21
.5 ก 3 (ก) 4 8 AD = AB + BC + CD = - u + v - w FD = FE + ED = - u + v BD = BC + CD = v - w FC = FA + AB + BC = w - u + v 6 9 = NM NC + CB + BM = 1 2 AC + CB + 1 2 BD = 1 2 ( AB + BC29 + CB + 1 2 ( BC + CD29 = 1 2 AB + 1 2 BC + CB + 1 2 BC + 1 2 CD = NM 1 2 AB + 1 2 CD (129 = NM NA + AD + DM = 1 2 CA + AD + 1 2 DB = 1 2 ( CB + BA29 + AD + 1 2 ( DA + AB29 = 1 2 CB + 1 2 BA + AD + 1 2 DA + 1 2 AB = NM 1 2 CB + 1 2 AD (229 (1) + (2) : 2 NM = 1 2 AB + 1 2 CD + 1 2 CB + 1 2 AD 2 NM = 1 2 ( AB + CD + CB + AD29 4 NM = AB + CD + CB + AD AB + AD + CB + CD = 4 NM R = 4 A B C D E F v u w A B C D N M 1

Ma5 vector-u-s54

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Page 1: Ma5 vector-u-s54

��������� �.5�������� ������ ��ก���� 3 ����(������ �ก!"�#���$%&���������'��#)

��� 4 ���� 8

�� ��� AD = AB + BC + CD = − u + v − w

FD = FE + ED = − u + v

BD = BC + CD = v − w

FC = FA + AB + BC = w − u + v

��� 6 ���� 9

�� ��� ��ก��� =NM NC + CB + BM

= 1

2AC + CB + 1

2BD

= 1

2(AB + BC) + CB + 1

2(BC + CD)

= 1

2AB + 1

2BC + CB + 1

2BC + 1

2CD

=NM1

2AB + 1

2CD (1)

��ก��� =NM NA + AD + DM

= 1

2CA + AD + 1

2DB

= 1

2(CB + BA) + AD + 1

2(DA + AB)

= 1

2CB + 1

2BA + AD + 1

2DA + 1

2AB

=NM1

2CB + 1

2AD (2)

(1) + (2) : 2NM = 1

2AB + 1

2CD + 1

2CB + 1

2AD

2NM = 1

2(AB + CD + CB + AD)

����� �������4NM = AB + CD + CB + AD AB + AD + CB + CD = 4NM

∴ R = 4

A

B

C

D

E

F

vu w

A

B

CD

NM

1

��������� �� ������� ����������������

Page 2: Ma5 vector-u-s54

��� 3 ���� 10

�� ��� ��กก$%�& cosine b2 = a2 + c2 − 2ac cos 60

b2 = 102 + 52 − 1001

2 = 75

b = 75 = 5 3 km

-�./. ��กก$ :θ cos ine a2 = b2 + c2 − 2bc cos θ

52 = (5 3 )2 + 102 − 2(5 3 )(10)cos θ

cos θ = 3

2→ θ = 30

∴ 0%����1�2-2�&��ก�/�0�3�.4��0�5���1�6�& 7�638 5 3 km 060

��� 5 ���� 14 ��� 2�� ��� ��ก��� =AH AF + FH

= 2FD + 1

2CF

= 2u + 1

2v

��� 6 ���� 15 * :ก�0;<16=�>��7�-��&?�1:ก� 0�5� *OC OE

��� 8 ���� 15 ��� 1�� ��� ��ก��� u = AB + BM (1)

v = 1

3AB + 5BM (2)

(1) × 5 : 5u = 5AB + 5BM (3)

(3) − (2) : 5u − v = 14

3AB

AB = 3

14(5u − v)

AB = 15u

14− 3v

14

B

A

C

a = 5 km

b

c = 10

km30 + 0 = 60

30H

150H

30H 30H

0H

B D C

A

EH F2

221 1

1

AH

D

A B

CN1 2

4

1

v

u

5BM

AB13

M

2

��������� �� ������� ����������������

Page 3: Ma5 vector-u-s54

��� 9 ���� 16

�� ��� ��ก��� =EF EC + CF

= 2

3u + (−1

3)(u + v)

= 2

3u − 1

3u − 1

3v

= 1

3u − 1

3v

=EF1

3(u − v)

��� 11 ���� 16

�� ��� ��ก��� =BF AF − AB

= 1

2AE − u = 1

2v + 1

2u

− u

= 1

2v + 1

4u − u = 1

2v − 3

4u

��� 13 ���� 17

�� ��� ��ก��� ?�1:42<�����%�& ∆AFD ∼ ∆BFE

1�L0�5� 2 062�%�& ∆AFD ∆BFE

��ก>�4��%L� = AF1

3(AD + 2AB) = 1

3(u + 2v)

