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/Introduction to Geometry (20222), Coursework, Spring 2012

Solutions1

a) Let (Xl, .r2, x3 ) be coordinates of the vector x, and ( : v I , y 2 , : v 3 ) be coordinates of the vector y u:R3.Does the [ormula (x. y) = Xl yl + x2y2 + x2y3 + x3y2 + X3y3 define a scalar product on R3? Justify your

answer.Show that this formula does not define a scalar product: (x, y) = xlyl + (x 2 + X3 ) ( y 2 + y 3 ) . Hence for

non-zero vector x =(0,1,-1) (x,x) =O. This contradicts to the condition of positive definiteness .b) Let x,y be two vectors in the Euchdean space E2 such that the length of the vector x is equal to 1,

the length of the vector y is equal to 29 and scalar product of these vectors is equal to 21.Find a vector e in E2 [express it ihrouqh. the vectors x and y) such that the following corulitions hold

i) an ordered pair {e, x} is an orthonormal basis m E2,ii) the vector e has an obtuse angle unih. the uecior y.

Notice that scalar product (y.y) = 292 and (x,y) = 2l. Let e = ax+by. Since {e,x} is orthonormalbaSIS, we have that (e,x) = and (e,e) = 1.

(e,x) = 0 = (ax + by, x) = a(x, x) + b(x, y) =a + 21b. Hence a = -2lb. On the other hand(e,e) = 1 = (ax+by,ax+by) = a2(x,x) +2ab(x,y) +b2(y,y) = a2 +42ab+292b2 = 1. Since a = -21b wecome to the equation 1 = a2 + 42ab + 292b2 = (212 - 42 . 21 + 292)b 2 = 1. Note that (212 - 42 . 21 + 292) =(212 - 2.212 + 292) =(292 - 212) = 8 50 a 2 =400. Hence 400b 2 =1 and b =~ To choose a sign notethat (e,y) = (ax + by, y ) < 0 since the angle is obtuse. We come to (e, y) = (ax + b y, y) = 21a + 292b =(292 - 212)b = 400b < O. Hence b =-f a and a = g. Finally we have that e =ax - b y = 21;~Y. II

. . ( cos o - sin e )(c) Consider the matrix A = . e eSIn cosCalculate the matrix A 18 in the case ~fe = ~.Find all 2 x 2 orthoqotial matrices A such that 2A 3 = (~ ~A) .

(COS e - sin ()) . . . n ( cos nB - sin n e )A = . iJ () IS rotation matrix. A . = . () {Jsin {) cos . sin n cos nil

If ()= 1'. then A 2 = (cos 2 ~ - sin 2 % ) =(cos 7i - sin 7 f ) = ( - 1 ~1 ) .2 81112 ~ cos 2} sintt cos tt If e = .".then A 18 = (cos 18~ - sin 18~) = (cos 3 7 f - sin 3 7 f ) = (-01 ~1 ) .6 S1l118~ cos 18~ sinSn cosSn

(COS 38 - sin 3 f! ) .. 3 _ ( 4We have that A3 = sin 3e cos 3e . On the other hand.A - ~

Calculate the mairi A2 in the case if e = ~ .

1 )' We have1 1 .3 = (cos 3f !. sin 38 -Sin3e) = ( V ; .cos 38 12 -~) (cos.".- 6l' - sin ~ - sin i )cos %

We see that 38 = ~ + 21fk, and e = * + 2~k. Hence we have three solutions:l.A = (cos j -sin7f~) (for () = * + 27fm, m = 0, 1 2, ... ),2. A = ( : ~ : ! i ~o.:i~7Y) (for e = % + 2311"+ 27Tm , m = O.l 2, .. ),sin9 cosg" .

(COS 13K _ sin 1 31 1" )and 3. A = 'lf1 1 " 1 3~ (for e = * + 43"1f + 2 7 f m , m =0, l 2, ... ) ..sin -9- cosg- III

2a) Consider vector a =2e x + 3ey + Ge , in E3.Show that the angle e between vectors a and ez belongs to the inierual (i, ~).

