document15
TRANSCRIPT
�
15 Vectors
1. (a) Scalar(b) Vector(c) Scalar(d) Vector
2. (a)B A
5 cm �cm:�0N(b)
v~
5 cm
�cm:20ms–�
3. (a) (i) u~=5×�0 =50ms–�
u~totheeast
(ii) v~=5×�0 =50ms–�
v~tothewest
(b) (i) –u~=50ms–�
–u~tothewest
(ii) –v~=50ms–�
–v~totheeast
4. (a) No(b) Yes(c) No
5. P→Q=2u~
R→S=–u~
A→B=� �—
2u~
C→D=– �—
2u~
6.
u~
–2u~
(a) (b)
(c)
u1–2 ~
u3– –2 ~
7. (a) A→B=2u~
R→S=–4u~
=–2(2u~)
=–2A→B
Therefore,A→BandR
→Sareparallel.
(b) R→S =–4u~
=–4(3P→Q)
=–�2P→Q
Therefore,R→SandP
→Qareparallel.
8. A→B=3u~,B
→C=–7u~
A→B=31– �—
7B→C2
A→B=– 3—
7B→C
Therefore,A,BandCarecollinear.
9. (a) (2k–�)u~=(2+ �—3
h)v~
2k–�=0and 2+ �—3
h=0
�—3
h=–2
h=–6
k= �—2
2
(b) 61 �—3
k+22u~–(3h–2)v~=0~
61 �—3
k+22=0 and 3h–2=0
h= 2—3 �—
3k=–2
k=–6
10. 2a+3b–�=0................................... 1 2–4a+b=0................................... 2
1×2,4a+6b–2=0..................... 3
2+3, 7b=0 b=0
Substituteb=0into1,2a+3(0)–�=0
a= �—2
Therefore,a= �—2
andb=0.
11. (a) A→B=D
→C,D
→A=C
→B
(b) S→R=P
→Q,S
→P=R
→Q
(c) H→E=G
→F,F
→E=G
→H
12.
a~ b~
a + b~ ~ ~ ~b + a
a~+b~=b~+a~
13. (a) 2P→Q+3R
→S=2(3u~)+3(5u~)
=6u~+�5u~ =2�u~
(b) 2P→Q+3R
→S=2�u~
=2�u~ =2�×2 =42units
14. (a) B→C+D
→E=6v~+ 2—
3B→C
=6v~+ 2—3
(6v~)
=6v~+4v~ =�0v~
(b) B→C+D
→E=�0v~
=�0v~ =�0×5 =50units
15. (a)
u
B
A
P
Q
~2u~
AB – PQPQ – AB(ii)(i)
→→→→
P→Q–A
→B=–(A
→B–P
→Q)
(b)
vDC
R S~
v~
CD – RS(ii)(i)
1–2
RS – CD→→→→
C→D–R
→S=–(R
→S–C
→D)
16. (a) (i) A→B–C
→D=5r~–3r~
=2r~
(ii) C→D–A
→B=3r~–5r~
=–2r~
(iii) A→B–2C
→D =5r~–2(3r~)
=5r~–6r~ =–r~
(b) (i) A→B–C
→D =2r~
=2r~ =2×2 =4units
(ii) C→D–A
→B =–2r~
=2r~ =2×2 =4units
(iii) A→B–2C
→D=–r~
=r~ =2units
3
17. (a) A→D= 3—
2B→C
= 3—2
s~
(b) A→D–B
→C = 3—
2s~–s~
= �—2
s~
(c) B→C–A
→D=– �—
2s~
= �—2
s
= �—2
(5)
= 5—2
units
18.
p~
q~
–q~ q
~
q~
p~
p~
p – q~ ~
~ ~p~
p + q
q~ 2
(a) (b) (c)
– –1
p~2
– –1
19.