= 1

3u + 2

3v

��� 14 ���� 17 ��� 2�� ��� ��ก��� ∆AEF ∼ ∆CDF

����� AE = 3DC CF = 3FA

AD = 3

2a − 4b

2

2

1

1

C

E

F

v

EF

u

u + v

D

A B

C

F

E

D C

BA

F E

v

u 2

1

D C

BA

F

E1 2

ba

a32

a12

3b

3

��������� �� ������� ����������������

Page 4: Ma5 vector-u-s54

��� 4 ���� 18 ��� 2�� ��� AF = mAP → BF = nBQ

= AF mAP

= m(AB + BP)

= m(AB + 4

7BC)

=AF mAB + 4

7mBC..........(1)

= = = = BF nBQ n(BC + CQ) n(BC + 4

9CD) n [ BC + 4

9(−AB) ]

= = = BF nBC − 4

9nAB → −FB nBC − 4

9nAB → FB

4

9nAB − nBC .......(2)

= (1) + (2) ; AF + FB mAB + 4

7mBC + 4

9nAB − nBC

= AB mAB + 4

7mBC + 4

9nAB − nBC

= AB − mAB − 4

9nAB

4

7mBC − nBC → AB(1 − m − 4

9n) = BC(4

7m − n)

0����&��ก ∴ >.ก���=M��0�5���3&0.��� :<� AB //BC 1 − m − 4

9n = 0

4

7m − n = 0

:ก�>.ก����� :<� ∴ m = 63

79n = 36

79m + n = 99

79

��� 16 ���� 19 ��� 9�� ��� O3���P� ∆ABC

��ก>�4� >�.&2�. �����BD = 1

3 + 1(1BC + 3BA)

BD = 1

4BC − 3

4AB

��ก��� DF = BF − BD = 1

3BC −

1

4BC − 3

4AB

= 3

4AB + 1

12BC

∴ a

b=

34

112

= 9

B C

A D

FQ

4

5

4 3P

3

1 1

2

C

A B

FBF = BC1

3

a a

4

��������� �� ������� ����������������

Page 5: Ma5 vector-u-s54

��� 1 ���� 23 ��� 2�� ��� =2

5u + (6 − 3x2)v 100u + 2

3v

=(2

5− 100)u (2

3− 6 + 3x2)v

=u (3x2 − 163

)

(25

− 100)

v

0����&��ก .=S<<�OTU0�5�<V ��&��M� 2

5− 100 3x2 − 16

3< 0

�W� 3 -��4<��>.ก������� x2 − 16

9< 0

x2 − 4

3

2< 0

(x − 4

3)(x + 4

3) < 0 x ∈ (−4

3,

4

3)

��� 3 ���� 24 ��� 3�� ��� ��ก?�61U =x2v − xv 2u

=(x2 − x)v 2u → u = (x2 − x)2

v

0����&��ก ก�V .=6384�&%��.ก�� �����L2�u v

x2 − x

2< 0 → x2 − x < 0 → x(x − 1) < 0

0%=1�X2L&%�&�W�4�V���0�5� 0 < x < 1

��� 5 ���� 24 ��� 3�� ��� 1. au + v = 3v − u → au + u = 2v → (a + 1)u = 2v

�.2>�.��Z>�/����L2� ก�V .=6384�&%��.ก��u v

2. 3u − 2v = 4v − 2u → su = 6v → u = 6

5v

∴ ก�V .=6380�=1Lก��u v

3. 3

2v = v − u → 1

2v = − u → v = − 2u

∴ ก�V .=6384�&%��.ก��u v

4. ∴ ก�V .=6380�=1Lก��2v = u + v → u = v u v

4��&0�5�VLก0O��� u ก�V v .=6380�=1Lก��

34

34

0 1

5

��������� �� ������� ����������������

Page 6: Ma5 vector-u-s54

��� 1 ���� 25

�� ��� 8x + r(y − x) = 2s(y + x)

8x + ry − rx = 2sy + 2sx

8x − rx − 2sx = 2sy − ry

(8 − r − 2s)x = (2s − r)y

0.��� // �����L2� :<� x y 8 − r − 2s = 0 − (1) 2s − r = 0 − (2)

:ก�>.ก�� (1) :<� (2) ��� r = 4 :<� s = 2 ∴ rs = (4)(2) = 8

��� 5 ���� 27

�� ���

=ZB mYB = m(AB − 2

5v) = mu − m

2

5v ..........(1)

