Chapter Given three points, A(4, 3, 2), B( 2, 0, 5), and C(7, 2, 1): a) Specify the vector A extending from the origin to the point A.

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1 Chapter Given the vectors M = 1a x +4a y 8a z and N =8a x +7a y a z, find: a) a unit vector in the direction of M +N. M +N =1a x 4a y +8a z +16a x +14a y 4a z = (6, 1, 4) Thus (6, 1, 4) a = =(.9,.6,.14) (6, 1, 4) b) the magnitude of 5a x + N M: (5,, )+(8, 7, ) (, 1, 4) = (4, 5, ), and (4, 5, ) =48.6. c) M N (M + N): ( 1, 4, 8) (16, 14, 4) (, 11, 1) = (1.4)(1.6)(, 11, 1) =( 58.5, 19, 9) 1.. Given three points, A(4,, ), B(,, 5), and C(7,, 1): a) Specify the vector A extending from the origin to the point A. A =(4,, ) = 4a x +a y +a z b) Give a unit vector extending from the origin to the midpoint of line AB. The vector from the origin to the midpoint is given by M =(1/)(A + B) =(1/)(4, +, +5)=(1, 1.5,.5) The unit vector will be m = (1, 1.5,.5) =(.5,.8,.89) (1, 1.5,.5) c) Calculate the length of the perimeter of triangle ABC: Begin with AB =( 6,, ), BC =(9,, 4), CA =(, 5, 1). Then AB + BC + CA = =. 1.. The vector from the origin to the point A is given as (6,, 4), and the unit vector directed from the origin toward point B is (,, 1)/. If points A and B are ten units apart, find the coordinates of point B. With A =(6,, 4) and B = 1 B(,, 1), we use the fact that B A = 1, or (6 B)a x ( B)a y (4 + 1 B)a z =1 Expanding, obtain 6 8B B +4 8 B B B B = 1 or B 8B 44=. ThusB = 8± = (taking positive option) and so B = (11.75)a x (11.75)a y + 1 (11.75)a z =7.8a x 7.8a y +.9a z 1

2 1.4. given points A(8, 5, 4) and B(,, ), find: a) the distance from A to B. B A = ( 1, 8, ) =1.96 b) a unit vector directed from A towards B. This is found through a AB = B A =(.77,.6,.15) B A c) a unit vector directed from the origin to the midpoint of the line AB. a M = (A + B)/ (, 1, ) = =(.69,.,.69) (A + B)/ 19 d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z =. Note that the midpoint, (, 1, ), as determined from part c happens to have z coordinate of. This is the point we are looking for A vector field is specified as G =4xya x + 1(x +)a y +18z a z. Given two points, P (1,, 1) and Q(, 1, ), find: a) G at P : G(1,, 1) = (48, 6, 18) b) a unit vector in the direction of G at Q: G(, 1, )=( 48, 7, 16), so a G = c) a unit vector directed from Q toward P : ( 48, 7, 16) =(.6,.9,.88) ( 48, 7, 16) a QP = P Q (, 1, 4) = =(.59,.,.78) P Q 6 d) the equation of the surface on which G = 6: We write 6 = (4xy, 1(x +), 18z ), or 1 = (4xy, x +4, z ), so the equation is 1=16x y +4x 4 +16x +16+9z For the G field in Problem 1.5, make sketches of G x, G y, G z and G along the line y =1,z =1, for x. We find G(x, 1, 1) = (4x, 1x +4, 18), from which G x =4x, G y =1x + 4, G z = 18, and G =6 4x 4 +x + 5. Plots are shown on page Given the vector field E =4zy cos xa x +zy sin xa y + y sin xa z for the region x, y, and z less than, find: a) the surfaces on which E y =. With E y =zy sin x =, the surfaces are 1) the plane z =, with x <, y < ; ) the plane y =, with x <, z < ; ) the plane x =, with y <, z < ; 4) the plane x = π/, with y <, z <. b) the region in which E y = E z : This occurs when zy sin x = y sin x, or on the plane z = y, with x <, y <, z < 1.