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Find a unit vector b such that it is orthogonal to vectors a and eZ! and the angle between ueciors bandex 7 :S acute. Show that the ordered triple {a, b, ez} is a basis and this basis has orientation opposite to theorientation of the basis {ex, ey, e;}.Solution. Using the formula for scalar product calculate

cose ' = (a,g) = (a,g)lallgl j(a,a)

6 6-V4 + 9 + 36 7The vector a is in the positive octant. Hence 0 $ e $~,We have that

'IT J2 6 J3 tt 1 ( 6 ) 2 36 3 tt t;COS - = - < - < - = cos - because - < '" = - < - =? - < e < - ,4 2 7 2 6 2 { 49 4 6 4, Consider vector x = a x g = (2e + 3f + 6g) x g = -2f + Se. Following properties of vector product thisvector is orthogonal to vectors a and g. The length of this vector equals to V4 + 9=vII Hence the vectorb = J h - is the unit vector which is orthogonal to vectors a and ez,

Now we have to fix b = Jh or b = - J h ' Following the last axiom of vect.or product the basis {a, g, a xg} ={a,g,x} has the same orientation as the basis {e.fvg}. Note that the transformation {a,g,x} -t{a, x, g} changes orientation (swapping vectors) and the transformation {a, x, g} --t {a, - J h - , g} changesorientation too (mulitplicatin of vector on negative number). Hence their composition, the transformation{a, g, x} --t {a, - J h - , g} does not change orientation. Hence the basis {a, ~ J h ) g} as well as the basis{a,g,x} has the same orientation as the basis {e,f,g}, 1 , 7 I l e see that b = - Jh = 2~e,

Scalar product of two units vectors band e equals jh-, The angle between these vectors equalsf) = arccos \ )i 3 ' II

b) In oriented Euclidean spaceE3 consider the following [unction of three vectors: F(X, Y, Z)(X, Y x Z) , where (, ) is the scalar product and Y x Z is the vector product zn E3

Show that F(X, X, Z) = 0 for arbitrary vectors X and Z,Deduce that F(X, Y, Z) = -F(Y, X, Z) for arbitmry vectors X, Y, Z,What is the geometrical meaning of the function F?F(X, X, Z) = (X, X x Z) = 0 since the vector X is orthogonal to the vector X x Z,Using the fact that F(X, X, Z) = 0 for arbitrary vectors X, Z and linearity of the function F(X, X, Z)

with respect to the vectors X, Y and Z we have: 0 = F(X + Y, X + Y, Z) = F(X, X, Z) + F(X, Y, Z) +F(Y,X, Z)+F(Y, Y, Z) = F(X,X, Z)+F(X, Y, Z)+F(Y, X, Z)+F(Y, Y, Z) = F(X, Y, Z)+F(Y, X, Z) =0, i.e. F(X, Y, Z) = -F(X, Y, Z),

Consider parallelepiped formed by vectors X, Y, Z, The length of the vector Y x Z is equal to area ofparallelogram spanned by these vectors, The vector Y x Z is orthogonal to the plane of this parallelogram,The scala.r product (X, Y x Z) equals to length of the vector Y x Z (= area of parallelogram) multiplied onthe length of the vector X and the cosinus of the angle between these vectors i.e. height of the parallelepipedmulitplied on the area of the base =voluem of the parallelepiped, II

c) Give an example of at least one paraLlelogram in the plane such that ~ts all vertices have inteqercoordinates, its area is equal to 1 and one of the sides has length greater than 2012,

Describe all parallelograms ABCD in the plane which obey the following conditions: their area is equalto 1 and all their ueriices have integer coordinates such that vertex A is Q ,t the origin and vertex B hascoordinates (2,3),

There are plenty examples, Kg, consider parallelogram ABCD such that A =(0,0), B =(1 ,0) ,C =(M, 1) and D = (/1.1+ 1,1) This paralleogram evidently has area 1 and its edges Be and AD havelength V I +M2 If we take M enough big (e.g. M = 2012) then the length of these edges will be biggerthan 2012,

Sure one can consider many other examples using the fact that area if parallelogram formed by twovectors is equal to to the the modulus of the determinant formed by the vectors, Kg. consider an arbitrary

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parallelogram with vertices in integer coordinates. Without loss of generality suppose that one of vertices isat the origin. Let other vertices be at points (p , q ), (m , 1 1 , ) and (m + p , q + n), The area of this parallelogramis equal to S =Iclet (:~ ~) I =p 'l l, - qml In particular if one of vertices is at the point 2,3 then we cometo the equation

Idet (~. ~.) I = 12n - 3ml = 1Find all integer pairs obeying this condition.

