Vetors MC Qld-3 49 Chapter 3 Vetors Exerse 3A Revew of vetors a d e f e a x + y omponent: x a os(θ 6 os(80 + 39 6 os(9.4 omponent: y a sn(θ 6 sn(9 0. a.4 0. f a x + y omponent: x a os(θ 5 os( 5 3.6 omponent: y a sn(θ 5 sn( 5 6.3 a 3.6 6.3 g a x + y omponent: x a os(θ 5 os(80 54 5 os(6 3.5 omponent: y a sn(θ 5 sn(6 6.6 a 3.5 + 6.6 h a x + y omponent: x a os(θ 3 os(35.6 omponent: y a sn(θ 3 sn(35 a.6.6.6 3 a Let r a + a 6 5 a x + y omponent: x a os( θ 0 os(40 7.7 omponent: y a sn(θ 0 sn(40 6.4 a 7.7 + 6.4 d a x + y omponent: x a os(θ 9 os(0 4.5 omponent: y a sn(θ 9 sn(0 7.8 a 4.5 + 7.8 a 5os(0 + 5sn(0 3.49+ 8.55 5os(55 + 5sn(55 8.60 +.9 r a + 3.49 + 8.55 + 8.60 +.9 3.09 + 0.84
MC Qld-3 50 Vetors r 38.3 (3.09 + (0.84 tan(θ 0.84 3.09 θ 33 The resultant vetor has a magntude of 38.3 n a dreton of 33 north of east. Let r a + tan(θ 9.6 3.08 θ 7. The resultant vetor has magntude 3.5 n a dreton of 7. south of east. d Let r a + a 5os(45 + 5sn(45 0.6 + 0.6 0os(30 + 0sn(30 7.3 + 0 r a + 0.6 + 0.6+ 7.3 + 0 7.93 + 0.6 r 9.3 + 7 9.3 7 os(4 08.30 r 4.4 sn( α sn(4 7 4.4 sn(α 7sn(4 4.4 sn(α 0.40 α 3.7 The resultant vetor has magntude 4.4 n a dreton of 3.7 north of east. e Let r a + r 34.7 (7.93 + (0.6 tan(θ 0.6 7.93 36.4 The resultant vetor has a magntude of 34.7 n a dreton of 36.4 north of east. Let r a + a 5.3.os(80 43 +.sn(80 43.os(37 +.sn(37 8. + 7.57 r a + 5.3 8. + 7.57 7.8 + 7.57 a os(60 + sn(60 + 9.05 35os( 55 + 35sn( 55 0.08 8.67 r a + + 9.05 + 0.08 8.67 3.08 9.6 r 0.4 (7.8 + (7.57 tan(θ 7.57 7.8 θ 46.5 The resultant vetor has magntude 0.4 n a dreton 46.5 north of east. f Let r r 3.5 (3.08 + ( 9.6
Vetors MC Qld-3 5 4 r + 9 9 os(74 64.57 r 4.8 sn( α sn(74 9 4.8 sn(α 9sn(74 4.8 α 47.5 30 + α + θ 90 θ 90 30 α 90 30 47.5.5 The resultant vetor has magntude 4.8 n a dreton.5 south of east. g r r 3 + 4 3 4 os(40 479.78 r 68.7 sn( α sn(40 4 68.7 sn(α 4sn(40 68.7 α 3. 50 + α + θ 90 θ 90 50 α 90 50 3. 