 = (1

r )k ⇢

(Draw yourslf a picture: v1 and v2 are the points of intersection of the line through k and 0 with the circle C.) As before, vi 2 C and so 1 = ||v1 || = ||v2 ||, which is impossible unless ⇢ = 0.

To see that r = 1, suppose v 2 C. Then 1 = ||v||2 = ||v k||2 = r2 .

33

That v 2 K \ S2 implies v 2 C is now clear.

Solution: We show that f1 f2 (P ) = f2 f1 (P ) for all P 2 S2 . First note that if c 2 C, then f1 (c) 2 C and f2 (c) = c, while f1 (±a) = ±a, and f2 (±a) = ⌥a.

Moreover, both f1 and f2 are of the form fi (v) = Ai (v) for some Ai 2 O(3), so that both are linear functions.

So let P 2 S2 . Then, we may write P = (P · a)a +(P (P · a)a), where the vector P 0 = (P (P · a)a) is perpendicular to a. If P 0 = 0, then P = (P · a)a, so that (P ·a)2 = 1, in which case P = a or P = a. In that case, f1 f2 (±a) = f1 (⌥a) = ⌥a, while f2 f1 (±a) = f2 (±a) = ⌥a, so f1 f2 (P ) = f2 f1 (P ) in this case.

P0 Now we may assume that P 0 6= 0, so ||P 0 || 2 C. Hence, we can write P = sa + tc 0

P 0 where s = P · a , c = ||P 0 || 2 C and t = ||P ||. Then,

f1 f2 (P ) = f1 f2 (sa + tc) = f1 (sf2 (a) + tf2 (c)) = f1 ( sa + tc) =

sa + tf1 (c).

On the other hand, f2 f1 (P ) = f2 f1 (sa + tc) = f2 (sf1 (a) + tf1 (c)) = f2 (sa + tf1 (c)) = sf2 (a) + tf2 f1 (c) =

sa + tf1 (c)

= f1 f2 (P ),

(because f1 (c) 2 C)

as required.

2

0 2 2 4 38. Let g : S ! S be defined by g(v) = Av where A = 1 0 a) Show that g is an orientation reversing isometry of S2 .

0 0 1

3 1 0 5.

0

34

b) Find a great circle C, a point a 2 S2 and ✓ 2 [0, 2⇡) such that g is rotation by ✓ about a followed by reflection in C. c) Check your answer in (b) by an explicit computation.

Solution: a) The columns of A are clearly an orthonormal basis of R3 , so g is an isometry. A short calculation shows that det A = 1, so g reverses orientation. b) Since we are told implicitly that g is a spin reflection, we begin bypsolving Av = v and obtain v = s(1, 1, 1) for s 2 R. Hence, we can take a = 33 (1, 1, 1) and so C = {v 2 S2 | v · a = 0} = {(x, y, z) 2 S2 | x + y + z = 0}.

c)

p To find the angle ✓, we take w = 22 (1, 1, 0) 2 C, and compute cos ✓ = w·g(w) = 1 1 ⇡ 5⇡ 2 (1, 1, 0) · (0, 1, 1) = 2 . Thus, ✓ = 3 or 3 . To see which, we compute w ⇥ g(w) = 12 (1, 1, 0) ⇥ (0, 1, 1) = 12 ( 1, 1, 1), which is anti-parallel to a, so ✓ = 5⇡ 3 . p 2(v·a) 1 1 3 We know that RC (v) = v 3 a and that ⇢✓,a (v) = 2 (v · a)a + 2 v 2 (a ⇥ v).

Hence, (where by a previous result we may reverse the order for convenience of calculation),

⇢✓,a RC (v) = ⇢✓,a (v

as required.

2(v · a)a)

p 1 1 3 = ((v 2(v · a)a) · a)a + (v 2(v · a)a) (a ⇥ (v 2 2 2 p 3 1 3 = (v · a)a + v (a ⇥ v) 2 2 2 1 1 1 = (v · (1, 1, 1))(1, 1, 1) + v ((1, 1, 1) ⇥ v) 2 2 2 = ( z, x, y) 2 32 3 0 0 1 x =4 1 0 0 54y 5, 0 1 0 z

2(v · a)a))

35

2

0 0 2 2 4 39. Let f : S ! S be defined by f (v) = Bv where B = 1 0 0 1 a) Show that f is an orientation preserving isometry of S2 .

3 1 0 5.

0

b) Find a point b 2 S2 and ' 2 [0, 2⇡) such that f is rotation by ' about b. c) Check your answer in (b) by an explicit computation.

Solution: a) The columns of B are clearly an orthonormal basis of R3 , so f is an isometry.