1. 2n - 3m = 1. Hence n =1+ ; m . . Since n is integer m has to be odd. Hence we come to the pairs(m,n) such that 771 = 2 k + 1 , 1 1 , = k + 2, where k = 0,l, 2, ....2. 2n - 3m = -1. Hence ti = -1t3m. Since n is integer m has to be odd. We come to the pairs (m, n)such that 171 = 2 k + 1 , 1 1 ,= 3 k + 1 , where k : = 0 , l , 2,. . . .

3a) Let P be a linear orthogonal operator acting 'in E3 .. such that it pr'eserves th e orietuoiior: of E3 andthe foUowing r e l a t i o n s hold: P ( e) =cos e e + sin ef, P ( g ) = Eg, where 6 1 i s an arbitrary angle and e = 1.

Write down the matrix of operator P in the basis {e, f, g}. (You have to consider both cases e = 1 ande=-1.)

Let (~~; ~ D boa matrix of operator P in the basis {e, f , g) .The condition that P orthogonal operator means that all rows of this matrix have length 1 and they are

orthogonal to each other. V>lecome to z =, X ' cos 1 1+ y sin e = and x2 + y2 +z2 = 1. Solving these equationswe come to two solutions: x = - sin e , y =cos 6 1 , Z = 0 or x = sin 6 1 , y =-cos e , Z = O . Determinant of thematrix P equals to in the first case and it is equa.l to -E in the second case. Since P preserves orientationhence

( cos e ' - sine 0 ) ( " " - sine n= sJ~e cos 6 1 o = sin e cosB P(f) = -sinBe+cosBf.0 t: 0 0in the case if E =1and

( ' 0 " sinB ! ) C O " sine :) ,= si~e - cos 6 1 Si~B - cos 6 1 P(f) = sin Be - cos tif .0 0 -1in the case if E = -1. II

b) We know that due to the Euler Theorem l inear operator P considered abov e 1 ,5 rotation operator.Find the axis of this rotation.(You. have to consider both cases t:=1 and e = -1.)Consider first the case [ =1. In this case P is given by the rela.tion (*). Due to the Euler Theorem

to find the axis of rotation we have to find eigenvector with eigenvalue 1. This is the case for vector g:P g = g = g. The axis is directed along the vector g.

Second case. Let N =xe + yf + zg i' be eigenvector with eigenvalue 1: P(N) = N, i.e.P(xe + yf + zg) = zi(cos e'e + sin ef ) + y(sinee - cos ef) = (xcose + ysin8)e + (xsine - ycose)f - zg.

{X(l-COSe)=YSine . "\""e come to simultaneous equations: y(l + cos ti) = x sin fj. Vlfesee that O f = 1SIn e D = 1+.C005e = cos ~ ., y - cos o SIn . s~n2'z=OHence solutions of these equations are x = A cos ~,y = A sin ~,z = 0, where A is an arbitrary constant.

Respectively eigenvector with eigenvalue 1 is equal to n =cos ~e + sin ~g or any non-zero vector Xn directeda.long this vector. Axis of rotation is the line in the plane e,-f which- form the angle ~ with axis directedalong e. III

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c) Let P be a lineor operator acting in E3! such that P(e) = f, P(f) =g- and P(g) = e. Show that Pis a rotation operator.Find the ans and the angle of the rotation.Operator P sends orthonormal basis {e,f,g} to the new basis {f,g,e}. This new basis is orthonormal

and it has the same orientation as initial basis. (We have 2 swappings {e,f,g} -t {f,e,g} - - - + {f,g,e}).Hence P isan orthogonal operator and it preserves orientation. Due to Euler Theorem it isrotation operator.The vector N = = e + f + g is eigenvector with eigenvalue 1:

P(N) = = P( e + f + g) = f + g = = e = N.Hence axis of rotation is directed along the vector N= e + f + g.

To find an angle of rotation note that if P is operator of rotation on the angle e , then P" rotates onthe angle ni). On the other hand it is evident that p 3 is identity operator (p3(e) = p2( f ) = P(g) = e, thesame for vectors f and g.) , We see that 3e = 2", e = 23". IIAnother solution: Consider an arbitrary vector x = ae + bf + cg and vector y=P(x). These bothvectors have the same length. Let the angle between these vectors be equal to e . We have that

(x.y ) = (x,Px) = (ae+bf+cg,af+bg + ce) = a c+ ab+ bc= Ixl2cosB = (a 2 +b 2 + c2)cose.Hence cos() = ~itbt~~i= !~~!t;~~:1). In the case if vector x is orthogonal to the axis (x, N)(x,e+f+g)=a+b+c=Oandcose=-~,i.e. e = 2 ; .