6.9 The resultant vetor has magntude 68.7 n a dreton 6.9 south of east. R 00 + 400 00 400 os(47 863799 R 436 N 5 a a 4 and + 3 4 + + 3 6+ 6 + 36 + 4 40 6.3 a 4 and 3 4 + 3 6 4 6 + ( 4 6 36+ 6 5 7. a + + and + 3 + + + + 3 3 + 4+ 3 + 4 + 9+ 6+ 4 9 5.4 d a 4 and + 4 + + 6 6 6 e a 4+ and + 3+ 3 4 + + + 3 + 3 4 + 5 4 + ( + 5 6 + + 5 4 6.5 Let R a 400os(8 + 400sn(8 353. + 87.8 600 R 353. + 87.8 600 353. 4. R (353. + ( 4. 543N 7 a Let the two vetors e a and. a a os( θ 8 0 os(30 69.3 Let the two vetors e a and. a a os( θ 8 0 os(50 38.6 Let the two vetors e a and. a a os( θ 0 8 os(80 30 60 80 os(90 0
MC Qld-3 5 Vetors 8 a a ( + ( 3 + 3 6 4 a ( 3( 3 + + + 3 3 4 + 9 3 a (3 + ( + 3 3 + 3 6 + 6 0 9 The par of vetors n 8( are perpendular eause ther dot produt s equal to zero. 0 a a 3 + +, + 3 a (3 + + ( + 3 3 + 3+ 0 6 + 6 + 0 0 a 3 + 3+, + 3+ 3 a (3 + 3 + ( + 3 + 3 3 + 3 3+ 3 6 + 9 + 3 6 a +, + 3+ 3 a ( + ( + 3 + 3 + 3 3 + 6 6 d a +, + 3 a ( + ( + 3 0 + 3+ 0 0 + 6 + 0 6 e a 4 + +, 3 a ( 4 + + ( 3 4 + 0+ 3 8 + 0 3 The par of vetors n 0(a are perpendular eause ther dot produt s equal to zero. a + 3 and n + are perpendular so ( + 3 ( n + 0 n + 3 0 n + 3 0 3 n n.5 n and n + 8 are perpendular so ( n ( n + 8 0 n n + 8 0 n 6 0 n 6 n ±4 n + 3 and + are perpendular so ( n + 3 ( + 0 n 0 + 3 0 n + 6 0 6 n n 3 d + + and 3 + n+ 3 are perpendular so ( + + (3 + n+ 3 0 3 + n+ 3 0 3 + n + 3 0 n + 6 0 n 6 3 a Let the 3 fores e, and 3 and the resultng fore e R R + + 3 5os(30 + 5sn(30.65 +.5 35 3 0os(80 + 40 + 0sn(80 + 40 0os(0 + 0sn(0 5.3.86 R + + 3.65 +.5 + 35 5.3.86 (.65 + 35 5.3 + (.5.86 4.33 0.36 R 4.33 + ( 0.36 4.3N 0.36 tan(θ 4.33 θ 0.5 The resultng fore has magntude 4.3N n the dreton of 0.5 south of east. Let the 3 fores e, and 3 and the resultng fore e R. R + + 3 It s evdent from the dagram that of the fores have equal omponents and equal and opposte omponents, so + 35os(45 49.50 3 5os(80 + 50 + 5sn(80 + 50 5os(30 + 5sn(30 6.07 9.5 R + + 3 49.50 6.07 9.5 33.43 9.5 R 33.43 + ( 9.5 38.5N 9.5 tan(θ 33.43 θ 9.8 The resultng fore has magntude 38.5N n a dreton 9.8 south of east.
