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��������� �.5�������� ������ ��ก���� 3 ����(������ �ก!"�#���$%&���������'��#)
��� 4 ���� 8
�� ��� AD = AB + BC + CD = − u + v − w
FD = FE + ED = − u + v
BD = BC + CD = v − w
FC = FA + AB + BC = w − u + v
��� 6 ���� 9
�� ��� ��ก��� =NM NC + CB + BM
= 1
2AC + CB + 1
2BD
= 1
2(AB + BC) + CB + 1
2(BC + CD)
= 1
2AB + 1
2BC + CB + 1
2BC + 1
2CD
=NM1
2AB + 1
2CD (1)
��ก��� =NM NA + AD + DM
= 1
2CA + AD + 1
2DB
= 1
2(CB + BA) + AD + 1
2(DA + AB)
= 1
2CB + 1
2BA + AD + 1
2DA + 1
2AB
=NM1
2CB + 1
2AD (2)
(1) + (2) : 2NM = 1
2AB + 1
2CD + 1
2CB + 1
2AD
2NM = 1
2(AB + CD + CB + AD)
����� �������4NM = AB + CD + CB + AD AB + AD + CB + CD = 4NM
∴ R = 4
A
B
C
D
E
F
vu w
A
B
CD
NM
1
��������� �� ������� ����������������
��� 3 ���� 10
�� ��� ��กก$%�& 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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�$?��@!��:!�/� -�� a + b + c = 0
b × c = b × (−a − b) = − b × a − b × b = a × b
c × a = c × (−b − c) = − c × b − c × c = b × c
∴ a × b = b × c = c × a
?. ��� /7�"�7��"40/� -�� a × b = a × c b = c
��� 6 ���� 101 ��� 2 ������ a ⋅ x = a ⋅ b × c
a ⋅ b× c= a ⋅ b × c
a ⋅ b × c= 1 a ⋅ x + 2b ⋅ y + 3c ⋅ z = 2
2b ⋅ y = 2b ⋅ a × c
a ⋅ b× c= − 2
a ⋅ b× c
a ⋅ b× c= − 2
3c ⋅ z = 3c ⋅ a × b
a ⋅ b× c= 3
a ⋅ b× c
a ⋅ b× c= 3
oo
o o
o
o
A C
B
a b
c
17
��������� �� ������� ����������������
��� 7 ���� 102
������ 1 : u = 2i − j + 0k → u = 22 + (−1)2 + 02 = 5
v = 2i + j + k → v = 22 + 12 + 12 = 6
= u × v =i j k
2 −1 0
2 1 1
i j
2 −1
2 1
= − i − 2j + 4k u × v = (−1)2 + (−2)2 + 42 = 21
��8���!�ก u × v = u v sin θ → 21 = 5 ⋅ 6 sin θ → sin θ = 21
30
������ 2 ��8���!�ก u ⋅ v = u v cos θ
=(2)(2) + (−1)(1) + (0)(1) 5 6 cos θ
=cos θ 3
30
!�ก�0 sin θ = 21
30
��� 9 ���� 103
������ u ⋅ v = 2(−1) + (−1)(1) + 1(−2) = − 5 u ⋅ v = 5
u × v =
2
−1
1
×
−1
1
−2
=
1
3
1
u × v = 12 + 32 + 12 = 11
�-ก� �:���7�?�������ก�� ,$�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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������
��� 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
��������� �� ������� ����������������