p~
p~
p~
p~p + q
~ ~
p – q~ ~
q~
q~
q~
q~
–q~
(a) (b) (c)
1– –
2
p~
1– –2
20. (a) A→B=r~–s~
(b) P→Q=r~+s~
(c) U→V=s~+2r~
21. (a) p~+q
~=B
→D+D
→A
=B→A
(b) (i) A→C+C
→D=A
→D
=–q~
(ii) B→A+A
→C=B
→C
=2p~
(iii) C→B+B
→A=C
→A
=C→D+D
→A
=–p~+q
~
22. (a) r~+s~=A→B+B
→C
=A→C
(b) A→C=A
→D+D
→C
=t~+ �—2
(A→B)
=t~+ �—2
r~
(c) s~– �—2
r~=B→C+C
→D
=B→D
23.p~
p~
p~
p~
p + r + s~ ~ ~
~ ~ ~p – q + r
p + r + s + q~ ~ ~ ~
~s~
s~s~
q~
q~
–q~
(a) (b)
(c)
r
~r
~r
~r
24. (a) R→Q=R
→S+S
→Q
=–4v~+5u~
(b) (i) P→Q–S
→Q=2v~–5u~
(ii) R→S–Q
→S=–4v~–(–5u~)
=5u~–4v~
(c) P→M=P
→Q+Q
→M
=2v~+Q→S+S
→M
=2v~+(–5u~)+ �—4
S→R
=2v~–5u~+ �—4
(4v~)
=3v~–5u~
4
25. (a) A→B=3A
→P
=3(2r~) =6r~
(b) B→C=B
→A+A
→C
=–6r~+4s~
(c) P→C=P
→A+A
→C
=–2r~+4s~
(d) P→Q=P
→B+B
→Q
=4r~+ 3—4
B→C
=4r~+ 3—4
(–6r~+4s~)
=4r~– 9—2
r~+3s~
=3s~– �—2
r~
(e) A→Q=A
→B+B
→Q
=6r~+ 3—4
B→C
=6r~+ 3—4
(–6r~+4s~)
=6r~– 9—2
r~+3s~
= 3—2
r~+3s~
26. (a) R→T=Q
→T–Q
→R
=4a~+hb~–(2a~+b~) =2a~+(h–�)b~
S→T=S
→R+R
→T
=(a~–b~)+2a~+(h–�)b~ =3a~+(h–2)b~
(b) Q→T=kR
→T
4a~+hb~=k[2a~+(h–�)b~] 4a~+hb~=2ka~+k(h–�)b~ 2k=4andh=k(h–�) k=2 =2(h–�) =2h–2 h=2
27. (a) E→D=E
→B+B
→D
=u~+ �—2
v~–u~
= �—2
v~
= �—2
A→C
Therefore,ACisparalleltoDE.
(b) DE:AC=�:2
(c) A→D=A
→B+B
→D
=2u~+ �—2
v~–u~
=u~+ �—2
v~
28. (a) (i) O→R=O
→Q+Q
→R
=q~+ �—
3Q→P
=q~+ �—
3(Q→O+O
→P)
=q~+ �—
3(–q
~+p
~)
=q~– �—
3q~+ �—
3p~
= �—3
p~+ 2—
3q~
(ii) R→S=R
→P+P
→S
= 2—3
Q→P+2q
~
= 2—3
(–q~+p
~)+2q
~
=– 2—3
q~+ 2—
3p~+2q
~
= 2—3
p~+ 4—
3q~
(b) R→S= 2—
3p~+ 4—
3q~
=21 �—3
p~+ 2—
3q~2
=2O→R
Therefore,O,RandSarecollinear.