=ZB XB − XZ = 3

5u − nXC = 3

5u − n(AC − AX) = 3

5u − n(v − 2

5u)

= 3

5u − nv + n

2

5u = (3

5+ 2

5n)u − nv ..........(2)

(1) = (2) : mu − m2

5v =

3

5+ 2

5n

u − nv

06=1V>�.���>36T3\OVL2� :<� m = 3

5+ 2

5n ..........(3) m

2

5= n ..........(4)

�W��2� n ��ก (4) :6�7� (3) ����� m = 3

5+ 2

5(m2

5) → m = 5

7

�W��2� m :6�7� (4) ����� n = 2

7

�W��2� m :<� n :6�7� (1) ����� ZB = 5

7u − 2

7v

��� 1 ���� 28 ��� 3�� ��� 7-� =w au + bv ..........(1)

=−7i − 14j a(i − 4j) + b(3i + 2j) = ai − 4aj + 3bi + 2bj = (a + 3b)i + (2b − 4a)j

��&��M� = :<� a + 3b −7 2b − 4a = − 14

:ก�>.ก����� ∴a = 2 , b = − 3 w = 2u − 3v

X

A

YZ

B C

6

��������� �� ������� ����������������

Page 7: Ma5 vector-u-s54

��� 3 ���� 29

�� ��� =w mu + nv

=−14i + 14j m(2i + 3j) + n(4i − j)

= 2mi + 3mj + 4ni − nj

=−14i + 14j (2m + 4n) i + (3m − n)j

����� :<�2m + 4n = − 14 ..........(1) 3m − n = 14 ..........(2)

:ก�>.ก�� (1) :<� (2) ��� m = 3 :<� n = − 5

∴ m2 + n2 − 25 = 32 + (−5)2 − 25 = 3

��� 3 ���� 34

�� ��� 0����&��ก0Lก04��U .=�L�.X�� ��� 3i − 4j −4

3

�L�.X��%�&0>��4�&6=�4�M&;�กก�V ��� 3i − 4j3

4

>.ก��0>��4�&6=�S2���/� :<�4�M&;�กก�V ���(−2, 1) 3i − 4j

���������0�5� y − 1 = 3

4(x − (−2)) 4x + 3y + 5 = 0

��� 4 ���� 35 ��� 4�� ��� Z�� u = 2i − 3j mu = − 3

2m u = 2

3

>.ก��0>��4�&6=�S2���/� :<�4�M&;�กก�V ���(2, −2) u

���������0�5� y − (−2) = 2

3(x − 2) 2x − 3y − 10 = 0

��� 6 ���� 35 ��� 3�� ��� AC = AB + AD

AC = (2i + 3j) + (5i − 2j)

AC = 7i + j

AC = 72 + 12 = 50 = 5 2

D

A

C

B

5i - 2

j

2i + 3j

7

��������� �� ������� ����������������

Page 8: Ma5 vector-u-s54

��� 10 ���� 37 ��� 3�� ��� ������� ��ก>�4�>�.&2�.∆ABC

CD = 1

2 + 32(3i − 2j) + 3(2i + 3j)

CD = 1

56i − 4j + 6i − 9j

CD = 1

512i + 5j

CD = 1

5122 + 52 = 1

5(13) = 13

5

��� 11 ���� 37 ��� 1�� ��� ������� ��ก>�4�>�.&2�.∆AOB

=OC1

33i − 2j + 2(2i + 5j)

= 1

33i − 2j + 4i + 10j

=OC1

37i + 8j

= ∴ OC1

3113 OC

2 = 113

9

��� 9 ���� 42 ��� 3�� ��� Z�� 0Lก04��U 1 -�2L1 6=�4�M&;�กก�V ��� u = i + 2j u ± (2i − j)

5

0����&��ก 0�5�0Lก04��U%��� -�2L1 .=6384�M&;�กก�V AB 5 u

��&��M� AB = ± 5 (2i − j)5

= ± (2i − j) = ±

2

−1

(1)

��ก?�61U A(1, 0) :<�7-� B .=O3ก�� (x, y) ����� AB =

x −1

y −0

(2)

(1) = (2) : ±

2

−1

=

x −1

y − 0

:ก�>.ก����� (x, y) 2 X/� ��� :<� (−1, 1) (3, −1)

D

A

CB 2i + 3j

3i - 2j3

2

A

O B

C

2

1

2i + 5j

3i - 2j

8

��������� �� ������� ����������������

Page 9: Ma5 vector-u-s54

��� 10 ���� 42

�� ��� 7-� .=�/�0�3�.4����� (0, 0) ����� CA =

−7

14

A(−7, 14)