3 1.7. c) the region in which E = : We would have E x = E y = E z =,orzy cos x = zy sin x = y sin x =. This condition is met on the plane y =, with x <, z < Two vector fields are F = 1a x +x(y 1)a y and G =x ya x 4a y + za z. For the point P (,, 4), find: a) F : F at (,, 4)=( 1, 8, ), so F =8.6. b) G : G at (,, 4) = (4, 4, 4), so G =4.7. c) a unit vector in the direction of F G: F G =( 1, 8, ) (4, 4, 4) = ( 4, 84, 4). So a = F G ( 4, 84, 4) = =(.7,.9,.4) F G 9.7 d) a unit vector in the direction of F + G: F + G =( 1, 8, ) + (4, 4, 4) = (14, 76, 4). So a = F + G (14, 76, 4) = =(.18,.98,.5) F + G A field is given as 5 G = (x + y ) (xa x + ya y ) Find: a) a unit vector in the direction of G at P (, 4, ): Have G p =5/(9 + 16) (, 4, ) = a x +4a y, and G p =5. Thusa G =(.6,.8, ). b) the angle between G and a x at P : The angle is found through a G a x = cos θ. So cos θ = (.6,.8, ) (1,, )=.6. Thus θ =5. c) the value of the following double integral on the plane y =7: x + y (xa x + ya y ) a y dzdx = = G a y dzdx 4 [ tan 1 ( x dzdx = ) ] =6 4 5 x +49 dx 1.1. Use the definition of the dot product to find the interior angles at A and B of the triangle defined by the three points A(1,, ), B(, 4, 5), and C(,, 1): a) Use R AB =(, 1, 7) and R AC =( 1, 5, ) to form R AB R AC = R AB R AC cos θ A. Obtain = 59 5 cos θ A. Solve to find θ A =65.. b) Use R BA =(, 1, 7) and R BC =(, 6, 4) to form R BA R BC = R BA R BC cos θ B. Obtain = cos θ B. Solve to find θ B = Given the points M(.1,.,.1), N(.,.1,.), and P (.4,,.1), find: a) the vector R MN : R MN =(.,.1,.) (.1,.,.1) = (.,.,.4). b) the dot product R MN R MP : R MP = (.4,,.1) (.1,.,.1) = (.,.,.). R MN R MP =(.,.,.4) (.,.,.) = =.5.

4 1.11. c) the scalar projection of R MN on R MP : R MN a RMP =(.,.,.4) (.,.,.) =.5 = d) the angle between R MN and R MP : ( ) ( ) θ M = cos 1 RMN R MP = cos 1.5 =78 R MN R MP Given points A(1, 1, 6), B(16, 8, ), C(8, 1, 4), and D(, 5, 8), determine: a) the vector projection of R AB +R BC on R AD : R AB +R BC = R AC =(8, 1, 4) (1, 1, 6) = (, 11, 1) Then R AD =(, 5, 8) (1, 1, 6) = ( 1, 17, 14). So the projection will be: [ ] ( 1, 17, 14) ( 1, 17, 14) (R AC a RAD )a RAD = (, 11, 1) =( 6.7, 9.5, 7.8) b) the vector projection of R AB + R BC on R DC : R DC =(8, 1, 4) (, 5, 8) = (1, 6, 4). The projection is: (R AC a RDC )a RDC = [ ] (1, 6, 4) (1, 6, 4) (, 11, 1) =( 8., 5.,.) c) the angle between R DA and R DC : Use R DA = R AD = (1, 17, 14) and R DC = (1, 6, 4). The angle is found through the dot product of the associated unit vectors, or: ( ) (1, 17, 14) (1, 6, 4) θ D = cos 1 (a RDA a RDC ) = cos 1 = a) Find the vector component of F = (1, 6, 5) that is parallel to G =(.1,.,.): F G = F G (1, 6, 5) (.1,.,.) G = (.1,.,.) = (.9, 1.86,.79) G b) Find the vector component of F that is perpendicular to G: F pg = F F G = (1, 6, 5) (.9, 1.86,.79)=(9.7, 7.86,.1) c) Find the vector component of G that is perpendicular to F: G pf = G G F = G G F F F =(.1,.,.) 1. (1, 6, 5) = (.,.5,.6)

