(18A004)
Chapter 18
Introduction to Vectors
Teaching Examples
Teaching Example 18.4
In the figure, ABCD is a quadrilateral, E, F, G and H are the mid-points of AD, BD,
BC and AC respectively.
(a)
Express EF and HG in terms of AB only.
Hence, show that AB, EF and HG are parallel.
(b)
Express FG and EH in terms of DC only.
Hence, show that EFGH is a parallelogram.
(18A001)
Teaching Example 18.1
In the figure, ABCD is a rectangle, BEFC is a rhombus and CFGD is a parallelogram.
Prove that
(a)
(b)
AB FC DF ,
EF CD AC .
(18A005)
Teaching Example 18.5
It is given that AB ( s 1)a (3t 1)b and CD (3s 1)a (t 3)b , where a and b are two non-zero
and non-parallel vectors. If 2 AB CD , find the values of s and t.
(18A006)
Teaching Example 18.6
(18A002)
-1-
Teaching Example 18.2
In parallelogram ABCD, G is the mid-point of BC, and E is a point on DC such
It is given that the vectors AB (6a mb) 2b and CD 5a m(a b) are parallel, where a and b are
two non-zero and non-parallel vectors and m is a scalar. Find the values of m.
that DE : EC = 2 : 1.
(18A007)
(a)
(b)
Teaching Example 18.7
In the figure, A(–6, 3), B(–2, 1), C and D(–3, 5) are the vertices of a
parallelogram. Find the coordinates of C.
Express AE in terms of AB and AD .
Express GE in terms of AB and AD .
(18A003)
Teaching Example 18.3
In the figure, ABCD is a parallelogram. E is the mid-point of BC and
CF : FD 1 : 3.
(a)
(b)
Express AE and AF in terms of p and q.
If p 4 ,
(i)
find 2 AE AF ,
(ii)
find a unit vector which is parallel to 2 AE AF .
(18A008)
Teaching Example 18.8
Let A(4, 5), B(7, 2) and C(–1, 2) be three points on a coordinate plane.
(a)
(b)
Find AB .
If OQ AB , find the angle between OQ and the positive x-axis.
(c)
Find AB 4OC .
Hence, find a unit vector which has the opposite direction of AB 4OC .
(18A009)
(18A014)
Teaching Example 18.9
Let A(2, 3) and B(6, 7) be two points on a coordinate plane. The position vector of R
with respect to O is given by OR ( 1)OA ( 1)OB .
Teaching Example 18.14
(a)
(b)
Express OR in terms of , i and j.
If OR is parallel to 5i + 6j, find the value of .
In the figure, D and E are mid-points of AB and AC respectively. Let AB p
and AC q . BQ : QE m : n, CQ : QD m : 1.
(a)
(i)
Express AQ in terms of p, q, m and n only.
(b)
(ii) Express AQ in terms of p, q and m only.
Hence, find the values of m and n.
(18A010)
(18A015)
Teaching Example 18.10
Teaching Example 18.15
In each of the following, A and B are two points in a coordinate system and O is
the origin. Q is a point on AB such that AQ : QB 2 : 5.
(a) If the coordinates of A and B are (2, 3) and (9, −11) respectively,
find OQ .
(b) If the coordinates of Q and B are (−3, 5, 1) and (2, −5, 6) respectively,
find the coordinates of A.
It is given that OA 2i 3j k and OB 4i j k .
(a)
Find AB .
(b)
If AB 6 , find the values of .
(18A011)
Teaching Example 18.11
-2-
Three points A(1, m, n + 1), B(3, 2m + 1, 4n – 2) and C(7, 5m, 6n + 4) are given. If AB is parallel to
BC , find the values of m and n.
(18A016)
(18A012)
(a)
(b)
Find AB .
If OR 2 AB , find the angle between OR and the positive x-axis
correct to the nearest 0.1°.
(c)
Find OR 2OC .
Teaching Example 18.12
In the figure, ABCDHEFG is a right prism with parallelogram ABCD as its base.
(a)
(b)
Show that BA BC BF BH .
The coordinates of the vertices are A(4, 0, 3), B(5, 3, 4), C(1, 4, 5) and
F(4, 2, 8).
(i)
(ii)
Find the coordinates of H.
Find the lengths of BH and OH, where O is the origin.
(18A013)
Teaching Example 18.13
In the figure, ABCD is a parallelogram. E is a point on BC such that BE : EC 2 : 1.
F is a point on AE such that AF : FE 1 : 2. Let AB p and AD q .
(a)
(b)
Express DE in terms of p and q.
Express DF in terms of p and q.
Teaching Example 18.8 (Extra)
Let A(–1, 3), B(3, 5) and C(6, –1) be three points on the coordinate plane.
Hence, find a unit vector which has the same direction of OR 2OC .
(18A017)
Teaching Example 18.12 (Extra)
In the figure, ABCDHEFG is a prism with parallelogram ABCD as its base.
(a)
(b)
Show that CB CD CG CE .
It is given that the coordinates of the vertices are B(4, 6, 0), C(7, 12, 1),
D(5, 8, 3) and G(11, 12, 0).
(i)
(ii)
Find CE .
Find the coordinates of E and a unit vector in the same direction of OE .
Basic Questions
(18A018)
Teaching Example 18.14 (Extra)
In the figure, D and E are points on AB and AC respectively such that
AD : DB 2 : 1, and AE : EC 2 : 1. Suppose AB p , AC q ,
BQ : QE m : 1 and CQ : QD n : 1.
(a) (i) Express AQ in terms of p, q and m only.
(b)
(ii) Express AQ in terms of p, q and n only.
Hence, find the values of m and n.
§18.2 Operations and Properties of Vectors
(18B001)
In the figure, ABCD is a parallelogram, where AD p and
DC q . Express each of the following vectors in terms of p and q.
(a) CB
(c) AC
(b) AB
(d) DB
(18B002)
In the figure, ABCD is a rhombus and the two diagonals meet at E.
Express each of the following by a single vector.
(a)
(c)
AD CD
BC ED
(b)
(d)
AD CB
AB CE
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(18B003)
In the figure, ABCDEFGH is a regular octagon and O is the
intersection of the diagonals AE, BF, CG and DH. Express each of
the following by a single vector.
(a) AB ED
(c) CD OG
(b) AB EF
(d) HG ED OA
(18B004)
In the figure, ABCD is a rectangle, AD = 12 and AB = 5. Find
(a)
AD CB ,
(b)
AD AB .
(18B005)
In the figure, ABC is a triangle. D and E are the mid-points of AB and AC
respectively. BE and CD intersect at F. If AB p and AC q , express the
following in terms of p and q.
(a) BC
(b) BD
(c) CD
(d) BE
(18B006)
It is given that ( s 1)a 2tb 3a (t 1)b , where a and b are two non-zero and non-parallel vectors.
(18B013)
For each of the following, find the values of sin , cos and tan , where is the angle between the
Find the values of s and t.
vector OA and the positive x-axis.
(a) OA 3i j
(b) OA i 2 j
(18B007)
It is given that (2s t )a 3b 3a ( s t )b , where a and b are two non-zero and non-parallel vectors.
Find the values of s and t.
(18B014)
For each of the following position vectors, find its magnitude and the angle between the vector and the
positive x-axis. (Give your answers correct to 1 decimal place if necessary.)
(18B008)
(a) i 2 j
The vectors PQ 2a mb and RS (3m 10)a 4b are parallel, where a and b are two non-zero and
non-parallel vectors and m is a scalar. Find the values of m.
(18B009)
The vectors PQ (m 1)a 3b and RS 2ma 3mb are parallel, where a and b are two non-zero and
(b) i 3 j
(18B015)
For each of the following position vectors, find its magnitude and the angle between the vector and the
positive x-axis. (Give your answers correct to 1 decimal place if necessary.)
(a) 5i 2 j
(b) 2i 3 j
-4-
non-parallel vectors and m is a scalar. Find the values of m.
(18B016)
Let A(2, –1) and B(3, 4) be two points on the coordinate plane.
§18.3 Vectors in 2-dimensional Space
(18B010)
(a) Express AB in terms of i and j.
(b) Find a unit vector which has the same direction of AB .
For each of the following, express AB in terms of i and j.
(a) A(–1, 0), B(2, 3)
(b) A(2, –3), B(4, 7)
(18B017)
Let A(–2, 3) and B(–3, 5) be two points on the coordinate plane.
(18B011)
(a) Express AB in terms of i and j.
(b) Find a unit vector which has the same direction of AB .
For each of the following, express AB in terms of i and j.
(a) A(m + 1, n – 1), B(2m + 3, 3n + 2)
(b) A(2m + 3, 3m + n), B(4 – 2n, –3 – n)
(18B018)
(18B012)
For each of the following, find the values of sin , cos and tan , where is the angle between the
vector OA and the positive x-axis.
(a) OA 2i j
(b) OA 3i 2 j
It is given that a = –2i + j and b = 3i – 2j.
(a) Find a + 2b.
(b) If c + 2a = a – 2b, find c in terms of i and j.
(18B019)
It is given that a = 2i – 3j and b = 4i +3j.
(a) Find 2a – 3b.
(b) If 3c – 2a = 2b – a, find c in terms of i and j.
(18B020)
(18B027)
For each of the following, find OA .
(a) AB 2i 5j , B(–3, 2)
For each of the following, find the values of .
(b)
AB 3i 7 j , B(4, 1)
(18B021)
It is given that A(1, –2), B(–2, 5) and C(2, 4) are points on a coordinate plane. Express each of the
following in terms of i and j.
(a)
(c)
AB
BA BC
(b) BC
(d) 2 AC 3 AB
(18B022)
It is given that OA 2i j , OB 3i 2 j and OC 3i 4 j . Find
(a) AB ,
(c) 3CB 2BA ,
(b) BC ,
(d) 2OB 3 AC .
-5-
(18B023)
For each of the following, find the values of m and n.
(a) 2mi – 3nj = 4i + 9j
(b) (m + 2)i + (2n + 3)j = (2m – 3)i + (2 – 3n)j
(18B024)
For each of the following, find the value(s) of such that p is parallel to q.
(a) p = i + 2j, q = 4i – 6j
(b) p = i + 2j, q = i + 4j
§18.4 Vectors in 3-dimensional Space
(18B025)
For each of the following, express AB in terms of i, j and k.
(a) A(1, –1, 3), B(2, 0, –2)
(b) A(2, –3, 4), B(5, 1, –3)
(18B026)
For each of the following, express AB in terms of i, j and k.
(a) OA 2i 3j k , OB i 2 j 5k
(b) OA 3i 2 j 4k , OB 2i j 2k
(a)
AB i 2 j 3k and AB 22
(b)
AB 2i ( 1) j 5k and AB 33
(18B028)
For each of the following, find the values of .
(a) OA 2i 3j k , OB i j 3k and AB 2 6
(b) OA 3i j 2k , OB 5i 4k and AB 14 2
(18B029)
It is given that A(3, 2, –1), B(2, 4, 5) and C(1, –2, 3) are points in a 3-dimensional coordinate system.
Express each of the following in terms of i, j and k.