:<� .=�/�0�3�.4����� (0, 0) ����� CB =

0

−35

3

B(0, −35

3)

=CD CA + AD

= CA + 3

7AB

= CA + 3

7(CB − CA)

= 4

7CA + 3

7CB

= 4

7(−7i + 14j) + 3

7(0i − 35

3j)

=CD −4i + 3j

��&��M�0Lก04��U 1 -�2L1 6380�=1Lก�V =CDCD

CD= − 4i + 3j

5= − 4

5i + 3

5j

��� 6 ���� 49

�� ��� 7-� w = ai + bj

u ⋅ w = − 11 → 2a − 5b = − 11 (1)

v ⋅ w = 8 → 1a + 2b = 8 (2)

(2) × 2 : 2a + 4b = 16 (3)

(3) − (1) : 9b = 27 → b = 3

�W��2� b :6�7� (2) ����� a = 2 ∴ w = 2i + 3j

w = 22 + 32 = 13

0����&��ก =w ⋅ (5i + j) w 5i + j cos θ

=(2i + 3j) ⋅ (5i + j) 13 26 cos θ

=cos θ 1

2→ θ = 45

∴ tan θ + sin 2θ = tan 45 + sin 2(45 ) = 1 + 1 = 2

A(-7,14)

C(0,0)D(a,b)

4

3

B(0,- )353

9

��������� �� ������� ����������������

Page 10: Ma5 vector-u-s54

��� 12 ��� 51 ��� 1 ����� ก. ��ก��� �� ���กก�� ����� u v u ⋅ v = 0

��ก��� �� ���กก�� ����� u + v u − v (u + v) ⋅ (u − v) = 0

���������� ก. ��ก→ u 2 − v 2 = 0 u = v

����� = 0 u = v

�. �. ��ก(u + 2v) ⋅ (2u − 2v) = 2 u 2 + 3u ⋅ v − 2 v 2 = 0

��� 13 ��� 52

����� ก�� ����� u i − 2j mu = 1

2

m = − 2

����!" #�$ �"$% ���ก �� b > 0 �)�*#� ��!" u 2i + j −2i − j u = ai + bj u 2i + j

#�+�+��#���� ก�� ��",���ก�� ��� �� �+� dot ก�" u i + j

����� =u ⋅ (i + j) u i + j cos θ

=(2i + j) ⋅ (i + j) 2i + j i + j cos θ

=2(1) + 1(1) 5 2 cos θ

=cos θ 3

10

��ก��� �����∆ tan θ = 1

3

∴ 9 tan θ = 9

1

3 = 3

��� 14 ��� 52 ��� 2 ����� ��ก��� = 3 ����� b a b a ⋅ b = 0

= 3(−2p)2 + 22 + p2 (1)(−2p) +

1

2 (2) + (−3p)(p) = 0

= 35p2 + 4 −2p + 1 − 3p2 = 0 (1)

= 1 ",�3�� 4�"*" (1) �����p2 p2

−2p + 1 − 3(1) = 0 → p = − 1

∴ 3�� p �8%�,�*#� 49� a b b = 3

3$ :)%� ��*";��� −1 (−3

2, 0)

= 0

1

3

100

10

��������� �� ������� ����������������

Page 11: Ma5 vector-u-s54

��� 18 ��� 54 ��� 2 ����� =u + v 2 + u − v 2 2 u 2 + 2 v 2

=37 + 13 2( u 2 + v 2)

=u 2 + v 2 25 (1)

=u + v 2 − u − v 2 4u ⋅ v

=37 − 13 4 u v cos 60

=2 u v 24 (2)

(1) + (2) : = 25 + 24u 2 + 2 u v + v 2

= 49( u + v )2

= 7u + v

��� 23 ��� 56 ��� 2 ����� ��ก��� = ��ก��� x + 5y = 21u − v 2 u 2 + v 2

= = 21u − 2u ⋅ v + v 2 u 2 + v 2 5y

4+ 5y

= 0 y =−2u ⋅ v12

5

= 0 ",�3�� y 4�"*" (1) �����u ⋅ v

= 04x − 3y x = 3

4

12

5 = 9

5

x = ���"� "3y

4(1) u = xi + yj = 9

5i + 12

5j

∴ u ⋅ w = 9

5(2) + 12

5(1) = 6

��� 28 ��� 59

����� u + 4v = 3v − 2w → u = − v − 2w (1) : (1) × 3 : 3u = − 3v − 6w (3)