29. (a) –i~–3j~
(b) 4i~+5j~
(c) 6i~–5j~
30. (a) 1–�2–3
(b) 1425
(c) 1 6 2–5
31. (a) ABBBBBBBBB(–�)2+(–3)2
=ABB�0 units
(b) ABBBBB42+52
=ABB4�units
5
(c) ABBBBBBBB(6)2+(–5)2
=ABB6�units
32. (a) �——ABB�0
(–i~–3j~
)
(b) �——ABB4�
(4i~+5j~
)
(c) �——ABB6�
(6i~–5j~
)
33. (a) a~+b~=(2i~+3j~
)+(4i~+5j~
)
=6i~+8j~
(b) 2a~+c~=2(2i~+3j~
)+3i~ =4i~+6j
~+3i~
=7i~+6j~
(c) �—2
d~+c~+2a~
= �—2
(–4j~ )+3i~+2(2i~+3j
~)
=–2j~+3i~+4i~+6j
~ =7i~+4j
~
34. (a) p~+q
~
=1324+1 –�25
=1229
(b) 2p~+ �—
2r~+q
~
=21324+ �—
2 1024+1 –�25
=1628+1022
+1 –�25
=1 5 2�5
35. (a) 2a~–b~
=21425–1–32�
=1 8 2�0–1–32�
=1��29
(b) �—2
a~–3b~
= �—2 1425
–31–32�
=1 2 25—2
+1 9 2–3
=1 �� 2– �—2
36. (a) r~–s~=(3i~–4j~
)–(2i~+5j~
)
=3i~–4j~–2i~–5j
~ =i~–9j
~
(b) 4r~– �—3
s~ =4(3i~–4j~
)– �—3
(2i~+5j~
)
=�2i~–�6j~– 2—
3i~– 5—
3j~
= 34—–3
i~– 53—–3
j~
37. (a) r~–s~+t~=1–�22–1024
+1 3 2–5
=1–�–0+322–4–5
=1 2 2–7
(b) –2r~– �—2
s~+t~ =–21–�22– �—
2 1024+1 3 2–5
=1 2–0+3 2–4–2–5
=1 5 2–��
38. (a) 1h23+1�2k
=1428
1h+�23+k =1428
h+�=4and3+k=8 h =3 k=5
6
(b) 21h25+1–420
=162k
12h–42�0+0 =162k
2h–4=6and�0=k 2h=�0 k=�0 h=5
(c) �—2 152h
–31�22=21 k 2–�
15—2
–32h—2
–6
=12k2–2
5—2
–3=2kand h—2
–6=–2
h—2
=4
h=8
2k=– �—2
k=– �—4
39. (a) a~+b~=ABBBBB62+82 +ABBBBBBB32+(–4)2
=ABBB�00+ABB25 =�0+5 =�5units
(b) a~+b~=6i~+8j~+3i~–4j
~ =9i~+4j
~ a~+b~=ABBBBB92+42
=ABB97units
(c) a~–2b~=6i~+8j~–2(3i~–4j
~)
=6i~+8j~–6i~+8j
~ =�6j
~ a~–2b~=ABB�62
=�6units
40. Givenp~//q
~Letp
~=kq
~
1–�22=k1 a 2–2
–�=kaand2=–2k k=–�
Substitutek=–�into–�=ka,–�=ka–�=(–�)(a) a=�
41. a~–b~=1h24–1 3 2–�
=1h–325
Leta~–b~=ka~
1h–325 =k1h24
h–3=khand5=4k
k= 5—4
Substitutek= 5—4
intoh–3=kh,
h–3= 5—4
h
h– 5—4
h=3
– �—4
h=3
h=–�2
42. P→Q=1–224
=–21 � 2–2
=–2R→S
Therefore,P→QandR
→Sareparallel.
43. P→Q=P
→O+O
→Q
=1–�2–2+1–22–4
=1–32–6
Q→R=Q
→O+O
→R
=1224+1428
=1 6 2�2
=61�22
7
P→Q=1–32–6
=–31�22
=–31 �—6
Q→R2
=– �—2
Q→R
Therefore,P,QandRarecollinear.
44. P→Q=122�
P→O+O
→Q=122�
1 � 2–3+O
→Q=122�
O→Q=122�
–1 � 2–3
=1�24
Therefore,thecoordinatesofQare(�,4).