(a) AB
(c) 2OA 3BC
(b) CA
(d) OB 2 AC
(18B030)
It is given that A(2, 3, 6), B(–1, –2, 3) and C(1, 2, –1) are points in a 3-dimensional coordinate system.
Express each of the following in terms of i, j and k.
(a)
(c)
BC
AC 2OC
(b) 2 AB
(d) BA 3OB
(18B031)
Let A(3, 4, –1) and B(–2, 5, 2) be two points in a 3-dimensional coordinate system.
(a) Express AB in terms of i, j and k.
(b) Find a unit vector which has the same direction of AB .
(18B032)
Let P(–2, 3, 1) and Q(4, –1, –3) be two points in a 3-dimensional coordinate system.
(a) Express PQ in terms of i, j and k.
(b) Find a unit vector which has the same direction of PQ .
(18B033)
(18B040)
It is given that a = 2i – 3j – k, b = –i + 4j + 5k and c = 3i + 2j + k. Find, in terms of i, j and k, a unit
vector in the direction of each of the following vectors.
It is given that OA 2i j 3k and OB i 2 j 4k are two vectors in a 3-dimensional coordinate
(a) 2a + b
(c) b – 3a + 2c
system. If P is a point on AB such that AP : PB 1: 2 , express OP in terms of i, j and k.
(b) 3a – 2c
(18B041)
It is given that OA 3i 2 j k and OB 2i j 5k are two vectors in a 3-dimensional coordinate
(18B034)
It is given that a = –3i + 5k, b = 2i – 4j + 2k and c = –6i – 2j – k. Find, in terms of i, j and k, a unit vector
in the direction of each of the following vectors.
(a) 3a + c
(c) 2a + 3b – c
(b) 2b – 3c
system. If P is a point on AB such that AP : PB 3 :1 , express OP in terms of i, j and k.
(18B042)
In the figure, P is a point on line segment AB such that AP : PB 5 : 3 .
3
3
If OA 9i j and OP i j , find OB .
2
2
(18B035)
It is given that a = 3i + 2j – k and b = –2i + 3k.
(a) Express a – 2b in terms of i, j and k.
(b) If a 2b – 3c = 4i – j, find c.
(18B043)
It is given that M and N are two points in a coordinate system and O is the origin.
P is a point on line segment MN such that MP : PN 1: 2 . If the coordinates of M
and P are (2, 9, 12) and (2, 7, 9) respectively, find ON .
(18B036)
-6-
It is given that a = –2i + 3j + 5k and b = 3i – 2j – 4k.
(a) Express 2a + 3b in terms of i, j and k.
(b) If 2a + 3b + 2c = 3i 4j, find c.
(18B044)
In the figure, P is a point on line segment CD such that CP : PD 1: k , where k is
a positive constant. If OC i 9 j , OD 3i j and OP 7 j , find the value of k.
(18B037)
For each of the following, find the values of the unknowns such that p is parallel to q.
(a) p = 3mi + 10j + 4nk, q = 6i – 5j + 2k
(b) p = (m + n)i + (n – m)j + 3k, p = (m – 3)i – 2j + 3k
§18.5 Division of a Line Segment
(18B038)
It is given that OA i j and OB 4i 5j are two vectors on a 2-dimensional coordinate plane. If P is
a point on AB such that AP : PB 2 :1 , express OP in terms of i and j.
(18B039)
It is given that OA 3i 2 j and OB 4i j are two vectors on a 2-dimensional coordinate plane. If P
is a point on AB such that AP : PB 3 : 2 , express OP in terms of i and j.
(18B045)
It is given that R and S are two points in a coordinate system and O is the origin.
P is a point on line segment RS such that RP : PS m : n . If the coordinates of R,
1
7
S and P are , 0, 3 , 0, 7, and (1, 2, 2) respectively, find the values of m
2
5
and n.
Section A Questions
§18.2 Operations and Properties of Vectors
(18C001)
The figure shows △ABC. D, E and F are points on AB, BC and CA
(18C006)
In the figure, ABCD is a square. P and Q are the mid-points of CD and AD
respectively. It is given that AD a and AB b .
(a) If QP ma nb , find the values of m and n.
(b) If BP sa tb , find the values of s and t.
respectively, where AD : DB = 1 : 3, BE : EC = 2 : 3 and AF : FC = 2 : 1.
If AB p and BC q , express the following vectors in terms of p and q.
(a)
(c)
(b) DE
(d) FE
AF
DF
(18C002)
The figure shows a parallelogram ABCD. P, Q, R and S are points on AB,
BC, CD and DA respectively, where AP : PB = 1 : 1, BQ : QC = 1 : 2,
CR : RD = 1 : 3 and DS : SA = 2 : 1. If AD p and DC q , express the
following vectors in terms of p and q.
(a)
SR
(b) QR
(c)
PS
(d) QP
(18C007)
In the figure, APQC is a diagonal of parallelogram ABCD. If AP = CQ,
prove that BQDP is also a parallelogram.
(18C008)
In the figure, ABC is a triangle. E is the mid-point of BC. D and F are
points on AB and AC respectively such that AD : DB = AF : FC = 1 : 3.
AE and DF intersect at G. Let AB 4a , AC 4b and AG k AE .
(a) Express CB , AE and FD in terms of a and b.
(b) Hence, find the value of k if G is the mid-point of DF.
-7-
(18C009)
(18C003)
In the figure, ABCD is a rhombus. E is the mid-point of AB and
In the figure, ABCD is a parallelogram, AB 3 , AD 3 and
BF : FC = 2 : 1. It is given that AB a and AD b .
ABC 150 .
(a) Express EF and DF in terms of a and b.
(b) If a 3 , find
(a) Find AB AD .
(b) If AB p and AD q , find, in terms of p and q, a unit vector
(i)
EF 2DF ,
(ii) a unit vector which has the same direction of EF 2DF .
which is in the direction of AC .
(18C004)
(18C010)
In the figure, is the angle between two non-zero and non-parallel vectors a and b.
It is given that a and b are two non-zero and non-parallel vectors. Find the values of m1 and m2 in each of
the following.
(a) Prove that a b a b 2 a b cos .
2
2
2
(b) Hence, find the value of a b if a 3 , b 7 and
3
.
(18C005)
In the figure, ABCD is a quadrilateral. P, Q, R and S are the mid-points of
AB, BC, CD and DA respectively. Prove that
(a) 2SP DB ,
(b) PQRS is a parallelogram.
(a) (m1 + m2)a + (m12 – m22)b = a + 5b
(b) (m1 – 2m2)a + m1m2b = –10a – 12b
(18C011)
It is given that the vectors AB (m 2)a (1 2m)b and CD (4 m)a 5mb are parallel, where a
and b are two non-zero and non-parallel vectors.
(a) Find the values of m.
(b) If m < 0, find AP such that 3 AP 7 BA .
(18C018)
§18.3 Vectors in 2-dimensional Space
(18C012)
Find the values of m and n such that (m – 2n)i – (2m + 3n)j is parallel to 11i + 6j and
(5m – n)i + (5m + 4n + 13)j is parallel to i + 2j.
Let a = 3i – 2j and b = –2i + 4j.
(a) Express i and j in terms of a and b.
(b) Express 8i – 7j in terms of a and b.
(18C019)
It is given that OP 2i 3j , OQ 3i j and OR 2i 3j .
(18C013)
(a) If OP OQ QR , find the value of .
It is given that AB 2i 5j and AC 3i 4 j .
(a) Express BC in terms of i and j.
(b) Find a vector which has the same direction as BC and with a magnitude
(b) If mOP nOQ OR , find the values of m and n.
53 .
(18C014)
Let A(1, –3), B(–6, 2) and C(–2, 5) be three points on the coordinate plane.
(a) Find AC .
(b) If OP AC , find the angle between OP and the positive x-axis, correct to the nearest 0.1.
(c) Find 2 AC OB . Hence, find a unit vector which has the same direction of 2 AC OB .
-8-
(18C015)
In the figure, the angle between the vector p and the positive x-axis is 30 and
the magnitude of p is 5.
(a) Express p in terms of i and j.
(b) If q is a vector with the same direction of p and the magnitude of q is 3,
express q in terms of i and j.
(18C020)
Let A(–2, 3) and B(1, –3) be two points on the coordinate plane. The position vector of C with respect to
O is given by OC 2OA ( 1)OB .
(a) Express OC in terms of , i and j.
(b) If OC is parallel to 2i – j, find
(i) the value of ,
(ii)
OC .
(18C021)
In the figure, A, B and C are three points on the coordinate plane. OABC is a
square of side 4 units. P and Q are the mid-points of AB and BC respectively.
Let OP p and OQ q .
(a) Express p and q in terms of i and j.
(b) Hence, express AC in terms of p and q.
(18C016)
A(2, 5) and B are two points on the coordinate plane, and is the angle between AB and the line y = 5. If
AB 2 3 and sin
3
, find the coordinates of B.
2
(18C017)
It is given that OA 2i j , OB 3i 4 j and OC (m 1)i (2n 4) j are vectors on the coordinate
plane. If AC 3 AB , find the values of m and n.
(18C022)
It is given that OA mi nj , OB 2i j and OP 3i 4 j , where m and n are scalars.
(a) If AP 2 AB , find the values of m and n.
(b) Hence, find a unit vector which has the same direction of AB .
(c) Find a vector which is in the opposite direction of AB and has a magnitude of
8.
(18C023)
A(2, 5), B(3, 7) and C(2k, 6k + 2) are three points on the coordinate plane, where k is a scalar.
(a) (i) Express AB in terms of i and j.
(ii) Express AC in terms of k, i and j.
(b) If A, B and C are collinear, find
(i) the value of k,
(ii) AB : BC.
(b) Find PR .
(18C030)
A, B, C and D are points in the 3-dimensional system such that OAC and OBD are straight lines,
OA 2i j 3k and OB 3i 2 j k . It is given that OA : AC = 2 : 1 and OB : BD = 3 : 2.
(a) Express AB and CD in terms of i, j and k.
(b) Are AB and CD parallel? Explain your answer.
(a) Express AB and BC in terms of a, b, i and j.
(b) Determine whether A, B and C are collinear.
(18C031)
Let P(2, 3, –5) and Q(–3, 4, 1) be two points in the 3-dimensional coordinate system. R is a point on line
segment PQ such that PR : RQ = 3 : 1.
§18.4 Vectors in 3-dimensional Space
(18C025)
It is given that AB i 3j k and AC 2i 3j 2k .
(a) Find PQ and PR in terms of i, j and k.
-9-
30 .
(18C026)
It is given that OP 2i 4 j (m 1)k , OQ (2n 1)i 2 j 3k and OR 2i (r 2) j 5k . If
QR 2PQ , find the values of m, n and r.
(18C027)
Let A(–2, , 1) and B(2, 4, ) be two points in the 3-dimensional coordinate system. If O, A and B are
(a)
Express PQ and QR in terms of i, j and k.
(ii) Hence, find the coordinates of R.
It is given that A(a, b), B(b, a) and C(a + 1, b – 2) are three distinct points on a coordinate plane, where a
and b are real numbers.
collinear, find
Let P(1, 2, –5) and Q(–2, 3, 7) be two points in the 3-dimensional coordinate system. R is a point extended
from PQ such that 3PQ QR .