5 1.14. The four vertices of a regular tetrahedron are located at O(,, ), A(, 1, ), B(.5,.5, ), and C( /6,.5, /). a) Find a unit vector perpendicular (outward) to the face ABC: First find R BA R BC = [(, 1, ) (.5,.5, )] [( /6,.5, /) (.5,.5, )] =(.5,.5, ) ( /,, /) = (.41,.71,.9) The required unit vector will then be: b) Find the area of the face ABC: R BA R BC =(.47,.8,.) R BA R BC Area = 1 R BA R BC = Three vectors extending from the origin are given as r 1 = (7,, ), r = (, 7, ), and r =(,, ). Find: a) a unit vector perpendicular to both r 1 and r : a p1 = r 1 r (5, 5, 55) = =(.8,.41,.91) r 1 r 6.6 b) a unit vector perpendicular to the vectors r 1 r and r r : r 1 r =(9, 4, 1) and r r =(, 5, 6). So r 1 r r r = (19, 5, ). Then a p = c) the area of the triangle defined by r 1 and r : (19, 5, ) (19, 5, ) = =(.,.81,.5) (19, 5, ) 6.95 Area = 1 r 1 r =. d) the area of the triangle defined by the heads of r 1, r, and r : Area = 1 (r r 1 ) (r r ) = 1 ( 9, 4, 1) (, 5, 6) = Describe the surfaces defined by the equations: a) r a x =, where r =(x, y, z): This will be the plane x =. b) r a x =: r a x =(,z, y), and r a x = z + y =. This is the equation of a cylinder, centered on the x axis, and of radius. 5

6 1.17. Point A( 4,, 5) and the two vectors, R AM = (, 18, 1) and R AN =( 1, 8, 15), define a triangle. a) Find a unit vector perpendicular to the triangle: Use a p = R AM R AN (5,, 4) = =(.664,.79,.645) R AM R AN 57.5 The vector in the opposite direction to this one is also a valid answer. b) Find a unit vector in the plane of the triangle and perpendicular to R AN : a AN = ( 1, 8, 15) 89 =(.57,.46,.761) Then a pan = a p a AN =(.664,.79,.645) (.57,.46,.761) = (.55,.8,.77) The vector in the opposite direction to this one is also a valid answer. c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit vector in the required direction is (1/)(a AM + a AN ), where Now a AM = (, 18, 1) =(.697,.67,.48) (, 18, 1) 1 (a AM + a AN )= 1 [(.697,.67,.48) + (.57,.46,.761)] = (.95,.516,.7) Finally, a bis = (.95,.516,.7) =(.168,.915,.67) (.95,.516,.7) Given points A(ρ =5,φ=7,z = ) and B(ρ =,φ=,z = 1), find: a) a unit vector in cartesion coordinates at A directed toward B: A(5 cos 7, 5 sin 7, ) = A(1.71, 4.7, ), In the same manner, we find B(1.7, 1, 1). So R AB =(1.7, 1, 1) (1.71, 4.7, ) =(., 5.7, 4) and therefore a AB = (., 5.7, 4) =(.,.8,.57) (., 5.7, 4) b) a vector in cylindrical coordinates at A directed toward B: a AB a ρ =. cos 7.8 sin 7 =.77. a AB a φ =. sin 7.8 cos 7 =.8. Thus. a AB =.77a ρ.8a φ +.57a z 6