4u − 2w = 3v + 2w → 4 u = 3v + 4w (2) : (1) × 4 : 4u = − 4v − 8w (4)

(2) + (3) : 7u = − 2w → 7u = − 2w → 7 u = 2 w → 7 u = 2(14) → u = 4

(2) = (4) : 3v + 4w = − 4v − 8w → 7v = − 12w → 7v = − 12w → 7 v = 12 w

∴→ 7 v = 12(14) → v = 24 v 2 − w 2 + u 2 = 242 − 142 + 42 = 396

12

11

��������� �� ������� ����������������

Page 12: Ma5 vector-u-s54

��� 31 ��� 61

����� a + b + c 2 = a 2 + b 2 + c 2 + 2a ⋅ b + 2b ⋅ c + 2c ⋅ a = 0 (1)

",�3�� 49� 4�"*" (1) �����a + c 2 = − b 2 a , b c

a + b + c = 0 10 = b 2 9 + 10 + 25 + 2(a ⋅ b + b ⋅ c + c ⋅ a) = 0

a + b 2 = − c 2 a ⋅ b + b ⋅ c + c ⋅ a = − 22

25 = c 2

��� 32 ��� 61 ��� 3 ����� 1. DE� ����� (u ⋅ v)2 = u 2 v cos2θ = (u ⋅ u)(v ⋅ v)cos2θ

�"$% ���ก 0 ≤ cos2θ ≤ 1

���"� " (u ⋅ v)2 ≤ (u ⋅ u)(v ⋅ v)

2. DE� ����� ��� 49� u ≠ 0 , v ≠ 0 u v

����� u ⋅ v = 0 → (u ⋅ v)2 = 0

4�� ( u v )2 ≠ 0

∴ (u ⋅ v)2 ≠ ( u v )2

3. ��ก ����� =u + v + w 0

=u + v w

=u + v 2 w 2

=u 2 + 2u ⋅ v + v 2 w 2

=32 + 2u ⋅ v + 42 72

= 12u ⋅ v

4. DE� ����� =u − v 2 u 2 − 2u ⋅ v + v 2

�+$% ����+����� ก��� u v

�+���+������������� u − v 2 = u 2 + v 2

12

��������� �� ������� ����������������

Page 13: Ma5 vector-u-s54

��� 2 ��� 79

�����

��� 2 ��� 82

����� ��ก��� AB =

−2 − 1

3 − 2

1 − 0

=

−3

1

1

, AB = (−3)2 + 12 + 12 = 11

��ก�� � 3 #"�� �EF������+ 3$ AB −3AB

AB= −3

11

−3

1

1

��� 3 ��� 85 ��� 1 ����� ��ก��� r4 = ar1 + br2 + cr3

3

2

5

= a

2

−1

1

+ b

1

3

−2

+ c

−2

1

−3

�������� 3 = 2a + b − 2c (1)

2 = − a + 3b + c (2)

5 = a − 2b − 3c (3)

:(2) × 2 4 = − 2a + 6b + 2c (4)

:(1) + (4) 7 = 7b → b = 1

b (2) : 2 = − a + 3(1) + c → − 1 = − a + c (5)

b (3) : 5 = a − 2(1) − 3c → 7 = a − 3c (6)

:(5) + (6) 6 = − 2c → c = − 3

c (6) : 7 = a − 3(−3) → a = − 2

∴ a + b + c = (−2) + 1 + (−3) = − 4

4

31 y

x

zP(1, 3, 4)

13

��������� �� ������� ����������������

Page 14: Ma5 vector-u-s54

��� 2 ��� 87

����� a + c =

2 + d

4 + d + 1

−5 + d + 2

=

d + 2

d + 5

d − 3

b =

3

6

−2

����� �:��#���� �3��8% 4ก���� 3

d + 2= 6

d + 5= −2

d − 3d = 1

∴ c =

1

2

3

c = 12 + 22 + 32 = 14

��� 3 ��� 88

����� ��ก��� �����u ⋅ v = 2 1a + 2b + 3c = 2 (1)

�"$% ���ก w = ai + bj + ck // − 2

3i + 1

2j + 1

3k

����� a

− 23

= b12

= c13

→ − 3

2a = 2b = 3c (2)