(a) (i)
(18C024)
(a) Express BC in terms of i, j and k.
(b) Find a vector which is in the direction of BC and with a magnitude
(18C029)
(b) Find OR and the coordinates of R.
(c) If S is a point extended from PQ such that PS = 3PQ, find the coordinates of S.
(18C032)
P(1, –3, 4), Q(2, –1, 6) and R(m, 8n – 1, 5n – 3) are three points in the 3-dimensional coordinate system.
(a) Express PQ in terms of i, j and k.
(b) Express PR in terms of m, n, i, j and k.
(c) If P, Q and R are collinear, find
(i) the values of m and n,
(ii) PQ : QR.
AB ,
(b) AB : OB.
(18C028)
If cos i 2 sin j m k is parallel to 3i 6 j 2k , find the values of
(a) θ , where 0 θ 2π ,
(b) m.
(18C033)
Let A(2, –3, 1), B(1, 4, –2) and C(–2, 1, 2) be the vertices of △ABC.
(a) Find the median AP in terms of i, j and k.
(b) Find the coordinates of the following points.
(i) P
(ii) The centroid of △ABC
(18C034)
In the figure, A(1, –2, 4), B(5, 2, –3), C(6, 1, 2) and D are four points in the
3-dimensional coordinate system. If ABCD is a parallelogram,
(18C039)
(a) find the coordinates of D,
(b) express AC and BD in terms of i, j and k.
(a) Express OG in terms of p, OM and NB .
(b) Express OG in terms of q, OM and NB .
(c) Hence, find the values of p and q.
(18C035)
(18C040)
In the figure, OABC is a parallelogram. M and N are points on AB and BC
The coordinates of two points A and B are (2, 1, –3) and (0, 3, 4) respectively.
respectively such that AM : MB = CN : NB = r : s.
(a) Find a unit vector which has the same direction of AB .
(b) CD has the same direction of AB and CD 3 57 . If the coordinates of C are (3, –2, 6),
(i)
find the coordinates of D,
In the figure, M and N are the mid-points of OA and OB respectively. AN and
BM intersect at G such that AN p AG and BM q BG .
(a) Express OB in terms of OA , OM , r and s.
(b) Express OB in terms of OC , ON , r and s.
(c) Hence, find AC : MN in terms of r and s.
(ii) show that points A, B and D form a triangle.
§18.5 Division of a Line Segment
(18C036)
In the figure, OAD is a triangle. B and C are points on AD such that
- 10 -
AB : BC : CD = 1 : 3 : 1. Suppose OA u and OD v .
(a) Express OB in terms of u and v.
(b) Express OC in terms of u and v.
(c) Show that OA OD OB OC .
(18C037)
In the figure, OAB is a triangle. P and Q are points on AB and OB respectively such that AP : PB = 1 : 2
and OQ : QB = 2 : 3. Let OA a and OB b .
(a) Express OP in terms of a and b.
(b) Express AQ in terms of a and b.
(c) If OB 10 , find
3
OP AQ .
2
(18C038)
In the figure, OAB is a triangle. G and H are points on OA and OB respectively
such that OG : GA = 3 : 1 and OH : HB = 2 : 1. AH and BG intersect at M such
that AM : MH = 1 : r and BM : MG = 1 : s. Let OA a and OB b .
(a) Express OM in terms of r, a and b.
(b) Express OM in terms of s, a and b.
(c) Hence, find the values of r and s.
(18C041)
In the figure, ABCD is a quadrilateral. The diagonals AC and BD intersect at
E such that AE : EC = BE : ED = m : n, where m > n. With respect to a
reference point O, not shown in the figure, the position vectors of A, B, C
and D are a, b, c and d respectively.
(a) Express OE in terms of m, n, a and c.
(b) Express OE in terms of m, n, b and d.
(c) Is ABCD a parallelogram? Explain your answer.
(18C042)
In the figure, OAB is a triangle. S and T are points on AB and OB respectively
such that AS : SB = 1 : 3 and OT : TB = 3 : 2. AT and OS intersect at M such that
AM : MT = p : 1 and OS qOM . Let OA a and OB b .
(a) Express OS in terms of a and b.
(b) Express OM in terms of p, a and b.
(c) Hence, find the values of p and q.
(18C043)
In the figure, OCA and ADB are straight lines. C is the mid-point of OA
and AD : DB = 1 : 3. BC and OD intersect at E.
(a) Express OD in terms of OA and OB .
(b) If BE r BC and OE sOD ,
(i) find the values of r and s,
(ii) express AE in terms of OA and OB .
Section B Questions
(18D001)
In the figure, OACB is a rectangle. D and E are points on OA such that
OD : DE : EA = 3 : 4 : 2. BE and CD intersect at F such that BF : FE = 1 : m.
Let OA a and OB b .
(a) Express OE in terms of a.
(b) Express BF in terms of m, a and b.
(c) Find the value of m.
(d) Using the above results, express CF in terms of a and b.
(18D002)
In the figure, OACB is a parallelogram. D is a point on OA such that
OD : DA = 3 : 1. The diagonal AB intersects CD at M . N is a point on AB.
Let OA a and OB b .
(a) Express AD in terms of a.
(b) Find DM : MC.
- 11 -
(c) Express AM in terms of a and b.
(d) If DN is parallel to OB, express MN in terms of a and b.
(18D003)
In the figure, OACB is a parallelogram. D and E are points on AC and BC
respectively such that E is the mid-point of BC and AD : DC = 1 : 3. OD
and OE intersect AB at F and G respectively. Let OA a and OB b .
(a) Express OD and OE in terms of a and b.
(b) Suppose AG : GB = m : 1.
(i) Express OG in terms of m, a and b.
(ii) Hence, find the value of m.
(c) Find AF : FG : GB.
(18D004)
In the figure, OAB is a triangle. C and D are points on OA and OB respectively
such that OC : CA = OD : DB = 2 : 3. CD is produced to E such that
(18D007)
In the figure, OAB is a triangle. C and D are trisection points of OA. E is a
point on AB such that AE : EB = 2 : 1. OE intersects BC and BD at P and Q
CD : DE = 4 : 1. AE and OB intersect at F. Let OA a and OB b .
(a) (i) Express AB and CD in terms of a and b.
(ii) Hence, prove that AB and CD are parallel.
(b) Find AB : DE.
respectively. Let OA a and OB b .
(a) Express OE in terms of a and b.
(c) Hence, express AF in terms of a and b.
(18D005)
In the figure, OACB is a square of side 5 units. D and E are points on BC and
AC respectively such that E is the mid-point of AC and BD : DC = 1 : 2. AD
and OE intersect at G such that OG OE and AG AD . CG is produced
(b) (i)
r
If BP r BC , show that OP a (1 r )b .
3
(ii) Hence, find the value of r.
(c) If QD s BD , find the value of s.
(d) Find OP : PQ : QE.
(e) Find area of △OCP : area of △BPQ.
(18D008)
- 12 -
to meet OA at F. Let a be the unit vector with the same direction of OA ; b be
the unit vector with the same direction of OB .
(a) Find OG in terms of , a and b.
(b) Find OG in terms of , a and b.
(c) Find the values of and .
(d) Hence, find the lengths of OG and OF.
(18D006)
In the figure, ABC is a triangle. E and F are points on BC and AC respectively
such that E is the mid-point of BC and AF : FC = 1 : 3. AE and BF intersect at G
and CG is produced to meet AB at D. With respect to the reference point O, not
shown in the figure, the position vectors of A, B, C, E, F and G are a, b, c, e, f and
g respectively.
(a) Express e and f in terms of a, b and c.
(b)
(c)
(d)
(e)
Let AG : GE = 1 : . Express g in terms of , a, b and c.
Let BG : GF = : 1. Express g in terms of , a, b and c.
Express g in terms of a, b and c.
Find AD : DB.
The figure shows a quadrilateral ABCD. AC and BD intersect at E such that EC
OA 5i 2 j , OB 5i j and OE i .
(a) (i) Find BA and BE .
(ii) Find BC . Hence, find the coordinates of C.
Suppose OD 7i hj and DE : EB = 2 : k, where h and k are constants.
(b) (i) Find CE in terms of h, k, i and j.
(ii) Hence, find the values of h and k.
(c) Show that ABCD is a trapezium.
1
AC . It is given that
3
(18D009)
M2V3 Chapter 18 Quiz
Introduction to Vectors
Name: ______________________
In the figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : m. The coordinates of A,
1.
B and C are (2, 1), (6, 3) and (3, 6) respectively.
(a) (i)
Find AB , AC and BC .
(ii) Find AP in terms of m, i and j.
(iii) Hence, express the coordinates of P in terms of m.
(b) If AP BC , find
(i) the value of m,
(ii) the area of parallelogram ABCD.
Class: ____________ (
)
Result: ____________
In the figure, PQRS is a parallelogram, PQ u and PS v .
Express each of the following vectors in terms of u and v.
(a) RQ
(b) RS
(c) PR
(d) QS
It is given that the vectors AB 4a (2m 1)b and CD (3m 2)a 3b are parallel, where a and b
are two non-zero and non-parallel vectors and m is a scalar. Find the values of m.
3.
It is given that A(2, –3), B(4, 2) and C(–2, 4) are points on a coordinate plane. Express each of the
following in terms of i and j.
- 13 -
2.
(a) AB
4.
(c)
CB AB
Let a = 2i + 3j – 2k, b = –3i + j – k and c = –4i + 3k be three vectors in a 3-dimensional system. In
each of the following, find a unit vector which has the same direction of the given vector.
(a) 2a – b
5.
(b) BC
(b) 3b – 2c
Let OA 3i 4 j 2k and OB 2i 3j k be two vectors in a 3-dimensional system. P is a point
on line segment AB. Find OP in each of the following cases.
(a) P is the mid-point of AB.
(b) AP : PB = 2 : 3
6.
In the figure, OAB is a triangle. C and D are points on OA and OB respectively
such that OC : CA = 3 : 2 and OD : DB = 2 : 1. AD and BC intersect at E such
that AE : ED = 1 : m and BE : EC = 1 : n. Let OA a and OB b .
(a) Express OE in terms of m, a and b.
(b) Express OE in terms of n, a and b.
(c) Hence, find the values of m and n.
~ End of Quiz ~
- 14 -
Full Solutions
Chapter 18 Introduction to Vectors
Teaching Examples
(18A001)
(a) ∵ ABCD is a rectangle.
∴ AB = DC and AB // DC
∴
1
BC
2
ABCD is a parallelogram.
GC
∴
∵
BC AD
1
GC AD
2
E is a point on DC such that DE : EC = 2 : 1.
1
EC DC
3
1
AB (proved in (a))
3
∴
∴
(property of rectangle)
∵
AB DC
∴
Then AB FC DC ( FC )
DC CF
DF
(b)
GE GC CE
∵
BEFC is a rhombus.
GC ( EC )
∴
∵
EF BC
ABCD is a rectangle.
