1. ABCDEF is a regular hexagon with centre O.
= a and = b
(a) Find expressions, in terms of a and b, for
(i)
Answer ......
(1)
(ii)
Answer ......
(1)
(iii)
Answer ......
(1)
(b) The positions of points P and Q are given by the vectors
= a – b = a + 2b
(i) Draw and label the positions of points P and Q on the diagram.
(2)
(ii) Hence, or otherwise, deduce an expression for .
...... ………......
...... ………......
Answer ......
(1)
(Total 6 marks)
2. PQRSTU is a regular hexagon and O is the centre of the hexagon.
= p and = q
Express each of the following vectors in terms of p and q
(a)
......
(1)
(b)
......
(1)
(c)
......
(2)
(Total 4 marks)
3. The diagram shows two sets of parallel lines.
Vector = a and vector = b
= and =
(a) Write the vector in terms of a and b
......
Answer ......
(1)
(b) Write the vector in terms of a and b
......
Answer ......
(1)
(c) Find two vectors that can be written as 3a – b
......
Answer ...... and ......
(2)
(Total 4 marks)
4. In the diagram OACD, OADB and ODEB are parallelograms.
(a) Express, in terms of a and b, the following vectors.
Give your answers in their simplest form.
(i)
......
(1)
(ii)
......
(1)
(iii)
......
Answer......
(1)
(b) The point F is such that OCFE is a parallelogram.
Write the vector in terms of a and b.
......
......
Answer......
(2)
(c) What geometrical relationship is there between the points O, D and F? Justify your answer.
......
......
......
......
(2)
(Total 7 marks)
5. OACB is a parallelogram and M is the mid-point of BC.
= a and = b
(a) Express the following vectors in terms of a and b
(i) Answer ......
(1)
(ii) Answer ......
(1)
(b) AM is extended to N, where .
Show that = b
......
......
......
(2)
(c) What does this tell you about the position of N?
......
......
......
(1)
(Total 5 marks)
6. = –4a + 5b and = 5a – b.
R is a point on such that PR : RQ = 1 : 2.
(a) Express in terms of a and b.
......
......
......
Answer ......
(3)
(b) = a + 4b
Express in terms of a and b.
......
......
Answer ......
(2)
(c) What two facts do and indicate about the points O, R and S?
Give a reason for each of your answers.
......
......
......
(2)
(Total 7 marks)
7. In the diagram = 4a, = a, = 5b, = 3b and =
Not drawn accurately
(a) Find, in terms of a and b, simplifying your answers,
(i)
......
......
Answer ……………………………………………
(1)
(ii)
......
......
......
Answer ……………………………………………
(2)
(b) Show clearly that points P, Q and R lie on a straight line.
......
......
......
......
......
(3)
(Total 6 marks)
8.
OAB is a triangle where M is the mid-point of OB.
P and Q are points on AB such that AP = PQ = QB.
= a and = 2b
(a) Find, in terms of a and b, expressions for
(i)
......
......
Answer ......
(1)
(ii)
......
......
......
Answer ......
(2)
(iii)
......
......
......
Answer ......
(2)
(b) What can you deduce about quadrilateral OMQP?
Give a reason for your answer.
......
......
......
(2)
(Total 7 marks)
9. OPQR is a parallelogram.
M is the midpoint of the diagonal OQ.
= 2p and = 2r
(a) Express in terms of p and r.
......
......
Answer ......
(1)
(b) Use vectors to show that M is the midpoint of PR.
......
......
......
......
......
......
(3)
(Total 4 marks)
1. (a) (i) a + b B1
(ii) b – a B1
(iii) a + 2b B1
or equivalent eg. b + a + b
(b) (i) P in correct position B1
Q in correct position B1
‘by eye’ judgement ... tolerance of ±2mm in any direction (if no labels then B1, B0)
(ii) PQ = 3b B1
Accept PQ = PO + OQ = b – a + a + 2b = 3b
[6]
2. (a) – p + q oe B1
(b) 2p B1
(c) SQ = SO+ OQ M1
or SR + RQ or SP + PQ
p + q A1
[4]
3. (a) a + 2b B1
oe
(b) 2b –3a B1
oe
Note: and correct scores SC1
(c) SR B1
UT B1
[4]
4. (a) (i) a + b b + a B1
(ii) 2a + b 2a + b B1
(iii) b – a – a + b B1
(b) CF = OE M1
a + 2b a + b + b oe A1
(c) Straight line because
OD = a + b B1
3 times bigger because
OF = 3a + 3b B1
[7]
5. (a) (i) – b + a or a – b B1
(ii) b – a B1
oe
(b) M1
oe
= a + b – a A1
(c) = 2 B1
or OBN a straight line
or BN = OB
or B is midpoint of ON
[5]
6. (a) ((4a – 5b) + (5a – b)) M1
(– 4a + 5b) + their (3a – 2b) M1 dep
– a + 3b A1
or 3a – 2b
(b) (– 4a + 5b) + (a + 4b) M1
– 3a + 9b A1
(c) ORS is a straight line B1
oe
OS is 3 times the length of OR B1
oe
[7]
7. (a) (i) –5a + 5b B1
(ii) a + their(–5a + 5b) M1
condone one error (eg, missing brackets)
2b – a A1
(b) = –4a + 5b + 3b M1
= – 4a + 8b or 4(2b – a) A1
alternatively: = (5a + 5b) +3b
= 6b – 3a = 3(2b – a)
is a multiple of A1
hence points are co-linear
is a multiple of
hence points are co-linear oe
[6]
8. (a) (i) BA = a – 2b B1
or equivalent
(ii) MQ = MB + BA = b + (a – 2b) M1
Attempt to set up a route, must include substitution of a and b, condone one error
MQ = a + b or (a + b) A1
Need not be simplified
(iii) OP = OA + AB = a + (2b – a) M1
or OP = OB + BA = 2b + (a – 2b)
OP = a + b A1
M1 as above, need not be simplified for A1
(b) OP = 2 × MQ B1
Trapezium B1
only if accompanied by a sound reason
[7]
9. (a) p + r B1
(b) PM = – 2p + p + r M1
or MR = – (p + r) 2r or PR = – 2p + 2r
PM = – p + r A1
or MR = – p + r
PR = 2PM so M is mid-point A1 [4]
The Robert Smyth School 1