Vetors MC Qld-3 53 Exerse 3B Vetor produt a a a sn( θ n ˆ 9 7sn(40 nˆ 40.5nˆ 40.5 n a downward dreton (nto the page a 40.5 n an upward dreton (or out of the page a asn( θ nˆ 5 0sn(40 nˆ 3.nˆ 3. n an upward dreton (out of the page a 3. n a downward dreton (or nto the page a asn( θ nˆ 4 8sn(80 nˆ 0nˆ 0 a 0 a ( + + 3 + + 3 0+ 3 3 + ( 3+ ( + ( 3 + + 0+ + ( + ( 3 + ( 3 + + ( 3 0+ 3 4 6 6 + 3 4 3 a ( 3 + ( 3 + 3 3 3 3 + 3 3 (6 4 ( 4 + 6 + (4 9 5 ( + 3 + ( 3 3 3 3 3 + 3 3 ( 3 + ( + 3 + (4 + 9 5+ 3 (3 + 3 (3 + 4 3 3 3 4 3 3 3 3 + 4 3 3 4 ( ( 6 9 + ( + 3 0 + 5 + 5 4 a OA + + and OB 3 A a ( + + 3 ( + + 3 3 + 3 + 3 6 + 0 3 3 + 6 Therefore A 3 + 6 ( 3 + (6 9+ 36 45 3 5 3.45 square unts. OA + 3 and OB + A a ( + 3 ( + ( + 3 ( + 3 0 Therefore A 6 6 + 6 + + ( 6 36+ 4+ 36 76 4.36 square unts 5 3 + and r + 3 Torque r (3 + ( + 3 (3 + ( + 3 3 0 3 0 3 0 3 3 + 3 3 3 + 0 0 (6 0 ( 4 + + (0 6 6 + 6
MC Qld-3 54 Vetors ( 6 0 ( 9 0 + (6 4 6 + 9+ So torque 6 + 9+ ( 6 + 9 + 36+ 8+ 4 Nm 6 a Torque r r r sn( θ nˆ 0.8 50sn(80 40 30 nˆ 40sn(0 nˆ 37.6nˆ 37.6ˆ (out of the page So torque ˆ 37.6 (37.6 37.6 Nm Torque r r r sn( θ n ˆ 0.8 50sn(80 40 80 nˆ 40sn(60 nˆ 34.6nˆ 34.6ˆ (out of the page So torque (34.6 34.6 Nm 7 a ( ( ( 0 d ( ( 8 a a + and + Let û e the unt vetor perpendular to a and, so û a a a ( + ( + + + + + 0+ + So û ( + + ( + ( ( + 0.577( + a and + Let û e the unt vetor perpendular to a and, so û a a a ( + + + So û ( + ( + ( ( + 8 ( + ( + 0.707( + 9 a a ( ( + [4 ( ] ( + [4 + 4 ( ] ( + (0 + 8 ( + 8 8+ 8 6 + 0 6 Show that the result of the vetor produt a ( les n the plane of and Let a ( n + m 6 n(4 + m( 6 4n + m m 6 (4n+ m m Therefore, 6 m 0 4n + m m 8 0 4n + ( 8 6 4n n 4 Therefore, a ( 4 8, so a ( must le n the plane of and. Show that, n general, the vetor a ( les n the plane of and. The vetor ( s perpendular to the plane of and, so f a vetor s n the plane of and, t must e perpendular to ( and ve versa. Therefore, sne the vetor a (, wll always e perpendular to the vetor (, t wll always le n the plane of and. 0 Tae B as the orgn wth the x-, y- and z-axes taen as usual. BA 3 + AC 3 + 3 AC ( 3 + 3 + (