45. (a) P→Q–S
→R
=1–�22–1–523
=1 4 2–�
(b) P→Q–S
→R=ABBBBBBB42+(–�)2
=ABB�7 units
(c) Unitvector= �——ABB�7 1
4 2–�
46. (a) 2a~–3b~+c~=0~ 2[3i~+(�–k)j
~]–3(–4i~+5j
~)+hi~+7j
~=0~
6i~+2(�–k)j~+�2i~–�5j
~+hi~+7j
~=0~
(�8+h)i~+[2–2k–�5+7]j~=0~
(�8+h)i~+[–2k–6]j~=0~
�8+h=0and–2k–6=0 h=–�8 k=–3
(b) a~–2b~=3i~+(�–k)j~–2(–4i~+5j
~)
=3i~+(�–4)j~+8i~–�0j
~ =��i~–�3j
~
a~–2b~=ABBBBBBBBB��2+(–�3)2
=ABBB290 units
Unitvector= �——––ABBB290
(��i~–�3j~
)
(c) Let b~=mc~ –4i~+5j
~=m(hi~+7j
~)
=mhi~+7mj~
mh=–4.......................... 1 7m=5
m= 5—7
.......................... 2
Substitute2into1,
5—7
h=–4
h=–41 7—5 2
=– 28—–5
(d) a~=b~–c~ 3i~+(�–k)j
~=–4i~+5j
~–hi~–7j
~ =(–4–h)i~–2j
~ –4–h=3and�–k=–2 h=–4–3 k=�–(–2) =–7 k=3
(e) a~–2b~=3i~+(�–k)j~
–2(–4i~+5j~
)
=3i~+8i~+(–k–9)j~
=��i~+(–�–9)j~
=��i~–�0j~
a~–2b~=ABBBBBBBBB��2+(–�0)2
=ABBB22� units
(f) a~=ABBBBBBBBB32+(�–k)2 =5 32+(�–k)2=25 (�–k)2=�6 �–k=±4 k=�±4 =–3,5
(g) 2b~–c~=2(–4i~+5j~
)–(hi~+7j~
)
=–8i~+�0j~–hi~–7j
~ =(–8–h)i~+3j
~ 2b~–c~=ABBBBBBBBBB(–8–h)2+32 =ABBB298 (–8–h)2+32=298 (–8–h)2=298–9 =289 –8–h=±�7 h+8=±�7 =–25,9
8
47. 2a~ =3b~
21 m 2n–�=31 n+2 2–�–m
2m=3n+62m–3n=6............................ 1
2(n–�)=3(–�–m) 2n–2=–3–3m 2n=–�–3m2n+3m=–�......................... 2
1×2,4m–6n=�2............ 32×3,6n+9m=–3........... 4
3+4,�3m=9
m= 9—–�3
Substitutem= 9—–�3
into2,
2n+31 9—–�3 2=–�
2n=–�– 27—–�3
=– 40—–�3
n=– 20—–�3
Therefore,m= 9—–�3
andn=– 20—–�3
.
1. (a) O→P=–3i~+5j
~
(b) O→R=–2i~–7j
~
=1–22–7
2. (a) O→P=5i~+�2j
~
(b) O→P=ABBBBBB52+�22
=ABBB�69 =�3units
Unitvector= �—–�3
(5i~+�2j~
)
3. (a) O→A=8i~–6j
~
(b) A→B=A
→O+O
→B
=–O→A+O
→B
=(–8i~+6j~
)+�0j~
=–8i~+�6j~
=1–82�6
4. A→C=A
→B+B
→C
=1 � 2–2+A
→O
=1 � 2–2+1–22–4
=1–�2–6
5. P→R+2Q
→R=5i~–7j
~ P
→O+O
→R+2[Q
→O+O
→R]=5i~–7j
~ –O
→P+O
→R+2Q
→O+2O
→R=5i~–7j
~ –O
→P+3O
→R–2O
→Q=5i~–7j
~–(–4i~+5j
~)+3(7i~+2j
~)–2(hi~+kj
~)=5i~–7j
~ 4i~–5j
~+2�i~+6j
~–2hi~–2kj
~=5i~–7j
~ (25–2h)i~+(�–2k)j
~=5i~–7j
~25–2h=5and�–2k =–7 2h=20 2k =8 h=�0 k =4
6. (a) P→R=P
→O+O
→R
=–O→P+O
→R
=–(3i~–4j~
)+6i~ =–3i~+4j
~+6i~
=3i~+4j~
(b) P→R=ABBBBB32+42
=5units
Unitvector= �—5
(3i~+4j~
)
7. r~=2p~–3q
~(h+k)a~–hb~=2(4a~–b~)–3(–2a~+3b~)
=8a~–2b~+6a~–9b~ =�4a~–��b~\h=��,h+k=�4 ��+k=�4 k=3
9
8. (a) Let A→B=mB
→C
3p~–2q
~=m[(�–k)p
~+4q
~]
=m(�–k)p~+4mq
~
4m=–2andm(�–k)=3
m=– �—2
Substitutem=– �—2
intom(�–k)=3,
1– �—2 2(�–k)=3
�–k=–6 k=7
(b) A→B=– �—
2B→C
AB= �—2
BC
AB—––BC
= �—2
AB:BC=�:2
9. (a) P→Q =P
→O+O
→Q
=–O→P+O
→Q
1426=–1225
+O→Q
O→Q =1426
+1225
=1 6 2��
Therefore,Q=(6,��).