",�3�� 2b , 3c ��ก (2) 4�"*" (1) �����1a +

−3

2a +

−3

2a = 2 → a = − 1

",�3�� a 4�"*" (2) ����� 3

2= 2b = 3c → b = 3

4, c = 1

2

∴ a + 4b + 2c = − 1 + 4

3

4 + 2

1

2 = 3

��� 2 ��� 90

����� AB =

2 −1

0 −2

−3 +1

=

1

−2

−2

, AB = 12 + (−2)2 + (−2)2 = 3

�3�:"4����EF���� ���ก�� � 3$ AB1

3,

−2

3,

−2

3

��� 3 ��� 93

����� = = , = = AB (3 + 1)i + (1 − 2)j + (−2 − 1)k 4i − j − 3k AB 42 + (−1)2 + (−3)2 26

= = , = = AC (5 + 1)i + (3 − 2)j + (−2 − 1)k 6i + j − 3k AC 62 + 12 + (−3)2 46

= = 32AB ⋅ AC (4)(6) + (−1)(1) + (−3)(−3)

��ก = 32 = AB ⋅ AC AB AC cos θ → 26 ⋅ 46 cos θ

= = = cos θ 16

13 23

16

299→ θ cos−1 16

299

14

��������� �� ������� ����������������

Page 15: Ma5 vector-u-s54

��� 5 ��� 94

����� ��ก�� �3���8%�� ���กก�" ������ dot ก�" D99��M�� ���!"F�"��ก��� , , u = i − 2j + 5k v = 7i − 2j + 2k w = 2i − 3j + 4k

∴ ก�� �+��� ���กก�"u ⋅ v = (1)(7) + (−2)(−2) + (5)(2) = 21 u v

∴ ก�� �+��� ���กก�"u ⋅ w = (1)(2) + (−2)(3) + (5)(−4) = -24 u w

∴ ก�� �� ���กก�"v ⋅ w = (7)(2) + (−2)(3) + (2)(−4) = 0 u w

��� 8 ��� 94

����� ก,�#"���� A, B 49� C ��!"��� �� ���+�#98%+���� ���+���+8�Eก�����"8 49� A(4, 9, 1), B(−2, 6, 3) C(6, 3, −2)

����� AB = (−2 − 4)i + (6 − 9)j + (3 − 1)k = − 6i − 3j + 2k

BC = (6 + 2)i + (3 − 6)j + (−2 − 3)k = 8i − 3j − 5k

AC = (6 − 4)i + (3 − 9)j + (−2 − 1)k = 2i − 6j − 3k

ก���8% ����!" +�+��ก������� �+8��ก�� � 1 3���8%�� ���กก�" ∆ ABC ∆

(dot ก�"49����� 0) ���"� "ก������� �ก����!" +�+��ก���� �9 �� ���ก�� ��8%9�3��∆

+� dot ก�"AB ⋅ BC = (−6)(8) + (−3)(−3) + (2)(−5) = − 49 ≠ 0

AB ⋅ AC = (−6)(2) + (−3)(−6) + (2)(−3) = 0

BC ⋅ AC = (8)(2) + (−3)(−6) + (−5)(−3) = 49 ≠ 0

��กก�� dot ��ก�� ������ +8��ก�� � 1 3�� dot ก�"49�����F�" 4������+8���"� � ∆

�8%�� ���กก�" ∴ ��!" +�+��ก∆ABC ∆

��� 10 ��� 95

�����

BC = (−1 − 2)i + (−6 + 1)j + (4 + 4)k = 3i − 5j + 8k

BC = (−3)2 + (−5)2 + 82 = 98 = 7 2

BA = (5 − 2)i + (−3 + 1)j + (2 + 4)k = 3i − 2j + 6k

BA = 32 + (−2)2 + 62 = 7

BA ⋅ BC = BA BC cos θ

(3)(−3) + (−2)(−5) + (6)(8) = (7)(7 2 )cos θ → cos θ = 1

2→ θ = 45

A(5, -3, 2)

B(2, -1, -4) C(-1, -6, 4)

15

��������� �� ������� ����������������

Page 16: Ma5 vector-u-s54

��� 2 ��� 98

����� = , = AB

1 − 3

0 − 1

−1 − 2

=

−2

−1

−3

AC

−2 − 3

2 − 1

3 − 2

=

−5

1

1

−5k 3 i 2 j

= = AB × AC

i j k

−2 −1 −3

−5 1 1

i j

−2 −1

−5 1

2i + 13j − 7k

− i + 15 j − 2k

��� 1 ��� 100

����� ก) ��ก�� � ��ก�� � ���D99��M��!"��ก�� �(u × v) × w ×

��ก�� � ��ก�� �

�) ��ก9�� ��ก�� � �+�+8"E�+���3RE�F����(u ⋅ v) × w ×

��ก9�� ��ก�� �

3) ��ก�� � ��ก�� � ���D99��M��!"��ก9��u ⋅ (v × w) ⋅

��ก�� � ��ก�� �

�) ��ก�� � ��ก�� � ���D99��M��!"��ก�� �u × (v × w) ×

��ก�� � ��ก�� ���� 2 ��� 100 ��� 1 ����� 1. ��ก ����� (u × v) ⋅ w = u ⋅ (v × w)