Vetors MC Qld-3 55 9 + 9 + 4 The magntude of the vetor s. So 00 ( 3 + 3 Torque r BA (3 + 00 ( 3 3 + 00 (3 + ( 3 + 3 (3 + ( 3 + 3 3 0 3 3 0 3 3 0 3 + 3 3 3 (0 6 ( 6 + 6 + (9 0 6 + 9 So torque 00 6 + 9 00 ( 6 + 9 00 36 + 8 00 7 30.6 Nm Tae A as the orgn wth the x- and y-axes taen as usual and the postve z-axs as out of the page AB.5 40os(60 + 40sn(60 0 + 34.64 Torque r AB (.5 (0 + 34.64 (.5 (0 + 34.64.5 0 0 34.64 0 0.5 0.5 + 34.64 0 0 0 0 34.64 (0 0 (0 0 + (.5 34.64 + 0 7 So torque 7 7 7 Nm Exerse 3C Salar trple produt a a + +,, + 3+ 4 a a a3 a 3 3 0 3 4 0 0 + 3 4 4 3 (8 + 3 (0 + (0 + + + 4 7 a, +, + + 3 a a a3 a 3 3 0 3 0 3 3 0 (3 + ( 4 5 + 6 a +, 3 +, + a a a3 a 3 3 0 3 3 3 0 + ( 4 ( 6 + 0 + 4 d a + +,, + 3+ a a a3 a 3 3 0 3 0 0 + 3 3 ( + 3 (0 + + (0 4 0 8 0 e a 4 + +,, + +
MC Qld-3 56 Vetors a a a3 a 3 3 4 4 + 4( + ( + + (4 + 0 6 + 6 0 f a + +,, 3 + 3+ a a a3 a 3 3 3 3 + 3 3 3 3 ( + 6 ( + 6 + (3 + 6 8 4 + 9 3 g a + 3+, +, + + a a a3 a 3 3 3 3 + ( + 3( + + ( 4 6 + 0 6 a a +, 3, + Volume a a 0 3 0 Volume a u unts a + 3, +, + 3 Volume a 3 a 3 0 9 Volume a 9 9 u unts a 3+, + +, 3 + 3+ 3 Volume a 3 a 3 3 3 0 Volume a 0 0 u unts d a, + +, + 3 Volume a a 3 0 4 Volume a 4 4 u unts 3 a a +, 3 + and n + 3 + 4 a 0 a 3 0 n 3 4 0 3 3 0 3 4 n 4 n 3 (0 3 ( n (9 0 3 + n 9 n 4 So n 4 0 n 4 a + n+, + +, 3+ a 0 n a 3 n + 3 3 ( + 3 n(4 + ( 6 5 n 8 n 3 So n 3 0 3 n n.5 a 3 +, 3 + n 3, + 4 a 0 3 a 3 n 3 4 n 3 3 3 3 n 3 4 4 3( 4n + 3 ( + 3 (3 n n + 9 + 8 3 + n 4 n
Vetors MC Qld-3 57 So 4 n 0 n 4 4 a Let the ottom left-hand orner e the orgn wth the x-, y- and z-axes defned as usual. Let a 8 0sn(40 + 0os(40 6.43 + 7.66 Volume a 0 0 a 8 0 0 6.43 7.66 0 0 0 8 0 8 0 0 0 7.66 0 6.43 0 6.43 7.66 0 0 (8 7.66 0 735.4 Volume a 735.4 735.4 u unts Let the ottom left orner e the orgn wth the x-, y- and z-axes defned as usual. Let a 9sn(44 + 9os(44 6.5 + 6.47 The dagram elow s the horzontal or xz-plane Let 4os(70 4sn(70 4.79 3.6 Volume a 6.5 6.47 0 a 0 Volume a 0 0 0 4.79 0 3.6 0 u unts. 0 0 5 a 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 ( 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 (0 + 0 0 0 d 0 0 0 0 e 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 (0 0 + 0 0 0 f 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 + 0 + ( 0 6 If a s a lowse orderng of,, and, the result s. If t s antlowse, t s. 7 a If a, and are oplanar, the volume of the parallelpped formed s zero. If a s n the same plane as and, then a s perpendular to. Therefore the salar produt wll e 0. Exerse 3D Applatons of the vetor produt a P (,,, Q (, 0, and R ( 3,, 0 PQ ( + (0 + ( QR ( 3 + ( 0 + (0 4 + n PQ QR 0 4 0 0 + 4 4 ( 0 ( 0 + ( 4 6 n (x a + n (y + n 3 (z 0 (x (y 0 6(z 0 x y 6z + 6 0 x y 6z + 5 0 x y 6z 5 P(, 0, 0, Q(0,, 0 and R(0, 0, PQ (0 + ( 0 + (0 0 +