(b) O→Q=ABBBBBB62+��2
=ABBB�57 units
Unitvector= �——–ABBB�57
(6i~+��j~
)
(c) LetA→B=kP
→Q
142h=k1426
4k=4andh=6k k=� =6(�) =6
10. (a) B→C =B
→A+A
→C
=–4p~+8q
~
(b) D→A=D
→B+B
→A
= �—3
C→B+(–4p
~)
= �—3
(4p~–8q
~)–4p
~
= 4—3
p~– 8—
3q~–4p
~
=– 8—3
p~– 8—
3q~
11. (a) (i) A→C=A
→B+B
→C
=–B→A+B
→C
=–(4x~+8y~
)+6x~ =2x~–8y
~
(ii) A→E=A
→B+B
→E
=–(4x~+8y~
)+ �—2
(6x~)
=–4x~–8y~+3x~
=–x~–8y~
(iii) B→D=B
→E+E
→D
=3x~+ �—3
E→A
=3x~+ �—3
(x~+8y~
)
=3x~+ �—3
x~+ 8—3
y~
= �0—–3
x~+ 8—3
y~
(b) B→F=B
→C+C
→F
=6x~+ �—2
C→A
=6x~+ �—2
(–2x~+8y~
)
=6x~–x~+4y~
=5x~+4y~
B→D= �0—–
3x~+ 8—
3y~
= 2—3
(5x~+4y~
)
= 2—3
B→F
Therefore,B,DandFarecollinear.
�0
(c) A→C =2x~–8y
~ 2x~=2x~ =2×3 =6units
8y~
=8y~
=8×2 =�6units
A→C=ABBBBBB62+�62
=ABBB292units
12. (a) P→R=P
→S+S
→R
=–3y~+6x~
(b) GivenS→T=2T
→R
S→T= 2—
3S→R
= 2—3
(6x~)
=4x~
GivenS→R= 3—
2P→Q
P→Q= 2—
3S→R
= 2—3
(6x~)
=4x~ Since S
→T = P
→Q , therefore PQTS is a
parallelogram.
(c) (i) P→A=P
→Q+Q
→A
= 2—3
S→R+nQ
→T
= 2—3
(6x~)+n(–3y~
)
=4x~–3ny~
(ii) GivenP,AandRarecollinear.
Let P→A=kP
→R
4x~–3ny~=k(–3y
~+6x~)
=–3ky~+6kx~
6k=4and3k=3n
k= 2—3
3n=31 2—3 2
=2
n= 2—3
13. (a) (i) O→S=O
→P+P
→S
=9x~+ 2—3
P→Q
=9x~+ 2—3
(P→O+O
→Q)
=9x~+ 2—3
(–9x~+3y~
)
=9x~–6x~+2y~
=3x~+2y~
(ii) Q→R=Q
→P+P
→R
=(9x~–3y~
)+ �—3
P→O
=9x~–3y~+ �—
3(–9x~)
=9x~–3y~–3x~
=6x~–3y~
(b) (i) Q→T=nQ
→R
=n(6x~–3y~
) =6nx~–3ny
~
(ii) T→S=mO
→S
=m(3x~+2y~
) =3mx~+2my
~
(c) Q→S=Q
→T+T
→S
=6nx~–3ny~+3mx~+2my
~ =(6n+3m)x~+(2m–3n)y
~
AlsoQ→S= �—
3Q→P
= �—3
(Q→O+O
→P)
= �—3
(–3y~+9x~)
=–y~+3x~
3x~–y~=(6n+3m)x~+(2m–3n)y
~ 6n+3m=3 2n+m=�................................. 1
2m–3n=–�............................... 2
1×2,4n+2m=2................... 3
3–2,7n=3
n= 3—7
Substituten= 3—7
into1,
21 3—7 2+m=�
m=�– 6—7
= �—7
Therefore,n= 3—7
andm= �—7
.