GC EC
∴
BC AD
EF AD
EF CD AD (CD)
1
1
AD AB
2
3
∴
- 15 -
AD DC
AC
(18A002)
(a) ∵ E is a point on DC such that DE : EC = 2 : 1.
∴
∴
∵
∴
∴
DE and DC have the same direction and DE
2
DC .
3
2
DC
3
ABCD is a parallelogram.
DE
AF AD DF
3
DC
1 3
3
AD AB
4
3
pq
4
AD
DC AB
2
DE AB
3
AE AD DE
AD
(b)
2
AB
3
∵
G is the mid-point of BC.
∴
GC and BC have the same direction and GC
(18A003)
(a) AE AB BE
1
AB BC
2
1
AB AD
2
1
p q
2
1
BC .
2
(b)
2 AE AF
(i)
1 3
2 p q p q
2
4
3
2p p (q q)
4
5
p
4
5
∴ 2 AE AF p
4
5
( 4)
4
5
The required unit vector
(ii)
∴
∴
(b)
- 16 -
1
1
AD DB
2
2
1
( AD DB )
2
1
AB
2
HG HC CG
1
1
AC CB
2
2
1
( AC CB)
2
1
AB
2
FG FB BG
1
1
DB BC
2
2
1
( DB BC )
2
1
DC
2
EH EA AH
2 AE AF
1
1
DA AC
2
2
1
( DA AC )
2
1
DC
2
2 AE AF
5
p
4
5
1
p
4
(18A004)
(a) EF ED DF
1
1
AB and HG AB
2
2
EF // AB and HG // AB
AB, EF and HG are parallel.
EF
∵
∴
1
1
DC and EH DC
2
2
FG EH
∴
FG // EH and FG EH
∴
EFGH is a parallelogram. (opp. sides equal and //)
∵
FG
(18A005)
∵
2 AB CD
2[( s 1)a (3t 1)b] (3s 1)a (t 3)b
∴
(2s 2)a (6t 2)b (3s 1)a (t 3)b
2s 2 3s 1
6t 2 t 3
By solving the above system of linear equations, we have
s 3 and t 1 .
(18A006)
AB (6a mb) 2b
6a (2 m)b
CD 5a m(a b)
(5 m)a mb
∵
∴
AB 4OC i 5 j
AB and CD are parallel.
6
2m
5 m
m
6m (5 m)( 2 m)
(1) 2 52
26
m 3m 10 0
2
∴
(m 2)( m 5) 0
The required unit vector
m 2 or 5
AB 4OC
AB 4OC
1
(i 5 j)
26
1
5
i
j
26
26
(18A007)
OC OB BC
OB AD
(2i j) {[ 3 (6)]i (5 3) j}
(2i j) (3i 2 j)
(18A009)
(a) OA 2i 3 j and OB 6i 7 j
i 3j
∴ The coordinates of C are (1, 3).
∴
( 1)( 2i 3 j) ( 1)(6i 7 j)
2( 1)i 3( 1) j 6( 1)i 7( 1) j
(18A008)
(a) AB (7 4)i (2 5) j
(2 2 6 6)i (3 3 7 7) j
(8 4)i (10 4) j
- 17 -
3i 3j
(b)
Let be the angle between OQ and the positive x-axis.
OQ 3i 3 j
∴ Coordinates of Q = (3, 3)
∴ lies in quadrant IV.
3
tan
1
3
∴ 315
∴
(c)
The angle between OQ and the positive x-axis is 315 .
(b)
∵
∴
OR is parallel to 5i 6 j .
8 4 10 4
5
6
48 24 50 20
2
(18A010)
(a) AB OB OA
(4i j k ) (2i 3j k )
2i 4 j ( 1)k
OC i 2 j
AB 4OC (3i 3 j) 4(i 2 j)
(3i 3 j) (4i 8 j)
OR ( 1)OA ( 1)OB
(b)
AB 6
(3 4)i (3 8) j
2i 4 j ( 1)k 6
i 5 j
2 2 4 2 ( 1) 2 6
4 16 ( 1) 2 36
( 1) 2 16
3 or 5
(18A011)
AB (3 1)i (2m 1 m) j [4n 2 (n 1)]k
2i (m 1) j (3n 3)k
Length of OH OH
(1) 2 0 2 8 2
65
BC (7 3)i [5m (2m 1)] j [6n 4 (4n 2)]k
4i (3m 1) j (2n 6)k
∵
∴
∴
AB is parallel to BC .
2 m 1 3n 3
4 3m 1 2n 6
3m 1 2(m 1) and
m3
(18A013)
(a) ∵ ABCD is a parallelogram.
DC AB p and BC AD q
1
1
EC BC q (∵ BE : EC = 2 : 1)
3
3
∴
2n 6 2(3n 3)
and
n3
DE DC CE
(18A012)
(a) With the notations in the figure,
BA BC BF ( BA BC ) BF
BD BF
BH (∵ HDBF is a rectangle.)
(b)
(i)
BA (4 5)i (0 3) j (3 4)k
DC ( EC )
H
D
DC EC
F
1
p q
3
B
(b)
- 18 -
i 3j k
∵
AF : FE 1 : 2
∴
DF
BC (1 5)i (4 3) j (5 4)k
4i j k
BF (4 5)i (2 3) j (8 4)k
i j 4k
∴ BH (i 3j k ) (4i j k ) (i j 4k )
(1 4 1)i (3 1 1) j (1 1 4)k
6i 3j 4k
OH OB BH
(5i 3j 4k ) (6i 3j 4k )
(5 6)i (3 3) j (4 4)k
i 8k
∴ The coordinates of H are (–1, 0, 8).
(ii)
Length of BH BH
(6) 2 (3) 2 4 2
61
(2)( DA) (1)( DE )
2 1
1
2(q) p q
3
3
1
1
2q p q
3
3
1
7
p q
3
9
(18A014)
(a)
(i)
1
1
AC q
2
2
∵ BQ : QE m : n
AE
(n)( AB) (m)( AE )
nm
n
m
p
q
nm
2(n m)
∴ AQ
(ii)
1
1
AB p
2
2
∵ CQ : QD m : 1
AD
( m)( AD) (1)( AC )
m 1
m
1
p
q
2(m 1)
m 1
∴ AQ
(b)
From the results of (a)(i) and (a)(ii), we have
n
m
m
1
p
q
p
q
nm
2(n m)
2(m 1)
m 1
- 19 -
m
n
n m 2(m 1) ...(1)
∴
1
m
...(2)
2(n m) m 1
(1) (2), we have
2n m
m 2
m2
n
...(3)
4
∴ Substituting (3) into (2), we have
m
1
2
m
m 1
2
m
4
2m
1
2
m 4m m 1
2
1
(∵ m 0)
m 4 m 1
2m 2 m 4
m2
22
1.
Substituting m 2 into (3), we have n
4
(18A015)
(a) OA 2i 3 j and OB 9i 11j
5(2i 3 j) 2(9i 11j)
∴ OQ
52
(10i 15 j) (18i 22 j)
7
1
(28i 7 j)
7
4i j
(b)
Let OA xi yj zk .
5OA 2OB
52
5( xi yj zk ) 2(2i 5j 6k )
3i 5 j k
∴
7
21i 35 j 7k (5 x 4)i (5 y 10) j (5 z 12)k
By Theorem 18.6(a), we have
21 5 x 4
35 5 y 10
7 5 z 12
By solving the above equations, we have x –5, y 9, z –1.
∴ The coordinates of A are (–5, 9, –1).
OQ
∵
(18A016)
(a)
AB [3 (1)]i (5 3) j
4i 2 j
(b)
Let be the angle between OR and the positive x-axis.
OR 2 AB
2(4i 2 j)
8i 4 j
∴ Coordinates of R = (8, 4)
∴ lies in quadrant I.
4 1
tan
8 2
∴ 26.6 (cor. to the nearest 0.1)
∴
The angle between OR and the positive x-axis is 26.6 .
(c)
OC 6i j
(ii)
(7i 12 j k ) ( i 10 j)
OR 2OC (8i 4 j) 2(6i j)
(7 1)i (12 10) j k
8i 4 j 12i 2 j
6i 2 j k
∴ The coordinates of E are (6, 2, 1).
(8 12)i (4 2) j
4i 6 j
OE 6i 2 j k
OR 2OC 4i 6 j
∴
OE OC CE
(4) 2 6 2
6 2 2 2 12
2 13
41
The required unit vector
∴ The required unit vector
OR 2OC
OE
OR 2OC
1
(4i 6 j)
2 13
2
3
i
j
13
13
- 20 -
(18A017)
(a) With the notations in the figure,
(18A018)
(a)
(i)
CB CD CG (CB CD) CG
CA CG
2
2
AC q
2 1
3
∵ BQ : QE m : 1
AE
(1)( AB) (m)( AE )
1 m
1
2m
p
q
1 m
3(1 m)
∴ AQ
CA AE
CE
(b)
(i)
CB (4 7)i (6 12) j (0 1)k
3i 6 j k
CD (5 7)i (8 12) j (3 1)k
2i 4 j 2k
CG (11 7)i (12 12) j (0 1)k
4i k
∴ CE (3i 6 j k ) (2i 4 j 2k ) (4i k )
(3 2 4)i (6 4) j (1 2 1)k
i 10 j
OE
(ii)
2
2
AB p
2 1
3
∵ CQ : QD n : 1
AD
(n)( AD) (1)( AC )
n 1
2n
1
p
q
3(n 1)
n 1
∴ AQ
1
41
6
41
( 6i 2 j k )
i
2
41
j
1
41
k
(b)
From the results of (a)(i) and (a)(ii), we have,
1
2m
2n
1
p
q
p
q
1 m
3(1 m)
3(n 1)
n 1
2n
1
1 m 3(n 1) ...(1)
∴
2m 1
...(2)
3(1 m) n 1
(1) (2), we have
3
2n
2m 3
9
n
...(3)
4m
∴ Substituting (3) into (2), we have
2m
1
9
3(1 m)
1
4m
2m
4m
3 3m 9 4m
9 4m 2(3 3m)
- 21 -
AB DC q
(c)
AC AD DC p q
(d) DB DC CB
q p (by (a))
(18B002)
(a) AD CD BC CD BD
(b)
AD CB AD AD 0
(c)
BC ED AD DE AE (or EC )
(18B003)
(a) AB ED FE ED FD (or HB )
(b)
AB EF AB AB 0
(c)
CD OG CD CO CD OC OD (or HO )
(18B004)
(a)
AD CB AD AD 0 0
(b)
AD AB AD BA BD 52 122 13
(18B005)
(a) BC BA AC AC AB q p
(b) BD
(c)
Basic Questions
(b)
AB CE AB EA EB (or DE )
(d) HG ED OA CD DE EO CO (or OG )
3
m
2
3
9
3
.
Substituting m into (3), we have n
2
3 2
4
2
(18B001)
(a) CB DA AD p
(d)
1
1
1
BA AB p
2
2
2
CD CB BD
BC BD
1
(q p) p
2
1
pq
2
(d) BE BC CE
BC 1 AC
2
(q p) 1 q
2
1
qp
2
(by (a) and (b))
(by (a))
(18B006)
s 1 3
2t t 1
By solving the above linear equations, we have
1
s 2 and t .
