Magnitude of a vector: IGCSE Mathematics questions with answers
Free Magnitude of a vector questions for IGCSE Mathematics (0580), Extended tier (Supplement content), part of Transformations & vectors. Each question has the answer and the marking points an examiner would look for.
O is the origin. The position vector of A is a and the position vector of B is b. P is the point on AB such that AP : PB = 4 : 1. M is the midpoint of OA. R is the point with position vector 43b.
Find, in terms of a and/or b, in its simplest form, the vector
(a)(i)
from A to B,
[1]
(ii)
from O to P,
[2]
(iii)
from M to P.
[2]
(b)
Show that M, P and R lie on a straight line, and find the ratio MP : PR.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)(i)
b − a
b − a B1
1
(ii)
15a + 45b
a + 45(b − a) M1 15a + 45b A1
2
(iii)
−310a + 45b
(vector OP) − ½a M1 −310a + 45b A1
2
(b)
vector MR = −½a + 43b = 53 × (vector MP); parallel with common point M, so collinear. MP : PR = 3 : 2
vector MR = −½a + 43b B1 vector MR = 53 × vector MP M1 parallel and share point M, so collinear A1 MP : PR = 3 : 2 B1
4
Question 2Structured · 6 marksA* challenge
OABC is a parallelogram. O is the origin, OA = a and OC = c. M is the midpoint of AB.
(a)
Find AB in terms of c.
[1]
(b)
Find OM in terms of a and c.
[2]
(c)
N is the point on OB such that ON : NB = 2 : 1. Find MN in terms of a and c.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
c
c B1
1
(b)
a + ½c
OA + AM M1 a + ½c A1
2
(c)
MN = −⅓a + ⅙c
ON = ⅔(a + c) M1 MN = ON − OM M1 −⅓a + ⅙c A1
3
Question 3Structured · 4 marks
a = −34 b = 1−1
(a)
Find 2a − b.
[2]
(b)
Find |a + b|.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
−79
One component correct B1 −79B1
2
(b)
3.61
√((−2)2 + 32) M1 3.61 oe A1
2
Question 4Structured · 3 marks
p = 2−2
(a)
Find |p|.
[2]
(b)
Find 3p.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
2.83
√(22 + (−2)2) M1 2.83 A1
2
(b)
6−6
6−6B1
1
Question 5Structured · 8 marksA* challenge
OAB is a triangle with OA = a and OB = b. P is the point on AB such that AP : PB = 1 : 2. Q is the point on OB such that OQ = 12b. The lines OP and AQ intersect at X.
(a)
Find OP in terms of a and b.
[2]
(b)
By writing OX = λOP and AX = μAQ, find the values of λ and μ.
[5]
(c)
Find the ratio OX : XP in its simplest form.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
23a + 13b
a + 13(b − a) M1 23a + 13b A1
2
(b)
λ = 34, μ = 12
OX = a + μ(−a + 12b) M1 equate coefficients of a and of b M1 two correct equations A1 λ = 34A1 μ = 12A1
5
(c)
3 : 1
3 : 1 B1
1
Question 6Structured · 12 marksA* challenge
Use vectors in this question.
(a)
In triangle OAB, M is the midpoint of OA and N is the midpoint of OB. Prove that MN is parallel to AB and half its length.
[4]
(b)
PQRS is any quadrilateral. Prove that the midpoints of its four sides are the vertices of a parallelogram.
[5]
(c)
Prove that this parallelogram is a rhombus if and only if the diagonals PR and QS of the original quadrilateral are equal in length.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
MN = ½b − ½a = ½(b − a) = ½AB
OA = a, OB = b, AB = b − a B1 MN = ½b − ½a M1 = ½AB A1 so parallel and half the length A1
4
(b)
each pair of opposite sides of the midpoint quadrilateral equals ½PR or ½QS
position vectors p, q, r, s and midpoints ½(p + q) etc. M1 side joining midpoints of PQ and QR = ½(r − p) A1 side joining midpoints of RS and SP = ½(p − r) A1 opposite sides equal and parallel (same for the other pair) M1 so a parallelogram A1
5
(c)
adjacent sides are ½PR and ½QS
adjacent sides of the parallelogram are ½(r − p) and ½(s − q) M1 rhombus ⇔ these lengths are equal ⇔ PR = QS A1 both directions argued A1
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