��
1. (a) O→B=4i~+3j
~(b) O
→B=ABBBBB42+32
=5units
Unitvector= �—5
(4i~+3j~
)
(c) B→C =1–22–�
B→O+O
→C =1–22–�
–O→B+O
→C=1–22–�
O→C=1–22–�
+O→B
=1–22–�+1423
=1222
Therefore,C=(2,2).
2. (a)
3 O
A
k 5
k=4
(b) A→B=A
→O+O
→B
=–O→A+O
→B
=–1–324+1220
=1 5 2–4
3. (a) A→B =�2i~–5j
~ A
→O+O
→B=�2i~–5j
~ A
→O =(�2i~–5j
~)–O
→B
=�2i~–5j~–(–4i~+3j
~)
=�6i~–8j~
O→A=–�6i~+8j
~ Therefore,A=(–�6,8).
(b) A→B=ABBBBBBBB�22+(–5)2
=�3units
Unitvector= �—–�3
(�2i~–5j~
)
4. C→D=�2
ABBBBBBB42+(–k)2 =�2 �6+k2=�44 k2=�28 k=±ABBB�28
5. P→Q=ABBBBBBBk2+(–4)2
\ABBBBBk2+�6=3k k2+�6=9k2
8k2=�6 k2=2
k=±AB2
6. 2p~–q
~=4i~+5j
~2(ni~+4j
~)–(6i~–mj
~)=4i~+5j
~ 2ni~+8j
~–6i~+mj
~=4i~+5j
~ (2n–6)i~+(8+m)j
~=4i~+5j
~2n–6=4and8+m=5 n=5 m=–3
7. (a) p~+q
~=(5i~–7j
~)+(4i~+3j
~)
=9i~–4j~
p~+q
~=ABBBBBBB92+(–4)2
=ABB97 units
(b) kp~
=(m+�)q~
k(5i~–7j~
)=(m+�)(4i~+3j~
)
5ki~–7kj~
=4(m+�)i~+3(m+�)j~
5k=4(m+�) =4m+4 5k–4m=4................................. 1
–7k=3(m+�) =3m+3 –7k–3m=3............................... 2
1×7,35k–28m=28............. 3 2×5,–35k–�5m=�5........... 4
3+4,–43m=43 m=–�
Substitutem=–�into1, 5k–4(–�)=4 5k=0 k=0
Therefore,k=0andm=–�.
�2
8. (a) r~+s~=2r~
1 k23+142n
=21 k23
1k+423+n=12k26
k+4=2kand3+n=6 k=4 n=3
(b) Letr~+s~=mr~
1k+423+n=m1 k23
k+4=mk................................... 1
3+n=3m
m= 3+n——–3
............................. 2
Substitute2into1,
k+4=1 3+n——–3 2k
3k+�2=3k+nk nk=�2
k= �2—–n
9. 2a~–3b~=4c~
2a~–3b~=21223–31 k2h
=14–3k26–3h
2a~–3b~=ABBBBBBBBBBBBBB(4–3k)2+(6–3h)2
4c~=4c~
=4ABBBBBBB(–4)2+32
=4×5 =20
ABBBBBBBBBBBBBB(4–3k)2+(6–3h)2 =20 (4–3k)2+(6–3h)2=400�6–24k+9k2+36–36h+9h2=400 9k2+9h2–24k–36h–348=0 3k2+3h2–8k–�2h–��6=0
10. (a) p~+2q
~=ki~+3j
~+2(3i~–4j
~)
=(k+6)i~–5j~
p~+2q
~=ABBBBBBBBBBB(k+6)2+(–5)2 =ABBB�25
(k+6)2+25=�25 (k+6)2=�00 k+6=±�0 k=±�0–6 =4,–�6
(b) p~
=ABBBBBk2+32 =ABBBBBk2+9
q~
=ABBBBBBB32+(–4)2 =5
Given p~
+2q~
=�5
ABBBBBk2+9 +2×5=�5
ABBBBBk2+9 =5 k2+9=25 k2=�6 k=±4
11. (a) B→C=B
→A+A
→C
=–x~+y~
(b) A→D=A
→B+B