3
(18B007)
2 s t 3
s t 3
By solving the above system of linear equations, we have
s 2 and t 1 .
cos
- 22 -
(18B009)
∵ PQ and RS are parallel.
m 1 3
∴
2m
3m
2
m m 0
m(m 1) 0
m 0 or 1
(18B010)
(a) AB [2 (1)]i (3 0) j 3i 3j
(b)
AB (4 2)i [7 (3)] j 2i 10 j
(18B011)
(a) AB [( 2m 3) (m 1)]i [(3n 2) (n 1)] j (m 2)i (2n 3) j
(b)
AB [( 4 2n) (2m 3)]i [( 3 n) (3m n)] j (1 2m 2n)i (3m 2n 3) j
(18B012)
1
(a) sin
2 1
2
2
cos
tan
2
2 2 12
1
2
1
5
2
5
3 (2)
3
2
32 (2) 2
tan
(18B008)
∵ PQ and RS are parallel.
2
m
∴
3m 10
4
3m 2 10m 8 0
(3m 4)( m 2) 0
4
m or 2
3
2
(b) sin
2
2
13
3
13
2
3
(18B013)
1
(a) sin
(3) 2 12
3
cos
tan
(b) sin
cos
tan
1
10
(3) 2 12
3
10
1
3
2
(1) (2)
1
2
2
(1) (2)
2
2
2
5
1
5
2
2
1
(18B014)
(a) The required magnitude = 12 22 5
Since i 2 j lies in quadrant I,
1
the required angle tan 2
63.4 (cor. to 1 d. p.)
(b) The required magnitude = 12 ( 3 ) 2 2
Since i 3 j lies in quadrant IV,
the required angle 360 tan 1 3
300
(18B015)
(a) The required magnitude = (5) 2 29
2
2
Since 5i 2 j lies in quadrant II,
the required angle 180 tan 1 2
5
158.2 (cor. to 1 d. p.)
(b) The required magnitude = (2) 2 (3) 2 13
Since 2i 3 j lies in quadrant III,
the required angle 180 tan 1 3
2
236.3 (cor. to 1 d. p.)
(b) 3c 2a 2b a
c 1 (a 2b)
3
1 [( 2i 3 j) 2(4i 3 j)]
3
1 [( 2 8)i (3 6) j]
3
10 i j
3
(18B020)
(a)
OA OB AB
3i 2 j (2i 5 j)
(3 2)i [2 (5)] j
5i 7 j
(18B016)
(a) AB (3 2)i [4 (1)] j i 5 j
(b)
AB OB OA
AB 12 52 26
(b) OA OB AB
The required unit vector =
AB
AB
1
5
i
j
26
26
4i j (3i 7 j)
[4 (3)]i (1 7) j
- 23 -
7i 6 j
(18B017)
(a) AB [3 (2)]i (5 3) j i 2 j
(b)
AB (1) 2 2 2 5
AB
1
2
The required unit vector =
i
j
5
5
AB
(18B021)
(a) AB (2 1)i [5 (2)] j 3i 7 j
(b) BC [2 (2)]i (4 5) j 4i j
(c)
BA BC AB BC
(3i 7 j) (4i j)
(18B018)
(a) a + 2b = –2i + j + 2(3i – 2j) = (–2 + 6)i + (1 – 4)j 4i 3j
(b) c 2a a 2b
c (a 2b)
(4i 3 j)
4i 3 j
(by (a) and (b))
(3 4)i (7 1) j
7i 8 j
(d) 2 AC 3 AB 2( AB BC ) 3 AB
(by (a))
AB 2 BC
(3i 7 j) 2(4i j)
(by (a) and (b))
(3 8)i (7 2) j
(18B019)
(a) 2a – 3b = 2(2i – 3j) – 3(4i +3j) = (4 – 12)i + (–6 – 9)j 8i 15 j
11i 9 j
(18B022)
(a) AB OB OA (3 2)i (2 1) j i 3j
(18B026)
(a) AB OB OA (1 2)i (2 3) j [5 (1)]k 3i 5 j 6k
(b) BC OC OB (3 3)i [4 (2)] j 6i 6 j
(c)
3CB 2 BA 3BC 2 AB
3(6i 6 j) 2(i 3j)
(by (a) and (b))
(b)
AB OB OA (2 3)i [1 (2)] j (2 4)k i 3j 6k
(18 2)i (18 6) j
(18B027)
(a)
20i 24 j
(d) 2OB 3 AC 2OB 3( AB BC )
2(3i 2 j) 3[(i 3j) (6i 6 j)]
2 2 2 (3) 2 22
2 13 22
2 9
3
(by (a) and (b))
(6 3 18)i (4 9 18) j
9i 5 j
- 24 -
(18B023)
2m 4
(a)
3n 9
By solving the above linear equations, we have
m = 2 and n = –3.
m 2 2 m 3
(b)
2n 3 2 3n
By solving the above linear equations, we have
1
m = 5 and n = .
5
AB 22
(b)
AB 33
(2) 2 ( 1) 2 (5) 2 33
( 1) 2 29 33
( 1) 2 4
1 2
3 or 1
(18B028)
(a) AB ( 2)i [1 (3)]j (3 1)k ( 2)i 2 j 2k
AB 2 6
(18B024)
(a) ∵ p is parallel to q.
∴
(b) ∵
∴
2
4 6
4
3
( 2) 2 2 2 2 2 2 6
( 2) 2 8 24
( 2) 2 16
2 4
6 or 2
p is parallel to q.
2
4
2 2
1
2
(b)
AB [5 (3)]i (0 ) j [4 (2)]k 2i j 6k
AB 14 2
(2) 2 ( ) 2 6 2 14 2
2 40 392
2 352
4 22
(18B025)
(a) AB (2 1)i [0 (1)] j (2 3)k i j 5k
(b)
AB (5 2)i [1 (3)] j (3 4)k 3i 4 j 7k
(18B029)
(a) AB (2 3)i (4 2) j [5 (1)]k i 2 j 6k
(b) CA (3 1)i [2 (2)] j (1 3)k 2i 4 j 4k
(c)
2OA 3BC 2(3i 2 j k ) 3[(1 2)i (2 4) j (3 5)k ]
(6 3)i (4 18) j (2 6)k
3i 14 j 8k
(d) OB 2 AC OB 2CA
2i 4 j 5k 2(2i 4 j 4k )
(by (b))
(2 4)i (4 8) j (5 8)k
6i 12 j 3k
(18B030)
(a) BC [1 (1)]i [2 (2)] j (1 3)k
2i 4 j 4k
(b) 2 AB 2[( 1 2)i (2 3) j (3 6)k ]
2(3i 5 j 3k )
6i 10 j 6k
(c)
3i 5 j 5k
- 25 -
(d) BA 3OB {[ 2 (1)]i [3 (2)] j (6 3)k} 3(i 2 j 3k )
(3 3)i (5 6) j (3 9)k
j 12k
(b) 3a – 2c = 3(2i – 3j – k) – 2(3i + 2j + k) = (6 – 6)i + (–9 – 4)j + (–3 – 2)k = –13j –5k
3a 2c (13) 2 (5) 2 194
13
5
j
k
The required unit vector =
194
194
(c) b – 3a + 2c = –i + 4j + 5k – 3(2i – 3j – k) + 2(3i + 2j + k)
= (–1 – 6 + 6)i + (4 + 9 + 4)j + (5 + 3 + 2)k
= –i + 17j + 10k
b 3a 2c (1) 2 172 102 390
1
17
10
i
j
k
The required unit vector =
390
390
390
AB (5) 2 12 32 35
The required unit vector =
AB
AB
5
1
3
i
j
k
35
35
35
(18B032)
(a) PQ [4 (2)]i (1 3) j (3 1)k 6i 4 j 4k
PQ 6 2 (4) 2 (4) 2 68 2 17
The required unit vector =
(18B034)
(a) 3a + c = 3(–3i + 5k) + (–6i – 2j – k) = (–9 – 6)i – 2j + (15 – 1)k = –15i – 2j + 14k
3a c (15) 2 (2) 2 142 5 17
3
2
14
i
j
k
The required unit vector =
17
5 17
5 17
(b) 2b – 3c = 2(2i – 4j + 2k) – 3(–6i – 2j – k) = (4 + 18)i + (–8 + 6)j + (4 + 3)k = 22i – 2j + 7k
(18B031)
(a) AB (2 3)i (5 4) j [2 (1)]k 5i j 3k
(b)
2a b 32 (2) 2 32 22
3
2
3
i
j
k
The required unit vector =
22
22
22
AC 2OC [(1 2)i (2 3) j (1 6)k ] 2(i 2 j k )
(1 2)i (1 4) j (7 2)k
(b)
(18B033)
(a) 2a + b = 2(2i – 3j – k) + (–i + 4j + 5k) = (4 – 1)i + (–6 + 4)j + (–2 + 5)k = 3i – 2j + 3k
PQ
PQ
3
2
2
i
j
k
17
17
17
2b 3c 222 (2) 2 72 537
22
2
7
i
j
k
The required unit vector =
537
537
537
(c) 2a + 3b – c = 2(–3i + 5k) + 3(2i – 4j + 2k) – (–6i – 2j – k)
= (–6 + 6 + 6)i + (–12 + 2)j + (10 + 6 + 1)k
= 6i – 10j + 17k
2a 3b c 62 (10) 2 172 5 17
The required unit vector =
6
2
17
i
j
k
5
5 17
17
(18B035)
(a) a – 2b = 3i + 2j – k – 2(–2i + 3k) = (3 + 4)i + 2j + (–1 – 6)k 7i 2 j 7k
(b) a 2b 3c 4i j
a 2b 4i j
c
3
7i 2 j 7k 4i j
3
i j 7k
3
(18B040)
2(2i j 3k ) (i 2 j 4k )
OP 2OA OB
i 4 j 2k
2 1
3
3
3
(by (a))
(3i 2 j k ) 3(2i j 5k ) 3
OP OA 3OB
i 1 j 7k
1 3
4
4
4
2
(18B036)
(a) 2a + 3b = 2(–2i + 3j + 5k) + 3(3i – 2j – 4k) = (–4 + 9)i + (6 – 6)j + (10 – 12)k 5i 2k
(b) 2a 3b 2c 3i 4 j
2a 3b 3i 4 j
c
2
5i 2k 3i 4 j
2
4i 2 j k
(18B041)
(18B042)
Let OB xi yj .
3OA 5OB
8
3 3
3(9i j) 5( xi yj)
∴ i j
2 2
8
12i 12 j (5 x 27)i (5 y 3) j
5 x 27 12
i.e.
5 y 3 12
By solving the above linear equations, we have x = 3 and y = 3.
∴ OB 3i 3j
∵
(by (a))
- 26 -
(18B037)
(a) ∵ p is parallel to q.
3m 10 4n
∴
6
5 2
By solving the above linear equations, we have
m = –4 and n = –1.
(b) ∵ p is parallel to q.
mn nm 3
∴
m3
2
3
m n m 3
i.e.
n m 2
By solving the above system of linear equations, we have
n = –3 and m = –1.