MC Qld-3 58 Vetors QR (0 + (0 0 + ( 0 + n PQ QR 0 0 0 0 + 0 0 ( 0 ( 0 + (0 + + + n (x a + n (y + n 3 (z 0 (x + (y 0 + (z 0 0 x + y + z 0 x + y + z P(3,,, Q(,, and R( 3,, PQ ( 3 + ( + ( QR ( 3 + ( + ( 4 + n PQ QR 0 0 4 0 0 0 0 0 + 0 4 0 4 0 0 + ( 0 n (x a + n (y + n 3 (z 0 0(x 3 + 0(y (z 0 (z 0 z 0 z d P(,,, Q(4,, 0 and R( 3,, 3 PQ (4 + ( + (0 3 QR ( 3 4 + ( + (3 0 7 + 3+ 3 n PQ QR 3 7 3 3 3 3 + 3 3 7 3 7 3 ( 3+ 3 (9 7 + (9 7 + n (x a + n (y + n 3 (z 0 0(x (y + + (z 0 (y + + (z 0 y + z 0 y + z 0 y + z e P(, 0,, Q(3, 3, and R(3,, PQ (3 + (3 0 + ( + 3 QR (3 3 + ( 3 + ( + n PQ QR 3 0 0 3 0 0 3 + 0 0 (3 0 ( 0 + ( 0 3 n (x a + n (y + n 3 (z 0 3(x (y 0 (z 0 3x 6 y z + 0 3x y z 5 0 3x y z 5 3x + y + z 5 f P(3,,, Q(,, and R(, 3, PQ ( 3 + ( + ( + QR ( + (3 + ( 3 + n PQ QR 0 3 0 0 + 3 3 (0 ( + 3 + ( 0 4 n (x a + n (y + n 3 (z 0 (x 3 4(y (z 0 x + 3 4y + 8 z + 0 x 4y z + 0 x 4y z x + 4y + z a n +, (,, n (x a + n (y + n 3 (z 0 (x + 0(y + (z 0 x + z 0 x + z 4 0 x + z 4 n + +, (0,, n (x a + n (y + n 3 (z 0 (x 0 + (y + (z 0 x + y + z 4 0 x + y + z 5 0 x + y + z 5 n +, (,, 0 n (x a + n (y + n 3 (z 0 (x + (y (z 0 0 x + y z 0 x + y z 4 0 x + y z 4 d n 3 +, (, 3, n (x a + n (y + n 3 (z 0 3(x (y 3 + (z 0 3x 3 y + 3 + z 0 3x y + z 0 3x y + z 3 n, (0, 0, 0 n (x a + n (y + n 3 (z 0 0(x 0 + (y 0 (z 0 0 y z 0 4 a AB AO + OB OA + OB
Vetors MC Qld-3 59 OA 3 OB 5 So AB (3 + 5 3 + + 5 5 3+ AC 5 n AB AC 5 3 5 0 0 3 5 5 3 + 0 0 5 0 5 0 0 (0 0 + (0 + 5 0 + 5 5( + 3 Any multple of n s normal to the ABC plane so + 3 s a normal. d n + 3, pont (0, 0, 0 n (x a + n (y + n 3 (z 0 0(x 0 + (y 0 + 3 (z 0 0 y + 3z 0 5 Pont (,, 3 Plane perpendular to y-axs: n n (x a + n (y + n 3 (z 0 0(x + (y + 0(z 3 0 y 0 y 6 a At P y 0 and at Q x 0 x + 3y 6 y 0 x + 3 0 6 x 6 x 3 P (3, 0 x 0 0 + 3y 6 3y 6 y Q (0, Pont (0, 0, and ontans lne x + 3y 6 Let R (0, 0, rom part (a, P (3, 0, 0 and Q (0,, 0 PQ (0 3 + ( 0 + (0 0 3 + QR (0 0 + ( 0 + (0 n PQ QR 3 0 0 0 3 0 3 + 0 0 ( 0 (3 0 + ( 6 0 3 6 n (x a + n (y + n 3 (z 0 (x 0 3(y 0 6(z 0 x 3y 6z + 6 0 x 3y 6z 6 x + 3y + 6z 6 7 3x + 4y + z 6 n 3 + 4+ ˆn n n (3 + 4 + 3 + 4 + (3 + 4 + 9 + 6 + 44 (3 + 4 + 69 (3 4 3 + + 8 z 3x + y 3x + y z 0 n 3 + ˆn n n (3 + 3 + + ( (3 + 9+ 4+ (3 + 4 9 m v 3 + 3 r 4 + L r mv (4 + (3 + 3 (4 + (6 + 6 4 0 6 6 0 0 4 0 4 + 6 0 6 0 6 6 0 0 + (4 0 a m 3 v + 3 P mv 3( + 3 3 + 9 r 4 L r P 4 ( 3 + 9 4 ( 3 + 4 9 0+ 36 36 a Show that v and r are perpendular r ros( θ + rsn( θ v vsn( θ + vos( θ r v ( os( sn( ( sn( os( r θ + r θ v θ + v θ ros( θ ( vsn( θ + rsn( θ vos( θ