→D
=x~+ 2—3
(B→C)
=x~+ 2—3
(–x~+y~
)
=x~– 2—3
x~+ 2—3
y~
= �—3
x~+ 2—3
y~
12. (a) P→T=P
→Q+Q
→T
=2x~+ 3—4
Q→R
=2x~+ 3—4
(4y~
)
=2x~+3y~
(b) P→T=2(2i~+3j
~)+3(i~–4j
~)
=4i~+6j~+3i~–�2j
~ =7i~–6j
~ P
→T=ABBBBBBB72+(–6)2
=ABB85units
13. (a) B→D=B
→A+A
→D
=–4x~+ 2—3
A→C
=–4x~+ 2—3
(A→B+B
→C)
=–4x~+ 2—3
(4x~+�2y~
)
=–4x~+ 8—3
x~+8y~
=– 4—3
x~+8y~
(b) A→B=4x~
=4x~ =4×3 =�2units
�3
B→C=�2y
~
=�2y~
=�2×2 =24units
12 unitsA B
C
D 24 units
A→C=ABBBBBBB242–�22
=ABBB432 units
A→D= 2—
3A→C
= 2—3
ABBB432 units
B→D=ABBBBBBBBB4—
9(432)+�22
=ABBB336 units
14. (a) (i) A→F= 2—
3A→B
A→B= 3—
2A→F
= 3—2
(4x~)
=6x~
(ii) D→C= 2—
3A→B
= 2—3
(6x~)
=4x~
A→C=A
→D+D
→C
=3y~+4x~
A→E= �—
2A→C
= �—2
(3y~+4x~)
=2x~+ 3—2
y~
(iii) D→F=D
→A+A
→F
=–3y~+4x~
=4x~–3y~
(b) A→F=4x~
F→B=2x~
\Areaof∆ACF=2×Areaof∆BCF =2×�2
=24unit2
15. (a) P→T=P
→S+S
→T
=(–2x~–4y~
)+(8x~+4y~
) =6x~
(b)PT—–TQ
=m—�
m=PT——TQ
=6x~—––2x~
=3—�
=3
(c) T→R=T
→S+S
→R
=–8x~–4y~+8x~
=–4y~
(d) S→P=2x~+4y
~
2x
4y
S O
P
~
~
S→O=2x~
=2x~ =2×3 =6units
P→O=4y
~
=4×4 =�6units
S→P=ABBBBBB�62+62
=ABBB292 units
16. (a) (i)
a
b
A
M
B
N
y
xO
~
~
�4
O→N= �—
2O→B
= �—2
b~
(ii) O→M=O
→A+A
→M
=a~+ �—2
A→B
=a~+ �—2
(A→O+O
→B)
=a~+ �—2
(–a~+b~)
=a~– �—2
a~+ �—2
b~
= �—2
a~+ �—2
b~
(b) (i) O→L=hO
→M
=h1 �—2
a~+ �—2
b~2 = �—
2ha~+ �—
2hb~
(ii) A→L=kA
→N
A→O+O
→L=k(A
→O+O
→N)
–a~+O→L=k1–a~+ �—
2b~2
O→L=–ka~+ �—
2kb~+a~
=(�–k)a~+ �—2
kb~
Compare(i)and(ii)
�—2
h=�–k
h=2–2k....................... 1
�—2
k= �—2
h
k=h............................... 2
Substitute2into1, h=2–2h 3h=2
h= 2—3
From2,k= 2—3
Therefore,h= 2—3
andk= 2—3
.
17. (a) (i) O→M=O
→A+A
→M
=u~+ �—2
A→B
=u~+ �—2
(A→O+O
→B)
=u~+ �—2
(–u~+v~)
=u~– �—2
u~+ �—2
v~
= �—2
u~+ �—2
v~
(ii) A→T= �—
4A→B
= �—4
(–u~+v~)
=– �—4
u~+ �—4
v~
(b) Since ∆AOT and ∆AOB has the base ratioAT :AB=� :4withOas thevertex, thereforetheareaof∆AOT
= �—4
×Areaof∆AOB
= �—4
×24
=6unit2
(c) u~–v~=B→A
u~–v~=B→A
M
B
O
30°
12–
~ ~(u – v)
10
sin30°= BM—––�0
BM=5
u~–v~=2×5 =�0units