7 1.18. (continued) c) a unit vector in cylindrical coordinates at B directed toward A: Use a BA =(,,.8,.57). Then a BA a ρ =. cos( )+.8 sin( )=.4, and a BA a φ =. sin( )+.8 cos( )=.71. Finally, a BA =.4a ρ +.71a φ.57a z 1.19 a) Express the field D =(x + y ) 1/ (xa x + ya y ) in cylindrical components and cylindrical variables: Have x = ρ cos φ, y = ρ sin φ, and x + y = ρ. Therefore D = 1 ρ (cos φa x + sin φa y ) Then D ρ = D a ρ = 1 ρ [cos φ(a x a ρ ) + sin φ(a y a ρ )] = 1 [ cos φ + sin φ ] = 1 ρ ρ and D φ = D a φ = 1 ρ [cos φ(a x a φ ) + sin φ(a y a φ )] = 1 [cos φ( sin φ) + sin φ cos φ] = ρ Therefore D = 1 ρ a ρ b) Evaluate D at the point where ρ =,φ =.π, and z = 5, expressing the result in cylindrical and cartesion coordinates: At the given point, and in cylindrical coordinates, D =.5a ρ. T o express this in cartesian, we use D =.5(a ρ a x )a x +.5(a ρ a y )a y =.5 cos 6 a x +.5 sin 6 a y =.41a x +.9a y 1.. Express in cartesian components: a) the vector at A(ρ =4,φ =4,z = ) that extends to B(ρ =5,φ = 11,z = ): We have A(4 cos 4, 4 sin 4, ) = A(.6,.57, ), and B(5 cos( 11 ), 5 sin( 11 ), ) = B( 1.71, 4.7, ) in cartesian. Thus R AB =( 4.77, 7., 4). b) a unit vector at B directed toward A: Have R BA =(4.77, 7., 4), and so a BA = (4.77, 7., 4) =(.5,.76,.4) (4.77, 7., 4) c) a unit vector at B directed toward the origin: Have r B =( 1.71, 4.7, ), and so r B = (1.71, 4.7, ). Thus a = (1.71, 4.7, ) =(.,.87,.7) (1.71, 4.7, ) 7

8 1.1. Express in cylindrical components: a) the vector from C(,, 7) to D( 1, 4, ): C(,, 7) C(ρ =.61,φ=.7,z = 7) and D( 1, 4, ) D(ρ =4.1,φ= 14.,z = ). Now R CD =( 4, 6, 9) and R ρ = R CD a ρ = 4 cos(.7) 6 sin(.7) = Then R φ = R CD a φ = 4 sin(.7) 6 cos(.7) =.77. So R CD = 6.66a ρ.77a φ +9a z b) a unit vector at D directed toward C: R CD =(4, 6, 9) and R ρ = R DC a ρ = 4 cos( 14.) + 6 sin( 14.) = Then R φ = R DC a φ =4[ sin( 14.)]+6 cos( 14.)=.4. So R DC = 6.79a ρ +.4a φ 9a z Thus a DC =.59a ρ +.1a φ.78a z c) a unit vector at D directed toward the origin: Start with r D =( 1, 4, ), and so the vector toward the origin will be r D = (1, 4, ). Thus in cartesian the unit vector is a =(.,.87,.44). Convert to cylindrical: a ρ =(.,.87,.44) a ρ =. cos( 14.) +.87 sin( 14.) =.9, and a φ =(.,.87,.44) a φ =.[ sin( 14.)] +.87 cos( 14.) =, so that finally, a =.9a ρ.44a z. 1.. A field is given in cylindrical coordinates as [ ] 4 F = ρ + (cos φ + sin φ) a ρ + (cos φ sin φ)a φ a z +1 where the magnitude of F is found to be: F = [ 16 F F = (ρ +1) + 4 ρ (cos φ + sin φ)+ +1 Sketch F : a) vs. φ with ρ = : in this case the above simplifies to F(ρ =) = Fa = [8 + 4(cos φ + sin φ)] 1/ b) vs. ρ with φ =, in which: [ 16 F(φ =) = Fb = (ρ +1) + 4 ρ +1 + c) vs. ρ with φ =45, in which [ F(φ =45 16 ) = Fc = (ρ +1) + 4 ρ +1 + Plots are shown on page The surfaces ρ =,ρ =5,φ = 1, φ = 1, z =, and z =4.5 define a closed surface. a) Find the enclosed volume: Vol = ρdρdφdz =6.8 ] 1/ ] 1/ ] 1/ 8