MC Qld-3 60 Vetors rvos( θsn( θ + rvos( θsn( θ 0 Sne the dot produt s zero, r and v must e perpendular. show that L 0 L r mv ( ros( θ + rsn( θ m( vsn( θ + vos( θ ( ros( θ + rsn( θ ( mvsn( θ + mvos( θ ros( θ rsn( θ 0 mvsn( θ mvos( θ 0 r sn( θ 0 mv os( θ 0 r os( θ 0 mvsn( θ 0 ros( θ rsn( θ + mvsn( θ mvos( θ 0 0 + r ( os( θ mvos( θ + rsn( θ mvsn( θ mvr(os ( θ + sn ( θ mvr So L ( mvr mvr, as requred. v s halved d v s douled e As her arms move to her sde, r dereases whh auses a orrespondng nrease n v due to onservaton of angular momentum. Ths auses her to spn faster. a The stuaton desred s shown n the dagram elow It an e seen from the dagram that d wll gve the shortest dstane from the orgn to the plane. nop n OP os( α d But os(α OP d OP os( α So nop nd d nop n Therefore the shortest dstane etween the plane and the orgn s gven y nop n x + 3y z n + 3 P (,, 7 (any pont that satsfes the equaton ( + 3 ( + 7 Shortest dstane + 3 + ( + 3 ( 7 4+ 9+ ( + 3 + 7 4 4 3. 3 a If n s normal to the frst plane and n normal to the seond, the lne of nterseton (whh s on oth planes must e normal to oth n and n, The ross produt n n fnds a vetor that s normal to oth n and n, so t must e parallel to the lne of nterseton of the two planes. Therefore, n n gves the dreton of the lne of nterseton of the two planes. x 3y + z 0 and 3x y + z 3x y z 0 Let n 3+ and n 3 n n 3 3 3 3 + 3 3 (3+ ( 3 + ( + 9 4 + 5+ 7 Chapter revew a Let the vetor e a x + y omponent: x os( a θ 0 os(40 5.3 omponent: y sn( a θ 0 sn(40.9 a 5.3 +.9 Let the vetor e a x + y omponent: x os( a θ 9 os( 50 5.8 omponent: y sn( a θ 9 sn( 50 6.9 a 5.8 6.9 Let the two vetors e a and, and r a 0os(40 + 0sn(40 5.3 +.86 4 os( 5 + 4 sn( 5 38.06 7.75 r 5.3+.86 + 38.06 7.75.74 4.89
Vetors MC Qld-3 6 r.74 + ( 4.89 3.3 4.89 tan(θ.74 θ. The resultant vetor has magntude 3.3 n a dreton of. south of east. 3 a 3+, 4 + 3 a ( 3 + (4 + 3 4 3 3+ ( 4 9 4 9 4 3 + 3+ and n + 3 are perpendular (3 + 3 + ( n + 3 0 3 n + 3 + ( 3 0 3n + 3 3 0 3n 0 n 0 5 Let the three fores e, and 3 and the resultng fore e R so R + + 3 5os(30 + 5sn(30.99 + 7.5 5 3 0os(80 + 60 + 0sn(80 + 60 0os(40 + 0sn(40 5 8.66 R + + 3.99 + 7.5 5 5 8.66 (.99 5 5 + (7.5 8.66 7.0.6 R ( 7.0 + (.6 7 N.6 tan(θ (3 rd quadrant 7.0 80 + 3.9 or 3.9 south of west The resultng vetor has magntude 7N n a dreton 3.9 south of west. 6 a a sn( θ n ˆ 7 sn(00 nˆ 8.7nˆ 8.7 out of the page. 7 a ( + + (3 + 4 3 3 4 3 4 3 + 3 3 3 4 ( 3 8 ( 3 6 + (4 3 + 9 + ( + 3 + (3 + + 4 3 3 4 3 3 + 4 3 4 3 ( 4 (8 6 + (4 9 8 5 8 A (,, 0, B (3,,, C (4,, AB (3 + ( + ( 0 + + BC (4 3 + ( + ( 4+ A a AB BC AB BC 4 4 + 4 ( + 4 ( + ( 4 5 5 So A 5 5 5 + ( 5 5 + 5 50 3.5 square unts. 9 + + 3 r + Torque r ( + ( + + 3 ( + ( + + 3 0 3 0 0 3 + 3 (6 0 (6 0 + (4 6 6+ So torque 6 6+ 6 + ( 6 + 36 + 36 + 4 76 8.7 Nm 0 60os(80 + 60sn(80 0.4 + 59.09 OP 0.6os( 40 + 0.6sn( 40 0.46 0.39
MC Qld-3 6 Vetors Torque r OP (0.4 + 59.09 (0.46 0.39 (0.4 + 59.09 (0.46 0.39 0.4 59.09 0 0.46 0.39 0 59.09 0 0.39 0 0.4 0 0.46 0 0.4 59.09 + 0.46 0.39 0 0 + ( 0.4 0.39 59.09 0.46 3. So torque 3. ( 3. 3. Nm Let ˆn e a unt vetor perpendular to a and If a + and 3 + + So ˆn a a a 3 + 3 3 ( + ( + 3 + ( 6 3 4 5 So ˆn (3 4 5 3 4 5 (3 4 5 3 + ( 4 + ( 5 (3 4 5 9+ 6+ 5 (3 4 5 50 3 4 5 50 50 50 0.4 0.57 0.7 Show that a ( a a where a +, 3 and LHS a ( 3 0 0 3 0 0 3 + 0 0 (6 0 ( 4 0 + (0 + 6 6 + 4+ 6 LHS a (6 + 4 + 6 ( + (6 + 4+ 6 6 4 6 + 4 6 6 6 6 4 ( + 4 (6 + 6 + (4 6 8 RHS a a ( ( + ( 3 ( ( 3 + ( 0 ( + ( 3 ( 3 0 + ( ( + ( 3 ( 6( 4( 3 + 4( 8 + 8 8 6 8 LHS, as requred. 3 a a + +, +, + 3+ 4 a 3 4 + 3 4 4 3 (8+ 3 (4 + + (3 5 + 9 a +,, + + 3 a 0 3 0 0 3 3 (6 + (0 + (0 4 7 + 8 3 4 a + +, 3 +, 3 + n+ 4 a 0 3 0 0 3 n 4 0 3 0 3 + n 4 3 4 3 n 0 (8 0 ( 0 + (3n 6 0 6 + 6n 0 6n 8 0 n 8 6 n 4 3 5 Let the orgn e at the ottom left-hand orner wth the x-, y- and z-axes defned as usual. V a
Vetors MC Qld-3 63 Let a 7 5sn(0 + 5os(0 5.303 + 4.0954 os(80 sn(80 3.803.6658 7 0 0 a 5.303 4.0954 0 3.80 0.6658 4.0954 0 7 0 0 0.6658 + 7[4.0954 (.6658] 59 u unts 6 The vetors are oplanar. 7 P (, 0,, Q ( 3,.3, R (, 4, 0 PQ ( 3 + ( 0 + (3 5 + QR ( 3 + (4 + (0 3 5 + 5 3 Let n e a normal to the plane ontanng P, Q and R. n PQ QR 5 5 5 3 5 5 + 5 3 5 3 5 5 (3 5 (5 5 + ( 5 + 5 0 0 n (x a + n (y + n 3 (z 0 (x 0(y 0 0(z 0 x + 4 0y 0z + 40 0 x 0y 0z + 44 0 x 0y 0z 44 x + 5y + 0z 8 n + +, pont (0, 3, n (x a + n (y + n 3 (z 0 (x 0 + (y 3 + (z 0 x + y 3 + z 0 x + y + z 5 0 x + y + z 5 9 x y + z 6 Let n e a normal to the plane and ˆn a normal unt vetor. n + ˆn n n + + ( + ( + + 4 + ( 6 + Modellng and prolem solvng a The magntudes of the horzontal omponents of the two fores are equal and opposte so T L os(30 T S sn(45 TS os(30 TL sn(45. T S 600N TS T. L T S T L. 600N. 490N W d os( θ where 80N, d 6m and θ 3 So W 80 6 os(3 407N 3 a Let the orgn e at the pvot of the seesaw wth the x- and y-axes taen as usual and the postve z-axs out of the page. OA. S 800 Torque r OA S. ( 800 960 960 960 Nm Steve s weght exerts a torque of 960 Nm Torque 960 Nm A 650 Let OB d where d s Andrea s dstane from the entre (orgn when she provdes a torque equal (and opposte to Steve s. Torque r Torque OB S 960 d ( 650 960 650d 960 ( 650 d 960 650d d.48 m Andrea needs to e.48 m from the entre. 