(18B038)
(i j) 2(4i 5 j)
OP OA 2OB
3i 11 j
1 2
3
3
(18B039)
2(3i 2 j) 3(4i j) 18 1
OP 2OA 3OB
i j
23
5
5
5
OP
(18B043)
Let ON xi yj zk .
2OM ON
3
2(2i 9 j 12k ) ( xi yj zk )
2i 7 j 9k
∴
3
6i 21j 27k ( x 4)i ( y 18) j ( z 24)k
x 4 6
i.e. y 18 21
z 24 27
By solving the above linear equations, we have x = 2, y = 3 and z = 3.
∴ ON 2i 3j 3k
∵
OP
(18B044)
k OC OD
k 1
k ( i 9 j) (3i j)
7j
k 1
7( k 1) j (3 k )i (9k 1) j
OP
3 k 0
i.e.
7(k 1) 9k 1
By solving the above linear equations, we have k = 3.
(18B045)
nOR mOS
mn
1
7
n i 3k m 7 j k
5
2
i 2 j 2k
mn
7
1
(m n)i 2(m n) j 2(m n)k ni 7mj 3n m k
5
2
m n 7 n
5
i.e. 2(m n) 7 m
1
2(m n) 3n m
2
By solving the above system of equations, we have m = 2 and n = 5.
OP
- 27 -
Section A Questions
(18C002)
(a) SR SD DR
2
3
2
3
AD DC p q
3
4
3
4
(b) QR QC RC
2
1
2
1
2
1
BC DC AD DC p q
3
4
3
4
3
4
(c)
1
1
1
1
1
1
AD AB AD DC p q
3
2
3
2
3
2
PS AS AP
1
1
1
1
1
1
(d) QP BQ PB BC AB AD DC p q
3
2
3
2
3
2
(18C003)
(a)
AB AD AC
2
2
AB AD 2 AB AD cos ABC
( 3 ) 2 3 2 2( 3 )(3) cos 150
21
(b) The required unit vector
AF
2
2
2
2
AC ( AB BC ) p q
3
3
3
3
(b) DE DB BE
(c)
pq
1
1
p
q
21
21
21
(18C004)
(a) Construct a parallelogram with a diagonal a – b as shown on the right.
3
2
3
2
AB BC p q
4
5
4
5
a b OC
2
2
2
2
OA OB 2 OA OB cos OAC
DF AF AD
AF 1 AB
4
2 p 2 q 1 p
3
3 4
5 p 2q
12
3
AC
AC
(18C001)
(a)
a b 2 a b cos
2
2
2
a b 3 2 7 2 2(3)(7) cos π
3
37
(d) FE DE DF
3 p 2 q 5 p 2 q
4
5 12
3
1p 4 q
3 15
2
a b 2 a b cos
(b) From (a), we have
(by (a))
(18C005)
(by (b) and (c))
(∵ BC AD )
SP SA AP 1 DA 1 AB 1 ( DA AB) 1 DB
2
2
2
2
∴ 2SP DB
(b) RQ RC CQ 1 DC 1 CB 1 ( DC CB) 1 DB
2
2
2
2
(a)
∴
2RQ DB
∴
From (a), 2SP DB
∴ SP RQ
∴
∴
SP // RQ and SP RQ
∴
PQRS is a parallelogram.
AG k AE
1
(a b) 2k (a b)
2
1
k
4
(opp. sides equal and //)
(18C009)
(18C006)
(a) QP QD DP
∴
1
1
1
1
1
1
AD DC AD AB a b
2
2
2
2
2
2
1
mn
2
1
1
1
(b) BP BC CP AD CD AD AB a b
2
2
2
1
∴ s = 1, t
2
(18C007)
BP BA AP
- 28 -
CD QC
( AP CQ)
QD
∴
BP // QD and BP QD
∴
BQDP is a parallelogram.
(opp. sides equal and //)
(18C008)
(a) CB AB AC 4a 4b
1
1
1
BC AB CB 4a (4a 4b) 2a 2b
2
2
2
1
1
FD AD AF AB AC a b
4
4
AE AB BE AB
(b)
AG AD DG
1
1
AB DF
4
2
1
1
AB FD
4
2
1
a (a b)
(by (a))
2
1
1
a b
2
2
From (a), AE 2a 2b
(a)
EF EB BF
1
2
1
2
1
2
AB BC AB AD a b
2
3
2
3
2
3
DF DC FC AB 1 BC AB 1 AD a 1 b
3
3
3
(b) (i)
1
2
1
a b 2 a b
2
3
3
5
a
2
5
a
2
5
3
2
15
2
EF 2 DF
5
a
EF 2 DF
1
2
(ii) The required unit vector =
a
15 3
EF 2 DF
2
(18C010)
m m 1
(a) 1 2 2 2
m1 m2 5
m1 m2 1
(m m )(m m ) 5
1
2
1
2
m1 m2 1
m1 m2 5
By solving the above system of linear equations, we have
m1 3 and m2 2 .
......(1)
m 2m2 10
(b) 1
......(2)
m1m2 12
From (1), m1 2m2 10 ……(3)
By substituting (3) into (2), we have
m2 (2m2 10) 12
m22 5m2 6 0
(m2 2)( m2 3) 0
m2 2 or 3
From (2), when m2 2 , m1 6 ; when m2 3 , m1 4
∴ m1 6 , m2 2 or m1 4 , m2 3
(18C011)
(a) ∵ AB and CD are parallel.
m 2 1 2m
∴
4m
5m
5m 2 10m 2m 2 7 m 4
7 m 2 3m 4 0
(7 m 4)( m 1) 0
4
m
or 1
7
(b) From (a), for m
4
18
15
, AB a b
7
7
7
- 29 -
3 AP 7 BA
7 18
15
AP a b
3 7
7
6a 5b
(18C012)
......(1)
a 3i 2 j
(a)
b 2i 4 j ......(2)
(1) 2 + (2): 2a b 4i
1
1
i a b
2
4
1
1
By substituting i a b into (1), we have
2
4
1
1
a 3 a b 2 j
4
2
1
3
j a b
4
8
1
1
1
3
(b) 8i 7 j 8 a b 7 a b
4 4
8
2
9
5
a b
4
8
(by (a))
(18C013)
(a) BC AC AB (3i 4 j) (2i 5 j) 5i 9 j
(b)
BC (5) 2 9 2 106
∴
5i 9 j
5 i 9 j
The required vector = 53 BC 53
106
2
2
BC
(18C014)
(a) AC (2 1)i [5 (3)] j 3i 8 j
(b) Since 3i 8 j lies in quadrant II,
8
the required angle 180 tan 1
3
110.6 (cor. to the nearest 0.1)
(c)
2 AC OB 2(3i 8 j) (6i 2 j)
14 j
14
The required unit vector = j
(18C015)
(a) Let p = ai + bj.
Since |p| = 5, we have
a
cos 30
5
5 3
a
2
b
sin 30
5
5
b
2
5 3
5
∴ p
i j
2
2
(b) q 3
p 35 3
5 3 3
3
i j
i j
p 5 2
2
2
2
(18C016)
Let the coordinates of B be (x, y). Then, AB ( x 2)i ( y 5) j .
∵ sin 0
∴ 0 θ 90 or 90 θ 180
By solving (1) and (2), we have
m 1 and n 4 .
3
Since AB 2 3 , we have
(18C019)
(a)
OP OQ QR
(2i 3 j) (3i j) [( 2 3)i (3 1) j]
i 2 j (i 2 j)
∴ 1
x2
2 3
x2 1
1
or
2
2 3 2
x 3 2 or 3 2
y 5
sin
2 3
y 5
3
2
2 3
y 8
cos
∴
Coordinates of B ( 3 2, 8) or ( 3 2, 8)
(18C017)
- 30 -
AC [(m 1) 2]i [(2n 4) (1)]j (m 3)i (2n 5) j
AB (3 2)i [4 (1)]j i 5j
AC 3 AB
(m 3)i (2n 5) j 3i 15 j
m 3 3
2n 5 15
By solving the above linear equations, we have
m = 6 and n = 5.
(18C018)
∵ (m 2n)i (2m 3n) j // 11i 6 j
m 2n 2m 3n
∴
11
6
6m 12n 22m 33n
4m 3n 0 ......(1)
∵ (5m n)i (5m 4n 13) j // i 2 j
5m n 5m 4n 13
∴
1
2
10m 2n 5m 4n 13
5m 6n 13 ......(2)
(b)
mOP nOQ OR
m(2i 3 j) n(3i j) 2i 3 j
(3n 2m)i (n 3m) j 2i 3 j
3n 2m 2
n 3m 3
By solving the above system of linear equations, we have
m = –1 and n = 0.
(18C020)
(a) OC 2 OA ( 1)OB
2 (2i 3j) ( 1)(i 3j)
[4 ( 1)]i [6 3( 1)] j
(5 1)i (9 3) j
(b) (i)
∵
∴
OC is parallel to 2i – j.
5 1 9 3
(by (a))
2
1
5
13
(ii) OC 5 5 1 i 9 5 3 j
13 13
12 i 6 j
13 13
2
∴
OC 12 6
13 13
2
6 5
13
(18C021)
(a) ∵ OABC is a square of side 4 units.
∴ The coordinates of A, B and C are (4, 0), (4, 4) and (0, 4) respectively.
∵ P and Q are the mid-points of AB and BC respectively.
∴
∴
The coordinates of P and Q are (4, 2) and (2, 4) respectively.
p 4i 2 j and q 2i 4 j
p 4i 2 j
(b) From (a), we have
q 2i 4 j
(1) 2 – (2): 6i 2p q
1
1
i p q
3
6
(2) 2 – (1): 6 j p 2q
j 1p 1q
6
3
(b) (i)
∴
......(1)
......(2)
(ii)
4(i j)
4 1 p 1 q 1 p 1 q
3
6 6
3
2p 2q
- 31 -
AB
AB
∵
A is a point between B and C.
∴
AB : BC 5 : (3 5 5 )
1: 4
BC (a 1 b)i (b 2 a) j (a b 1)i (b a 2) j
(b) Suppose that A, B and C are collinear.
Then, b a a b
a b 1 b a 2
2
(b a) 2(b a) (a b) 2 a b
2(a b) a b
ab 0
ab
i.e. A and B are two equal points.
∴ There is a contradiction.
∴ A, B and C are not collinear.
(b) From (a), AB 5i 5 j
5i 5 j
(5) 2 52
1
1
i
j
2
2
(c) The required vector 8 AB
AB
8 1 i 1 j
2
2
2i 2 j
(18C023)
(a) (i) AB (3 2)i (7 5) j i 2 j
(ii)
AB 12 2 2 5
(18C024)
(a) AB (b a)i (a b) j
(18C022)
(a)
AP 2 AB
(3 m)i (4 n) j 2[( 2 m)i (1 n) j]
(m 3)i (4 n) j 2(2 m)i 2(n 1) j
(m 3) 2(2 m)
4 n 2(n 1)
By solving the above linear equations, we have
m = 7 and n = –6.