2. DE� ����� (u × v) × w ≠ u × (v × w)

3. DE� ����� u ⋅ (v × w) = (u × v) ⋅ w

4�� u ⋅ (v × w) ≠ (u × v) ⋅ (u × w)

4. DE� ����� u × (v × w) ≠ (u × v) × (u × w)

16

��������� �� ������� ����������������

Page 17: Ma5 vector-u-s54

��� 3 ���� 100 ��� 3 ������ 1. ��ก ��� �����กก��������� ���� ������� A × B A A ⋅ (A × B) = 0

2. ��ก ��� =(A − B) × (A + B) A × A + A × B − B × A − B × B

= A × B + A × B = 2(A × B) = (2A) × B

3. ��� ��� =(A − B) × (A − B) A × A − A × B − B × A + B × B

= B × A − B × A = 0

4. ��ก ��� ��!��"��ก ��#$�� CROSS(A + B) ⋅ (A − B) × (A × B)

VECTOR VECTOR VECTORก��ก���,$ -.���/0 dot ก�� �-#� �"4� 56���78��"��ก ��#$�� CROSS ก��/� �$$�9:�0;� VECTOR �78��!��/0 dot ��ก�� �-#� �"4�!�/� �$$�9:�0;� SCALAR

��� 4 ���� 101 ��� 4 ������ 1. ��ก

2. ��ก : u ⋅ (v × r) = (u × v) ⋅ r = − (v × u) ⋅ r = − r ⋅ (v × u)

3. ��ก : � � ,$ - u = r v ⋅ (r × u) = v ⋅ (u × u) = v ⋅ 0 = 0

��� 5 ���� 101 ��� 1 ������ ก. ��ก !�ก�0����?��� �7? �7�$?��@!��:!�/� -�� a + b + c = 0

b × c = b × (−a − b) = − b × a − b × b = a × b

c × a = c × (−b − c) = − c × b − c × c = b × c

∴ a × b = b × c = c × a

?. ��� /7�"�7��"40/� -�� a × b = a × c b = c

��� 6 ���� 101 ��� 2 ������ a ⋅ x = a ⋅ b × c

a ⋅ b× c= a ⋅ b × c

a ⋅ b × c= 1 a ⋅ x + 2b ⋅ y + 3c ⋅ z = 2

2b ⋅ y = 2b ⋅ a × c

a ⋅ b× c= − 2

a ⋅ b× c

a ⋅ b× c= − 2

3c ⋅ z = 3c ⋅ a × b

a ⋅ b× c= 3

a ⋅ b× c

a ⋅ b× c= 3

oo

o o

o

o

A C

B

a b

c

17

��������� �� ������� ����������������

Page 18: Ma5 vector-u-s54

��� 7 ���� 102

������ 1 : u = 2i − j + 0k → u = 22 + (−1)2 + 02 = 5

v = 2i + j + k → v = 22 + 12 + 12 = 6

= u × v =i j k

2 −1 0

2 1 1

i j

2 −1

2 1

= − i − 2j + 4k u × v = (−1)2 + (−2)2 + 42 = 21

��8���!�ก u × v = u v sin θ → 21 = 5 ⋅ 6 sin θ → sin θ = 21

30

������ 2 ��8���!�ก u ⋅ v = u v cos θ

=(2)(2) + (−1)(1) + (0)(1) 5 6 cos θ

=cos θ 3

30

!�ก�0 sin θ = 21

30

��� 9 ���� 103

������ u ⋅ v = 2(−1) + (−1)(1) + 1(−2) = − 5 u ⋅ v = 5

u × v =

2

−1

1

×

−1

1

−2

=

1

3

1

u × v = 12 + 32 + 12 = 11

�-ก� �:���7�?�������ก�� ,$�7���F �����กก���������0�ก��� -�u ⋅ v

,$� .8� u v ± u ⋅ v (u × v)u × v

= ± 5

11

1

3

1

��� 10 ���� 103

������ w × v =

a

b

c

×

1

0

3

=

3b

c − 2a

−b

=

−15

−17

5

!�/� ,$� b = − 5 c − 3a = − 17 (1)

��8���!�ก u w: u ⋅ w = 0 → a + 2b + 4c = 0 → a + 2(−5) + 4c = 0

→ a + 4c = 10 (2)(2) × 3 : 3a + 12c = 30 (3)