A short calculation shows that det B = 1, so f preserves orientation. b) We first solve Bv = v and obtain v = s( 1, 1, 1) for s 2 R. So, we can take b = p p 3 2 3 ( 1, 1, 1). To find the angle ', we take w = 2 (1, 1, 0), which is perpendicular to b and compute cos ' = w · f (w) = 12 (1, 1, 0) · (0, 1, 1) = 12 . Thus, ' = 2⇡ 3 or 4⇡ 1 1 3 . To see which, we compute w ⇥ f (w) = 2 (1, 1, 0) ⇥ (0, 1, 1) = 2 (1, 1, 1), which again is anti-parallel to b, so ✓ = 4⇡ 3p.

3 c) We know that ⇢',b (v) = 32 (v · b)b 12 v 2 (b ⇥ v). Hence,

as required.

p 3 1 3 ⇢',b (v) = (v · b)b v (b ⇥ v) 2 2 2 1 1 = (v · ( 1, 1, 1))( 1, 1, 1) v 2 2 = ( z, x, y) 2 32 3 0 0 1 x = 4 1 0 0 54y 5, 0 1 0 z

1 (( 1, 1, 1) ⇥ v) 2

40. Suppose A 6= B 2 S2 , and let : [0, 1] ! S2 be the geodesic path from A to B. Show that if u = 0 (0), then u · B 0. (Hint: first reduce to the case where A = N = (1, 0, 0), and then use the formula we found in class for .)

Use only the cosine rule for the next two questions. (Final answers to 4 significant figures only.)

36

41. Solve the spherical triangle ABC if a) A = 75 , b = 80 , and c = 85 . b) a = 75 , b = 80 , and c = 85 .

4⇡ Solution: (a) We have the data: A = 75 = 5⇡ 12 , b = 80 = 9 , and c = 85 = 17⇡ 36 . Then, by the cosine rule (for sides)

a = arccos[(cos b cos c + sin b sin c cos A)], and

B = arccos[(cos b

cos a cos c)/(sin a sin c)],

C = arccos[(cos c

cos a cos b)/(sin a sin b)].

These yield a b c A B C

(radians) (degrees) (degrees, minutes) ⇡ 1.298 ⇡ 74.39 ⇡ 74 240 4⇡ = = 80 = 80 00 9 17 ⇡ = = 85 ⇡ 85 00 36 5⇡ = = 75 ⇡ 75 240 12 ⇡ 1.414 ⇡ 80.99 ⇡ 81 00 ⇡ 1.528 ⇡ 87.56 ⇡ 87 330

4⇡ (b) We have the data: a = 60 = ⇡3 , b = 70 = 7⇡ 18 , and c = 80 = 9 . Then, by the cosine rule (for sides)

and

A = arccos[(cos a

cos b cos c)/ sin b sin c],

B = arccos[(cos b

cos a cos c)/ sin a sin c],

C = arccos[(cos c

cos a cos b)/ sin a sin b].

These yield a b c A B C

= = = ⇡ ⇡ ⇡

(radians) 5⇡ 12 4⇡ 9 17 ⇡ 36

1.320 1.413 1.5286

= = = ⇡ ⇡ ⇡

(degrees) 75 80 85 75.62 80.97 87.46

= = = ⇡ ⇡ ⇡

(degrees, minutes) 75 00 80 00 85 00 75 370 80 580 87 270

37

42. A plane flies from Ottawa (45 240 N 75 430 W ) along a great circle to Bejing (39 550 N 116 250 E). a) Find its heading as it leaves Ottawa and its heading as it arrives in Bejing. b) If the radius of the earth is 6378km, and the plane averages 1000km per hour, how long will the trip take? (to the nearest minute.) c) What is the distance of closest approach to the north pole during the flight? (to the nearest kilometre.)

Solution: The latitude and longitude of Ottawa and Bejing are Latitude 45 240 N 39 550 N

Ottawa Bejing

Longitude 75 430 W 116 250 E

Consider a spherical triangle with vertices at N (the north pole), Ott, Ottawa and B, Bejing. As usual, we use small letters to indicate the arc (interior) angles and large letters to denote vertex angles (all of which are at most ⇡ = 180 .). Then, we know that N = 360

✓(B) + ✓(Ott) =

1259 ⇡ (radians) = 167 520 , 1350 223 ⇡ (radians) = 44 360 , 900

b = (ott) = 90

45 240 =

ott = (B) = 90

39 550 =

and

601⇡ (radians) = 50 50 .

260

Then, by the cosine rule (for sides) n = arccos[(cos b cos ott + sin b sin ott cos N )], B = arccos[(cos b

cos n cos ott)/ sin n sin ott],

and Ott = arccos[(cos ott

cos n cos b)/ sin n sin b].

38

These yield (radians) (degrees) (degrees, minutes) n ⇡ 1.640 ⇡ 93.99 ⇡ 94 00 233⇡ b = ⇡ 44.6 ⇡ 44 360 900 601 ⇡ ott = ⇡ 50.08 ⇡ 50 50 2160 1259 ⇡ N = ⇡ 167.9 ⇡ 167 520 1350 B ⇡ 1.485 ⇡ 8.508 ⇡ 8 300 Ott ⇡ 1.623 ⇡ 9.300 ⇡ 9 180 a) Thus, the initial heading of the plane will be N (Ott)W = N 9 180 W , and the final heading as it approaches Bejing will be S(B)W = S 8 300 W . b) Assuming the earth is a sphere of radius 6378km, the distance from Ottawa to Bejing is n ⇤ 6378 ⇡ 10460km. Therefore, at 1000 kmph, the journey will take approximately 10.47 hours, or 10 hours and 28 minutes. c) If X is the point of closest approach to N , the vertex angle X̂ = \N XO = ⇡/2.