The required unit vector =
A, B and C are collinear.
AB and AC are parallel.
2k 2 6k 3 (by (a))
1
2
k1
2
AC (3) 2 (6) 2 3 5
AC (0 4)i (4 0) j
∴
∵
∴
AC (2k 2)i [(6k 2) 5]j (2k 2)i (6k 3) j
(18C025)
(a) BC AC AB (2i 3j 2k ) (i 3j k ) (2 1)i (3 3) j [2 (1)]k 3i 6 j 3k
3i 6 j 3k
5i 2 5 j 5k
(b) The required vector 30 BC 30
BC
(3) 2 (6) 2 3 2
(18C026)
QR 2 PQ
[2 (2n 1)]i [(r 2) 2]j [5 (3)]k 2{[( 2n 1) (2)]i (2 4) j [3 (m 1)]k}
(3 2n)i (r 4) j 8k 2(2n 1)i 4 j 2(m 4)k
3 2n 2(2n 1)
(r 4) 4
8 2(m 4)
By solving the above linear equations, we have
1
m = –8, n
and r = 0.
6
QR 3PQ 3(3i j 12k ) 9i 3j 36k
(ii) OR OQ QR
(2i 3j 7k ) (9i 3j 36k ) (by (a))
11i 6 j 43k
∴ The coordinates of R are (–11, 6, 43).
(18C027)
(a) ∵ O, A and B are collinear.
∴
OA and OB are parallel.
2 1
∴
2
4
i.e. 4 and 1
∴
(b) PR PQ QR
12i 4 j 48k
PR (12) 2 4 2 48 2
∴
2464
AB [2 (2)]i [4 (4)] j (1 1)k
4 154
4i 8 j 2k
(b)
AB : OB
4 2 8 2 (2) 2 : 2 2 4 2 (1) 2
2 21 : 21
(18C030)
(a) AB OB OA (3 2)i (2 1) j [1 (3)]k 5i j 4k
2 :1
- 32 -
(18C028)
(a) ∵ cos i 2 sin j m k and 3i 6 j 2k are parallel.
cos 2 sin m
∴
3
6
2
3m
cos 2
i.e.
sin 3m
2
cos θ sin θ
tan θ 1
θ π or 5π
4
4
(b)
cos 2 sin 2 1
2
2
3m 3m
1
2 2
3m
2
2
2
2
m
3
(18C029)
(a) (i) PQ (2 1)i (3 2) j [7 (5)]k 3i j 12k
3
OA
2
5
OD OB
3
3
3
9
(2i j 3k ) 3i j k
2
2
2
5
10
5
(3i 2 j k ) 5i j k
3
3
3
CD OD OC (5 3)i 10 3 j 5 9 k 8i 11 j 37 k
3 2 3 2
6
6
OC
(b)
AB 5i j 4k
11 37
CD 8i j k
6
6
5 1
4
∵
8 11 37
6
6
∴ AB and CD are not parallel.
(18C031)
(a) PQ (3 2)i (4 3) j [1 (5)]k 5i j 6k
PR
3
3
15 3 9
PQ (5i j 6k ) i j k
4
4
4
4 2
(b) OR OP PR
9
15 3
(2i 3j 5k ) i j k
4
2
4
7 15 1
i j k
4
4
2
(by (a))
The coordinates of R are 7 , 15 , 1 .
4 4
2
(c) ∵ PS = 3PQ
∴
PS 3PQ
(18C033)
(a)
AP AB BP
AB
or PS 3PQ
When PS 3PQ ,
15i 3j 18k
OS OP PS
(2i 3j 5k ) (15i 3j 18k )
When PS 3PQ ,
PS 3(5i j 6k )
15i 3j 18k
OS OP PS
(2i 3j 5k ) (15i 3j 18k )
17i 23k
∴ The coordinates of S are (13, 6, 13) or (17, 0, 23).
- 33 -
(18C032)
(a) PQ (2 1)i [1 (3)] j (6 4)k i 2 j 2k
(b) PR (m 1)i [8n 1 (3)] j (5n 3 4)k (m 1)i (8n 2) j (5n 7)k
(c) (i)
∵
P, Q and R are collinear.
∴
PQ and PR are parallel.
1
2
2
∴
(by (a))
m 1 8n 2 5n 7
By solving the above system of linear equations, we have
m = –10 and n = –3.
(ii)
(b) (i)
∵
∴
P is a point between Q and R.
PQ : QR 3 : (33 3)
3 : 36
1 : 12
OP OA AP
5 11
( 2i 3 j k ) i j k
(by (a))
2
2
1 5
i j
2 2
1 5
∴ The coordinates of P are , , 0 .
2 2
(ii) Let M be the centroid of △ABC.
Then, M must lie on AP such that AM 2 AP .
3
∴ AM 2 5 i 11 j k
3 2
2
5 i 11 j 2 k
3
3
3
∴ OM OA AM
(2i 3j k ) 5 i 11 j 2 k
3
3
3
1i 2 j 1k
3
3
3
∴
PQ 12 2 2 2 2 9 3
PR (10 1) 2 [8(3) 2]2 [5(3) 7]2 1089 33
( P is the mid -point of BC .)
1
{(1 2)i [4 (3)] j (2 1)k} {( 2 1)i (1 4) j [2 (2)]k}
2
3
3
(i 7 j 3k ) i j 2k
2
2
5 11
i jk
2
2
PS 3(5i j 6k )
13i 6 j 13k
1
BC
2
The coordinates of the centroid of △ABC are 1 , 2 , 1 .
3 3 3
(18C034)
(a) Let the coordinates of D be (p, q, r).
∵ ABCD is a parallelogram.
∴
AB DC
(5 1)i [2 (2)] j (3 4)k (6 p )i (1 q ) j (2 r )k
4i 4 j 7k (6 p )i (1 q ) j (2 r )k
6 p 4
i.e. 1 q 4
2 r 7
By solving the above linear equations, we have p = 2, q = –3, r = 9.
∴ The coordinates of D are (2, –3, 9).
(b)
AC (6 1)i [1 (2)] j (2 4)k
OB OC 4 u 1 v 1 u 4 v
5
5 5
5
uv
OA OD
(18C037)
(a) OP
5i 3j 2k
BD (2 5)i (3 2) j [9 (3)]k
3i 5 j 12k
(b)
(18C035)
(a) AB (0 2)i (3 1) j [4 (3)]k 2i 2 j 7k
2i 2 j 7k
2 i 2 j 7 k
The required unit vector = AB
2
2
2
57
57
57
AB
(2) 2 7
(b) (i)
CD 3 57 AB
AB
- 34 -
3 57 2 i 2 j 7 k
57
57
57
6i 6 j 21k
(by (a))
OD OC CD (3i 2j 6k ) (6i 6 j 21k ) 3i 4j 27k
∴ The coordinates of D are (–3, 4, 27).
(ii)
AB 2i 2 j 7k
AD (3 2)i (4 1) j [27 (3)]k 5i 3j 30k
2 2 7
∵
5 3 30
∴ A, B and D are not collinear.
∴ A, B and D form a triangle.
(18C036)
(a) ∵ AB : BD = 1 : 4
∴
(b) ∵
∴
(c)
OB
4OA OD 4u v 4
1
u v
5
5
5
5
AC : CD = 4 : 1
OC
OA 4OD u 4 v 1
4
u v
5
5
5
5
OA OD u v
(c)
AQ
2OA OB 2a b 2
1
a b
3
3
3
3
3 AO 2 AB 3 AO 2( AO OB ) 2OB 5OA 2
ba
5
5
5
5
3 OP AQ 3 2 a 1 b 2 b a
2
23
3 5
a 1b 2ba
2
5
9b
10
9 10
10
9
(by (a) and (b))
(18C038)
2
r OA OB
r OA OH
r
2
3
a
b
(a) OM
1 r
1 r
1 r
3(1 r )
3
OA sOB
OG sOB 4
3
s
a
b
(b) OM
1 s
1 s
4(1 s )
1 s
(c) From (a) and (b), we have
r
2
3
s
a
b
a
b
1 r
3(1 r )
4(1 s)
1 s
3
r
......(1)
1 r 4(1 s)
2
s
......(2)
3(1 r ) 1 s
(1)
: 3r 3
(2)
2
4s
s 1
2r
By substituting s
(c) From (a) and (b), we have
rs
s
rs
s
OM OA
ON OC
r
r
r
r
s AC (r s) MN
1
into (2), we have
2r
2
1
3(1 r ) 2r 1
r 1
AC
∴
1
1
s
2(1) 2
MN
∴
(b) ∵
∴
OG
(18C041)
( p 1)OA ON ( p 1) 2OM NB 2( p 1)
1
OM NB
p
p
p
p
BG : GM = 1 : (q – 1)
OG
(q 1)OB OM (q 1) 2 NB OM 1
2(q 1)
OM
NB
q
q
q
q
- 35 -
(c) From (a) and (b), we have
2( p 1)
2(q 1)
OM 1 NB 1 OM
NB
p
p
q
q
2( p 1) 1 ......(1)
p
q
1 2(q 1) ......(2)
p
q
(1)
: q p
(2)
By substituting q = p into (1), we have p
∴
pq
3
2
(18C040)
sOA r OB
rs
rs
s
OB
OM OA
r
r
(a) OM
sOC r OB
rs
rs
s
OB
ON OC
r
r
(b) ON
rs
s
AC : MN (r s ) : s
i.e.
(18C039)
(a) ∵ AG : GN = 1 : (p – 1)
(a) OE nOA mOC na mc
mn
mn
(b) OE nOB mOD nb md
mn
mn
(c) From (a) and (b), we have
n a mc n b m d
mn
mn
n a n b md mc
n(a b) m(d c)
m
BA CD
n
BA CD
∴
3
2
(∵ m > n)
ABCD is not a parallelogram.
(18C042)
(a) OS
3OA OB 3a b 3
1
a b
4
4
4
4
3
3p
OA p OB a
b
OA pOT
1
3p
5
5
a
b
(b) OM
1 p
1 p
1 p
1 p
5(1 p)
(c)
OS qOM
3
1
q
3 pq
a b
a
b
4
4
1 p
5(1 p)
q
3
......(1)
4 1 p
1
3 pq
......(2)
4 5(1 p )
(by (a) and (b))
(1)
:
( 2)
5
3p
5
p
9
3
∴
By substituting p
5
into (1), we have
9
AE 6 AC AB
7
6 1 AO (OB OA)
2
7
3OA OB OA
7
4 OA 1 OB
7
7
3
q
4 1 5
9
7
q
6
Section B Questions
(18D001)
(18C043)
(a) OD
7
7
(a) OE OA a
9
9
(b) (i)
(b) BF
3OA OB 3
1
OA OB
4
4
4
∵ BE : EC = r : (1 – r)
1
r
OE r OC (1 r )OB r OA (1 r )OB OA (1 r )OB
2
2
OE sOD
r
1
3
OA (1 r )OB s OA OB
(by (a))
2
4
4
r
3s
s
OA (1 r )OB OA OB
2
4
4
r 3s
......(1)
2 4
s
1 r
......(2)
4
(1)
r
:
3 3r
( 2)
2
6
r
7
1
BE
1 m
1
(OE OB )
1 m
1 7
(by (a))
a b
1 m 9
7
1
a
b
9(1 m)
1 m
∴
- 36 -
By substituting r
6
4
into (1), we have s .