9 1.. (continued) NOTE: The limits on the φ integration must be converted to radians (as was done here, but not shown). b) Find the total area of the enclosing surface: 1 Area = ρdρdφ dφ dz dφ dz dρ dz =.7 c) Find the total length of the twelve edges of the surfaces: [ ] Length = π π 5 =.4 d) Find the length of the longest straight line that lies entirely within the volume: This will be between the points A(ρ =,φ = 1, z = ) and B(ρ =5,φ = 1, z =4.5). Performing point transformations to cartesian coordinates, these become A(x =.5, y =.95, z = ) and B(x =.1, y =.8, z =4.5). Taking A and B as vectors directed from the origin, the requested length is Length = B A = (.69,.88, 1.5) = At point P (, 4, 5), express the vector that extends from P to Q(,, 1) in: a) rectangular coordinates. R PQ = Q P =5a x 4a y 6a z Then R PQ = = 8.8 b) cylindrical coordinates. At P, ρ =5,φ = tan 1 (4/ ) = 5.1, and z =5. Now, R PQ a ρ =(5a x 4a y 6a z ) a ρ = 5 cos φ 4 sin φ =6. R PQ a φ =(5a x 4a y 6a z ) a φ = 5 sin φ 4 cos φ =1.6 Thus R PQ =6.a ρ +1.6a φ 6a z and R PQ = =8.8 c) spherical coordinates. At P, r = = 5=7.7, θ = cos 1 (5/7.7) = 45, and φ = tan 1 (4/ ) = 5.1. R PQ a r =(5a x 4a y 6a z ) a r = 5 sin θ cos φ 4 sin θ sin φ 6 cos θ =.14 R PQ a θ =(5a x 4a y 6a z ) a θ = 5 cos θ cos φ 4 cos θ sin φ ( 6) sin θ =8.6 R PQ a φ =(5a x 4a y 6a z ) a φ = 5 sin φ 4 cos φ =1.6 9

10 1.4. (continued) Thus R PQ =.14a r +8.6a θ +1.6a φ and R PQ = =8.8 d) Show that each of these vectors has the same magnitude. Each does, as shown above Given point P (r =.8, θ=,φ=45 ), and E = 1 ( r cos φ a r + sin φ ) sin θ a φ a) Find E at P : E =1.1a ρ +.1a φ. b) Find E at P : E = =.47. c) Find a unit vector in the direction of E at P : a E = E E =.45a r +.89a φ 1.6. a) Determine an expression for a y in spherical coordinates at P (r =4,θ=.π, φ =.8π): Use a y a r = sin θ sin φ =.5, a y a θ = cos θ sin φ =.48, and a y a φ = cos φ =.81 to obtain a y =.5a r +.48a θ.81a φ b) Express a r in cartesian components at P : Find x = r sin θ cos φ = 1.9, y = r sin θ sin φ = 1.8, and z = r cos θ =.4. Then use a r a x = sin θ cos φ =.48, a r a y = sin θ sin φ =.5, and a r a z = cos θ =.81 to obtain a r =.48a x +.5a y +.81a z 1.7. The surfaces r = and 4, θ = and 5, and φ = and 6 identify a closed surface. a) Find the enclosed volume: This will be Vol = where degrees have been converted to radians. b) Find the total area of the enclosing surface: Area = 6 5 (4 + ) sin θdθdφ + r sin θdrdθdφ = r(sin + sin 5 )drdφ 5 4 c) Find the total length of the twelve edges of the surface: Length = 4 4 =17.49 dr rdrdθ =1.61 (4+)dθ + (4 sin sin + sin 5 + sin )dφ 1