'Ne see that angle of rotation for an arbitrary vector is less or equal to 2 ; and it is equal to 2 ; forvectors orthogonal to axis.II

Another solution: Consider plane orthogonal to the axis Nand take an arbitrary orthonormal basis{x,y} in this plane preserving orientation. E.g. we may take x = 2e~-g, y = ~. One can see bystr aightforwar calculations that this basis rotates on the angle B = 23 7 [: P(x) = -~x + V ; y and P(y) =v' 3 1-2 - 2x.

4a) Given a vector field G =a1";" + b&8

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b) Let C be the upper' half of the circle with centre at the pouii (R,O) which is tangent to the y-axis.Write down an equation of this circle. Choose any parameterisation of this curve and calculate the integralfe w if i) w = x2dy and 1 : 1 ) a = x2dy + 2xydx.

Does answers depend on the chosen pammeterisation?Explain why I-form LV = x2y is not a.n exact form.The curve C may be defined by parametrisation r(t): { X ( ( t t ) ) = RR+ Rtcos ': for 0 < t ::; "if. Velocity vector. y . = Slll

v(t) =(-R~~~:~t) = -Rsint8x + Rcost8y. We have that w(v(t)) = x2(t)dy(-Rsin tax + R cos tay) =x2 (i)R cos t = (R + R cos t)2 R cos t. Hence

( W= r x2dy= r(w(v(t))dt= f " (R+Rcos t )2Rcos td t=7fR3.Jc)c)o ) 0l-forrn a = x2dy + 2xydy = d(x2y) is exatfrom. Hence calculations are much easier:

Show that LV = x2y is not exact form. Suppose it is exact form: w = dF = Fxdx + Fydy Then 0 = FIx2 =Fy. We see that FYI = 0 and Fxy = 2x. Contradiction since Fry = Fyx

Another solution Consider closed curve formed by semicircle C and its diameter. One can see thatintegral of form w over diameter equals to zero. Hence integral over this closed curve equals to 11 " ~3. Hencew is not exact Lform , since for exact l-form integral over closed curve is equal to zero. ~

c) Consider the curve in E2 defined by the equation r(2 - cos r p ) = 3 in polar coordinaies.Show that the sum of the distances between the points FI = (0,0) and F2 =2,0), and an arbitrary point

of this curue is constant, i. e. the curve is an ellipse and points P1, F 2 are its foci ..Find the integral of the I-form w = xdy - ydx o'ver this C11.rve. (Here as usual X,y are cartesian

coordinates x = r cos 'P, y = r sin 'P.)Take an arbitrary point P on the curve. Then due to the fact that r cos 'P = 21 ' - 3 we haveIPF} I + IPF21 =r + Jr2 + 4 - 4r cos 'P= + V 1 ' 2 + 4 - 4r cos rp = r + J1'2 - 8r + 16 = + [r - 4 1

On the other hand)' = 2-208'1' ::; 2 ' : : ' 1 = 3 . Hence IPFll + IPF21 = r + Ir - 4 1 = r + (4 - r) = 4. We provedthat sum of distances is constant. This is an ellipse.To find the integral of the form LV over this ellipse we rewrite the equation of ellipse in Cartesian

coordinates then take the integral'. Do it.Rewrite the equation of the ellipse in cartesian coordinates: 3 = r(2-cos rp ) = Zr=r cos 'P = 2Jx2 + y2_

X, j e. 2Jx2 + y2 = X + 3. Taking square we come to 4x2 + 4y2 =x2 + 6x + 9, i.e.(X-l)2 y2 .. {X=1+2cost4 +-3 = 1, parametric equation: /03 . t 0 < t < 2iT,Y = V o sin - v( t) =(-2sin t ) 8 In 8I3cost = -2sint x+v3cost" YHence for differentia'! form w = xdy - ydx we have w(v(t)) = x(t)}3cost - y(t)(-2sint) = }3(1 +2cos t) cos t + 2V3 sin2 t we have

1 lei = 1 0 27' (-/3(1 + 2 cost) cost + 2-/3sin2 t ) dt = 2J3 27f=4J37f 1 '1* \Ve rnay rewrite differential form in polar coordinates w = xdy - ydx = r2drp a.nd try to take the integralin polar coordinates

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