7
7
6 6
: 1
7 7
6 :1
(ii) BE : EC
(from (b)(i))
(c) ∵
∴
△BCF ~ △EDF (AAA)
BF BC
(corr. sides, ~△s)
EF ED
1 9
m 4
4
m
9
(d) From (b) and (c), BF
CF BF BC
9
7
a b a
13
13
6
9
a b
13
13
7
4
91
9
a
1
1
4
9
b
7
9
a b
13
13
(18D002)
1
1
(a) AD OA a
4
4
(b) ∵ △ADM ~ △BCM (AAA)
DM AD
∴
(corr. sides, ~△s)
MC BC
DM 1
MC 4
i.e. DM : MC 1 : 4
(c) From (b), DM : MC = 1 : 4
4 AD AC
∴ AM
5
4
1
AD OB
5
5
4 1 1
a b (by (a))
5 4 5
1
1
a b
5
5
(d) ∵
- 37 -
△DMN ~ △CMA (AAA)
AM MC
∴
(corr. sides, ~△s)
MN DM
AM
4
MN
i.e. AM : MN 4 :1
1
MN AM
4
1 1
1
a b
(by (c))
4 5
5
1
1
a b
20
20
(18D003)
1
1
1
AC OA OB a b
4
4
4
1
1
1
OE OB BE OB BC OB OA a b
2
2
2
(a) OD OA AD OA
(b) (i)
OG
(ii) ∵
OA mOB
1
m
a
b
1 m
1 m
1 m
OE and OG are parallel.
∴
1
2 1
From (a) and (b)(i),
1
m .
1 m 1 m
m2
1
2
(c) From (b), OG a b .
3
3
Let AF : FG = 1 : n.
1
2
a b na
OG nOA 3
1 3n
2
3
OF
a
b
1 n
1 n
3(1 n)
3(1 n)
∵
∴
∴
∵
∴
OD and OF are parallel.
1
1
4
(by (a))
1 3n
2
3(1 n) 3(1 n)
7
n
3
AF : FG = 3 : 7
AG : GB = 2 : 1
2 3 2 7 1
AF : FG : GB : : 3 : 7 : 5
3 10 3 10 3
(18D004)
(a) (i) AB OB OA b a
2
2
2
2
CD OD OC OD OC OB OA b a
5
5
5
5
2
2
(b a) AB
5
5
∴ AB and CD are parallel.
(b) From (a)(i), CD : AB = 2 : 5.
∵ CD : DE = 4 : 1
∴ AB : DE 10 : 1
(ii) From (a)(i), CD
(c) ∵
△ABF ~ △EDF (AAA)
BF
AB
(corr. sides, ~△s)
∴
FD DE
BF
10
(by (b))
FD
i.e. BF : FD = 10 : 1
∵ OD : DB = 2 : 3
2 3 1 3 10
∴ OD : DF : FB : : 22 : 3 : 30
5 5 11 5 11
i.e. OF : FB 25 : 30 5 : 6
6 AO 5 AB
∴ AF
11
6a 5(b a)
(by (a)(i))
11
5
a b
11
3
(d) OG OE
4
2
3 2 5
5 units
4
2
15
5 units
8
(18D005)
(a) OG OE
(OA AE )
1
OA AC
2
1
OA OB
2
5
5a b
2
5
5 a
b
2
∵
△AGF ~ △DGC (AAA)
∴
AF AG
(corr. sides, ~△s)
DC DG
AF 3
(by (c))
2 5 5
3
AF 2 units
OF (5 2) units 3 units
∴
(18D006)
bc 1
1
(a) e
b c
2
2
2
f
- 38 -
(b) OG OA AG
OA AD
OA ( AC CD)
2
OA OB CB
3
2
OA OB OA
3
2
51
a 5 b
3
(c) From (a) and (b), 5 a
(by (c))
5
2
b 51
a 5 b .
2
3
α 1 2 β
3
i.e.
α
β
2
By solving the above system of linear equations, we have
3
3
and .
8
4
(b) g
3a c 3
1
a c
4
4
4
a e
1
1
1
2
2
(by (a))
1
1
1
a
b
c
1
2(1 )
2(1 )
a b c
(c)
b f
1
1
3
b a c
4
4
(by (a))
1
3
1
a
b
c
4(1 ) 1
4(1 )
g
(d) From (b) and (c),
1
a
1
1
3
1
b
c
a
b
c.
2(1 )
2(1 )
4(1 )
1
4(1 )
3
1 4(1 )
1
1
i.e.
2(1 ) 1
1
2(1 ) 4(1 )
By solving the above system of linear equations, we have
3
and 4 .
2
3
1
1
∴ g a b c
5
5
5
(e)
CG OG OC
1
1
3
a b c c
5
5
5
3
1
4
a b c
5
5
5
Let AD : DB = 1 : r.
- 39 -
(18D007)
OA 2OB 1
2
a b
3
3
3
∵
∴
(ii) ∵
∴
∵
(by (d))
r CA CB r (OA OC ) (OB OC ) r (a c) (b c)
r
1
a
b c
1 r
1 r
1 r
1 r
1 r
∵ CG and CD are parallel.
1
4
5 5
∴
1
1
1 r
r 3
∴ AD : DB 1 : 3
(b) (i)
(c) ∵
∴
CD
(a) OE
∴
BP : PC r : (1 r )
1
r
OP r OC (1 r )OB r OA (1 r )OB a (1 r )b
3
3
OP and OE are parallel.
1
2
From (a) and (b)(i), 3 3
r 1 r
3
3
r
5
BQ : QD (1 s) : s
2
2(1 s )
OQ (1 s )OD sOB (1 s ) OA sOB
a sb
3
3
OQ and OE are parallel.
1
2
3
3
From (a),
2(1 s ) s
3
4
s
7
1
2
3
(d) From (a) and (b), OP a b OE .
5
5
5
2
4
6
From (a) and (c), OQ a b OE .
7
7
7
3 6
∴ OP : OQ : 7 : 10
5 7
i.e. OP : PQ 7 : 3
∵ OQ : OE = 6 : 7
i.e. OQ : QE = 6 : 1
6 7 6 3 1
: : 21 : 9 : 5
7 10 7 10 7
(e) From (b), BP : PC = 3 : 2
∴
OP : PQ : QE
From (d), OP : PQ = 7 : 3
∠OPC = ∠BPQ (vert. opp. ∠s)
1
OP PC sin OPC
Area of △OCP 2
Area of △ BPQ 1
PQ BP sin BPQ
2
OP PC
PQ BP
7 2
3 3
14
9
∴ Area of △OCP : area of △BPQ 14 : 9
(18D008)
(a) (i) BA OA OB (5i 2 j) (5i j) 10i j
(18D009)
(a) (i) AB (6 2)i (3 1) j 4i 2 j
BE OE OB i (5i j) 4i j
(ii) ∵
AC (3 2)i (6 1) j 5i 5 j
BC (3 6)i (6 3) j 9i 3j
AE : EC = 2 : 1
BE
∴
BA 2 BC
3
(ii) ∵
(10i j) 2 BC
4i j
3
∴
2 BC 2i 4 j
BC i 2 j
OC OB BC (5i j) (i 2 j) 4i j
∴ The coordinates of C are (4, 1).
(b) (i)
4m 5 2m 5
(2i j)
i
j
m 1
m 1
4m 5 2m 5
2
i 1
j
m 1
m 1
6m 3 3m 6
i
j
m 1
m 1
CD OD OC (7i hj) (4i j) 11i (h 1) j
∵ DE : EB = 2 : k
k CD 2CB
k2
k[11i (h 1) j] 2(i 2 j)
k2
2 11k
k (h 1) 4
i
j
k2
k2
CE
- 40 -
(ii) CE OE OC i (4i j) 3i j
From the result of (b)(i), we have
2 11k
k 2 3
k (h 1) 4 1
k 2
By solving the above system of linear equations, we have
h 2 and k 1 .
(c)
AD OD OA
(7i 2 j) (5i 2 j)
2i 4j
2 BC
∵
AD // BC but AD BC
∴
ABCD is a trapezium.
m AB AC
m 1
m(4i 2 j) (5i 5j)
m 1
4m 5
2m 5
i
j
m 1
m 1
AP
(iii) OP OA AP
CB i 2 j
∴
BP : PC = 1 : m
∴
(b) (i)
∵
∴
6m 3 3m 6
,
The coordinates of P are
.
m 1 m 1
AP BC
mAP mBC 1
3m 6
1
63
m 1
1
6m 3
2 36
m 1
2m 5
m 1 3 1
4m 5 9
m 1
2m 5
3
4m 5
2m 5 12m 15
m2
(ii) By substituting m = 2 into the result of (a)(ii), we have AP i 3j .
∴
5.
AP 12 32 10 units and
BC (9)2 32 90 units
∴
The area of parallelogram ABCD 10 90 sq. units 30 sq. units
(b) OP 3OA 2OB
5
3(3i 4 j 2k ) 2(2i 3j k )
5
18
8
i j k
5
5
Quiz
1.
(a)
RQ PS v
(b) RS PQ u
(c)
PR PQ PS u v
(d) QS QR RS v u
6.
2.
∵
∴
- 41 -
AB and CD are parallel.
4
(2m 1)
3m 2
3
2
6m 7 m 10 0
(m 2)(6m 5) 0
m 2 or
3.
(a)
4.
(c) From (a) and (b),
5
6
AB (4 2)i [2 (3)] j 2i 5 j
CB AB BC AB
(6i 2 j) (2i 5 j)
4i 7 j
(by (a) and (b))
(a) 2a – b = 2(2i + 3j – 2k) – (–3i + j – k) = 7i + 5j – 3k
2a b 7 2 52 (3) 2 83
2a b
7
5
3
i
j
k
The required unit vector =
2a b
83
83
83
(b) 3b – 2c = 3(–3i + j – k) – 2(–4i + 3k) = –i + 3j – 9k
3b 2c (1) 3 (9) 91
3b 2c
1
3
9
i
j
k
The required unit vector =
3b 2c
91
91
91
2
2
mOA OD
(a) OE
m 1
2
mOA OB
m
2
3
a
b
m 1
m 1
3(m 1)
3
OA nOB
OC nOB 5
3
n
a
b
(b) OE
n 1
n 1
5(n 1)
n 1
(b) BC (2 4)i (4 2) j 6i 2 j
(c)
(a) OP OA OB
2
(3i 4 j 2k ) (2i 3j k )
2
1 i 7 j 3 k
2
2
2
2
m
2
3
n
a
b
a
b
m 1
3(m 1)
5(n 1)
n 1
3
m
......(1)
m 1 5(n 1)
i.e.
2
n
......(2)
3(m 1) n 1
(1)
m 1
:
(2)
2 5n
2
n
5m
2
By substituting n
into (2), we have
5m
1
1
3(m 1) 2 5m
1
m
2
n
2
4
1 5
5
2
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