∴ (1) + (3) : 13c = 13 → c = 1 b + c = − 5 + 1 = − 4

+2k +0 - 2j

-i + 0 + 2k

3

2130

0

18

��������� �� ������� ����������������

Page 19: Ma5 vector-u-s54

��� 11 ���� 104 ��� 1 ������ AB × AC = AB AC sin θ

AB × AC

AB= AC sin θ (1)

��ก���� AB =

6 − 2

5 − 2

2 − 2

=

4

3

0

, AC =

−4 − 2

0 − 2

−3 − 2

=

−6

−2

−5

AB × AC =

4

3

0

×

−6

−2

−5

=

−15

20

10

= 5

−3

4

2

AB = 42 + 32 + 02 = 25 , AB × AC = 5 (−3)2 + 42 + 22 = 5 29

��ก�� sin θ = CD

AC→ CD = AC sin θ = AB × AC

AB= 5 29

5= 29

��� 12 ���� 104 ��� 1 ������ ����� �� � (1) �����u ⋅ v = u v cos θ a2

3a2 − 3b2 = 3 4a2 + 9b2 1

3 2(5 − b2) − 3b2 = 4(5 − b2) + 9b2

3a2 − 3b2 = 4a2 + 9b2 (1) 10 − 5b2 = 20 + 5b2

� ������ก = 3 �กก�� !�"���!#�"��$%&� �����u

= 3a2 + b2 + 22100 − 100b2 + 25b4 = 20 + 5b2

= 5 a2 + b2 25b4 − 105b2 + 80 = 0

=a2 5 − b2 (2) 25(b2)2 − 105b2 + 80 = 0

�ก�"(ก� ��� b = 1, 3.2

����� b (2) ����� a2 = 5 − 12 = 4 → a = 2

u × v =

a

b

2

×

2a

−3b

−5ab

=

6b

4a

−5ab

=

6

8

−10

C(-4, 0,-3)

A(2, 2, 2) B(6, 5, 2)0

D

19

��������� �� ������� ����������������

Page 20: Ma5 vector-u-s54

��� 2 ���� 106

������ AB = (2 − 1)i + (1 − 0)j + (−2 − 1)k = i + j − 3k

AC = (−1 − 1)i + (2 − 0)j + (0 − 1)k = −2i + 2j − k

2k 6i j

AB × AC =i j k

1 1 −3

−2 2 −1

i j

1 1

−2 2

= 5i + 7j + 4k

−i 6j 2k

5.�. ABCD = AB × AC = 52 + 72 + 42 = 90 = 3 10

5.�. (5.�. ABCD) = 7���8 �9�∆ABC = 1

2

1

2(3 10 ) = 3

210

��� 3 ���� 106

������ 5.�. ABCD = AB × AD

AB =

3 −2

0 −1

1 +1

=

1

−1

2

, AD =

0 −2

−4 −1

4 +1

=

−2

−5

5

−2k 10 i − 5 j

AB × AD =i j k

1 −1 2

−2 −5 5

i j

1 −1

−2 −5

= 5i − 9j − 7k

−5 i − 4 j − 5k

AB × AD = 52 + (−92) + (−7)2 = 155

∴ 5�# �:�":��8 :��(��� ; � ABCD = 155

C(-1, 2, 0) D

A(1, 0, 1) B(2, 1, -2)

A B

CD

20

��������� �� ������� ����������������

Page 21: Ma5 vector-u-s54

��� 4 ���� 106 ��� 2 ������ = 3 5�# �:� = 2u ⋅ v

= 3 = 2u v cos θ u × v

= 3 = 21 v cos θ u v sin θ

= = v 2cos2θ 9 (1) u 2 v 2sin2θ 4 (2)

(1) + (2) : = 9 + 4v 2cos2θ + v 2sin2θ

= 13v 2(cos2θ + sin2θ)

= 13v 2(1)

= 13v 2

��� 2 ���� 108

������ �8� u = AB =

−1 −1

0 −2

2 +1

=

−2

−2

3

v = AC =

2 −1

3 −2

4 +1

=

1

1

5

r = AD =

0 −1

−1 −2

2 +1

=

−1

−3

3

�>(�7;������":��8 :��(��� ; � ��� (u × v) ⋅ w

=−2 −2 3

1 1 5

−1 −3 3

−2 −2

1 1

−1 3

= − 26

∴ �>(�7�:�7���ก���� 8 �9�− 26 = 26

*************************

D

AB

Cv

ur

-6 10 -9

3 -30 6

21

��������� �� ������� ����������������