So, consider the spherical triangle with vertices Ott, N and X. To distinguish the interior angles in this new triangle from those in the previous one, we will denote the interior angles \X0Ott by nn and \X0N by oo. Draw a diagram and you will see that x = b (the same as in part (a)), so in OttN X, we know x, X and Ott. We wish to find oo in this case.

To unemcumber the calculations, we use the following notation: cos oo = , cos x = ↵, cos nn = , sin oo =

0

Then, since cos X = 0, the cosine rules for (1) and

, sin x = ↵0 , sin nn = ., cos Ott = k OttN X yield

= ↵ + ↵0 0 k ↵=

.

We know a and k, so we solve these equations for . Thus, in (1) yields s 2 p ↵ ↵2 2 = +k 1 ↵ 1 , 2

= ↵ and substitution

39

so 2

or 2

Either

so

2

p =↵ +k 1 2

p ↵ =k 1 2

↵2 ↵2

p p

2

↵2 ,

2

↵2 .

= ↵, which is impossible (since Ott < ⇡/2), or p p 2 2 ↵ = k 1 ↵2 ) 2 ↵2 = k2 (1

= ↵2 + k2 (1

↵2 ),

↵2 ) = ↵2 (1

k2 ) + k2 . Hence p oo = arccos( ) = arccos( ↵2 (1 k2 ) + k2 ) ⇡ .113714.

Thus, the distance of closest approach is oo ⇤ 6387 ⇡ 725.5km.

43. Let H be the plane {(x, y, z) | z = 1} and suppose f : S2 = {(x, y, z) 2 S2 | z < 0} ! H Prove that central projection does not always preserve angles.

Solution: Suppose f did preserve angles: Let ABC be a spherical triangle of positive area in S2 . Since f takes arcs of great circles to straight segments in H, the image of ABC is a Euclidean triangle f (A)f (B)f (C) in H. But we know that the sum of the angles in ABC exceeds ⇡ (by its area), while the sum of the angles in f (A)f (B)f (C) is exactly ⇡. Hence, f cannot preserve angles.

For example, for any point (0, 0, 1) 6= P 2 S2 , consider the line of longitude and the line of latitude that intersect at an angle of ⇡/2 at P . The images of these curves under f are respectively, a line through the origin and a circle centred on the origin, which will also intersect at f (P ) at an angle of ⇡/2. However, all other angles at the same point will not preserved - see ? below .

Moreover the central projection preserves all angles at (0, 0, 1): Any angle on the sphere there between 2 great circles C1 = H1 \ S2 , C1 = H1 S2 will be mapped to the (same) angle between the lines L1 = H1 \ H 1 , L2 = H2 \ H 1 in H 1 . (? Here’s an example: let K1 be the plane with equation x + y + z = 0, and K2 be the plane with equation y = 0. Then if C1 = K1 \ S2p and C2 = K2 \ S2 , 1) p then the acute angle between C1 and C2 at (1,0, is arccos( 33 ). However, f (C1 ) 2 is the line with equations x + y = 1, z = 1, while f (C2 ) is the line with equations p ⇡ y = 0, z = 1. Thus, the acute angle between f (C1 ) and f (C2 ) is 4 6= arccos( 33 ) ⇡ 0.304087⇡.)

40

44. Show that the images under the stereographic projection of great circles on S2 through N = (0, 0, 1) are lines, and describe these lines.

Solution: Suppose that K is a plane through the origin and N , so that C = K \ S2 is a great circle. If P 2 C {N }, then of course P 2 K. As K is a plane ! ! containing both N and P , the ray N P also lies in K. But f (P ) 2 N P \ H ⇢ K \ H = L, which is a line. Note that as N, 0 2 K, then N 2 K \ S2 , and f ( N ) = 0. Hence, f (C {N }) ⇢ L, and L ⇢ H contains the origin. To see that ! f (C {N }) = L, pick Q 2 L = H \ K. As before, Q, N 2 K ) QN ⇢ K, so ! f 1 (Q) 2 QN \ (S2 {N }) ⇢ K \ (S2 {N }) = C {N }. Hence, f (C {N }) = L. We now show that any line through the origin is the image of a line of longitude.

Suppose L ⇢ H is a line through the origin, let K be the plane through N which contains L. Note that as 0 2 L, K is a plane through the origin. Then, arguments as above show that f (K {N }) = L. Hence, the images under the stereographic projection f of great circles on S2 through N = (0, 0, 1) are lines through the origin, and in fact all such lines can be obtained in this way.