11 1.7. (continued) d) Find the length of the longest straight line that lies entirely within the surface: This will be from A(r =,θ =5,φ= )tob(r =4,θ =,φ=6 )or A(x = sin 5 cos,y = sin 5 sin,z = cos 5 ) to B(x = 4 sin cos 6,y = 4 sin sin 6,z = 4 cos ) or finally A(1.44,.5, 1.9) to B(1., 1.7,.46). Thus B A =(.44, 1.1,.18) and Length = B A = a) Determine the cartesian components of the vector from A(r =5,θ = 11,φ= )tob(r = 7,θ =,φ =7 ): First transform the points to cartesian: x A = 5 sin 11 cos = 4.4, y A = 5 sin 11 sin = 1.61, and z A = 5 cos 11 = 1.71; x B = 7 sin cos 7 =1., y B = 7 sin sin 7 =.9, and z B = 7 cos =6.6. Now R AB = B A =5.6a x +4.9a y +7.77a z b) Find the spherical components of the vector at P (,, 4) extending to Q(,, 5): First, R PQ = Q P =( 5, 5, 1). Then at P, r = =5.9, θ = cos 1 (4/ 9)=4., and φ = tan 1 ( /) = 56..Now So finally, R PQ a r = 5 sin(4 ) cos( 56. ) + 5 sin(4 ) sin( 56. ) + 1 cos(4 )=.9 R PQ a θ = 5 cos(4 ) cos( 56. ) + 5 cos(4 ) sin( 56. ) 1 sin(4 )= 5.8 R PQ a φ = ( 5) sin( 56. ) + 5 cos( 56. )= 1.9 R PQ =.9a r 5.8a θ 1.9a φ c) If D =5a r a θ +4a φ, find D a ρ at M(1,, ): First convert a ρ to cartesian coordinates at the specified point. Use a ρ =(a ρ a x )a x +(a ρ a y )a y.ata(1,, ), ρ = 5, φ = tan 1 () = 6.4, r = 14, and θ = cos 1 (/ 14)=6.7.Soa ρ = cos(6.4 )a x + sin(6.4 )a y =.45a x +.89a y. Then (5a r a θ +4a φ ) (.45a x +.89a y )= 5(.45) sin θ cos φ + 5(.89) sin θ sin φ (.45) cos θ cos φ (.89) cos θ sin φ + 4(.45)( sin φ) + 4(.89) cos φ = Express the unit vector a x in spherical components at the point: a) r =,θ = 1 rad, φ =.8 rad: Use a x =(a x a r )a r +(a x a θ )a θ +(a x a φ )a φ = sin(1) cos(.8)a r + cos(1) cos(.8)a θ +( sin(.8))a φ =.59a r +.8a θ.7a φ 11

12 1.9 (continued) Express the unit vector a x in spherical components at the point: b) x =,y =,z = 1: First, transform the point to spherical coordinates. Have r = 14, θ = cos 1 ( 1/ 14) = 15.5, and φ = tan 1 (/) =.7. Then a x = sin(15.5 ) cos(.7 )a r + cos(15.5 ) cos(.7 )a θ +( sin(.7 ))a φ =.8a r.a θ.55a φ c) ρ =.5, φ =.7 rad, z = 1.5: Again, convert the point to spherical coordinates. r = ρ + z = 8.5, θ = cos 1 (z/r) = cos 1 (1.5/ 8.5)=59., and φ =.7 rad = 4.1. Now a x = sin(59 ) cos(4.1 )a r + cos(59 ) cos(4.1 )a θ +( sin(4.1 ))a φ =.66a r +.9a θ.64a φ 1.. Given A(r =,θ =,φ=45 ) and B(r =,θ = 115,φ= 16 ), find: a) R AB : First convert A and B to cartesian: Have x A = sin( ) cos(45 )=7.7, y A = sin( ) sin(45 )=7.7, and z A = cos( )=17.. x B = sin(115 ) cos(16 )= 5.6, y B = sin(115 ) sin(16 )=9., and z B = cos(115 )= 1.7. Now R AB = R B R A =(.6,.,.), and so R AB =44.4. b) R AC, given C(r =,θ =9,φ =45 ). Again, converting C to cartesian, obtain x C = sin(9 ) cos(45 )=14.14, y C = sin(9 ) sin(45 )=14.14, and z C = cos(9 )=. So R AC = R C R A =(7.7, 7.7, 17.), and R AC =.. c) the distance from A to C on a great circle path: Note that A and C share the same r and φ coordinates; thus moving from A to C involves only a change in θ of 6. The requested arc length is then [ ( )] π distance = 6 =.9 6 1