4 Let the orgn e at A wth the x-, y- and z-axes taen as usual. AB AC 3 + A a AB AC ( (3
MC Qld-3 64 Vetors ( (3 + 0 3 0 0 0 + 0 3 0 3 (0 + (0 + 6 + (0 3 6 3 So A 6 3 + ( 6 + ( 3 4+ 36+ 9 49 7 3.5 m 5 Note the dagram elow s not to sale Let the postve dreton of the x-axs e along vetor OE and the postve dreton of the y-axs e along vetor OA. The postve dreton of the z-axs s perpendular to the x- axs and along the ground as shown n the dagram elow. 5 0.5 + 8.66 AC AO+ OC OA + OE + EC 3 + 0 + 0 AB AC 0 0 5 0.5 8.66 Therefore 0.5 8.66 5 8.66 5 0.5 + 0 0 0 0 (0 + 8.66 (0 86.6 + ( 5 + 5 8.66 + 86.6 A 8.66 + 86.6 8.66 + 86.6 43.5 m 6 a If a p, a p If a 4. a a ( a 4 The magntude remans onstant. A lowse orderng of a, and gves the same answer. An antlowse orderng of a, and gves the negatve of a. 7 m 3 v + r 3 L r mv ( 3 3( + ( 3 (3 + 6 3 0 3 6 0 A a AB AC Pont A s dretly aove pont O, pont B s dretly aove pont D and pont C s dretly aove pont E so OA 3 DB.5 EC And from the aove dagram OE 0 OD 0os(60 + 0sn(60 5 + 8.66 AB AO+ OB OA + OD + DB 3 + 5 + 8.66+.5 3 0 0 3 + 6 0 3 0 3 6 0 0 + ( + 9 8 Let AB,, AC AD d, et AC + CE AE AC + CB AE 3 ( + + AC CA AB AE 3 ( + AC AB AC AE 3 + ( 3 e + 3 3 e e + 3 3 []
Vetors MC Qld-3 65 Let AE dvde BD n a rato m:n m BI BD m+ n m ( d m+ n m m d + m+ n m+ n m ( m+ m+ n d + m+ n m+ n m n d + m+ n m+ n m n m+ n 3 + m+ n But les on the lne AE, therefore we an wrte or e AL AE Susttutng [] and []: m n + m+ n 3 m+ n + 3 3 n Equatng oeffents of results n m+ n 3 3n (m + n [3] Equatng oeffents of results n m 3 m+ n 3 m (m + n [4] [3] + 3 [4] : 3n + 3 m (m + n + 3 (m + n 6m + 6n 3(m + n + 4(m + n 6(m + n 3(m + n + 4(m + n 6 3 + 4 6 7 Therefore 6 7 e or We have ust shown that 6 CG C 7 6 AI 7 AE 6 7 e 7 6e 6 + 3 3 4+ CG 6 7 C 7CG 6C 7( g 6( f 7g 7 6 f 6 7g 6f + 6 7 6 AI AE 7 6 AI AE. In a smlar fashon 7 [] 6 + 3 + (4+ (7 Ths means that g AG AL But AI AG + GI AG AI + GI So AI GI AI GI AI 6 7 AE 3 7 AE 3 We have ust shown that GI 7 AE. In a smlar fashon 3 GH C 7 The area of ABC The area of GHI GI GH 3 AE 3 C 7 7 3 3 ( 7 7 e f 9 + 98 3 3 3 9 98 3 9 + 3 9 9 98 3 + 9 9 7 98 9 7 98 4 7 the area of ABC 7