Vectors

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Higher Mathematics
Vectors
Paper 1 Section B
1. The point Q divides the line joining P(−1, −1, 0) to R(5, 2, −3) in the ratio 2 : 1.
[SQA]
3
Find the coordinates of Q.
Part
Marks
3
Level
C
Calc.
NC
Content
G25
•1 pd: find vector components
•2 ss: use parallel vectors
•3 pd: process vectors
Answer
(3, 1, −2)
U3 OC1
2002 P1 Q2

6
−
→
•1 PR =  3 
−3
−→ 2 −
→
2
• PQ = 3 PR
•3 Q = (3, 1, −2)

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2. VABCD is a pyramid with a rectangular base ABCD.
x
y
Relative to some appropriate axes,
[SQA]
V
−→
VA represents −7i − 13 j − 11k
−→
AB represents 6i + 6 j − 6k
D
−→
AD represents 8i − 4 j + 4k .
A
K divides BC in the ratio 1 : 3.
−→
Find VK in component form.
Part
Marks
Level
Calc.
3
C
CN
3
C
1 K
B
3
Content
G25, G21, G20
Answer


1
 −8 
−16
U3 OC1
2000 P1 Q7
replacements
−→
−→
−→
−→
•1 VK
=
VA + AB + BK or
−→ −→ −→
VK = VB + BK
 
2
−
→
−
→
−
→
•2 BK = 14 BC or 14 AD or −1 or
1


−1
 −7 
−17


1
−
→
•3 VK =  −8 
−16
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•1 ss: recognise crucial aspect
•2 ic: interpret ratio
•3 pd: process components
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Page 1
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Higher Mathematics
3. (a) Roadmakers look along the tops of a set
of T-rods to ensure that straight sections
of road are being created. Relative to
suitable axes the top left corners of the
T-rods are the points A(−8,
−2),
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B(−2, −1, 1) and C(6, 11, 5).
OA
Determine whether or not the section of
x
road ABC has been built in a straight line. y
[SQA]
C
B
3
C
(b) A further T-rod is placed such that D has
coordinates (1, −4, 4).
B
Show that DB is perpendicular
AB.
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3
O
xA
y
Part
(a)
(b)
Marks
3
3
•1 ic:
•2 ic:
•3 ic:
Level
C
C
Calc.
CN
CN
Content
G23
G27, G17
Answer
the road ABC is straight
proof
−→
interpret vector (e.g. AB)
interpret multiple of vector
complete proof
 
6
−→
•1 e.g. AB = 9
3
−→
•4 ic: interpret vector (i.e. BD)
•5 ss: state requirement for perpend.
•6 ic: complete proof
•2 e.g.
•3
•4
•5
•6
−→
BC
=

8
12
4

U3 OC1
2001 P1 Q3
=
→
4−
3 AB
or
 
 
2
2
−
→
−→
AB = 3 3 and BC = 4 3
1
1
a common direction exists and a
common point exists, so A, B, C
collinear
 
3
−→  
BD = −3
3
−→ −→
AB.BD = 0
−→ −→
AB.BD = 18 − 27 + 9 = 0
or
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−→ −→
•5 AB.BD = 18 − 27 + 9
−→ −→
c SQA
marked
‘[SQA]’
•6 AB.BDQuestions
= 0 so AB
is at right
angles
to
c
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BD
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4.
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Paper 2
1.
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13. The vectors p , q and r are defined as follows:
[SQA]
p = 3i − 3 j + 2k , q = 4i − j + k , r = 4i − 2 j + 3k .
(a) Find 2 p − q + r in terms of i , j and k .
1
(b) Find the value of |2 p − q + r |.
2
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18.
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20.
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[SQA]
22. A cuboid measuring 11 cm by 5 cm by 7 cm is placed centrally on top of another
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cuboid measuring 17 cm by 9 cm by 8 cm.
Coordinates axes are taken as shown.
z
7
5
11
C
A
x
17
8
9
B
y
O
(a) The point A has coordinates (0, 9, 8) and C has coordinates (17, 0, 8).
1
Write down the coordinates of B.
6
(b) Calculate the size of angle ABC.
Part
(a)
(b)
Marks
1
6
•1 ic:
•2
•3
•4
•5
•6
•7
ss:
pd:
pd:
pd:
pd:
pd:
Level
C
C
Calc.
CN
CR
Content
G22
G28
interpret 3-d representation
know to use scalar product
process vectors
process vectors
process lengths
process scalar product
evaluate scalar product
Answer
B(3, 2, 15)
92·5◦
•1
•2
•3
•4
•5
•6
•7
U3 OC1
2000 P2 Q9


3
B= (3, 2, 15) treat  2  as bad form
15
−→ −→
BA
.
BC
b = −→ −→
cos ABC
|BA||BC|
 
−3
−→
BA =  7 
−7
 
14
−→  
BC = −2
−7
√
√
−→
−→
|BA| = 107, |BC| = 249
−→ −→
BA.BC = −7
b = 92·5◦
ABC
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25.
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Higher Mathematics
28. ABCD is a quadrilateral with vertices A(4, −1, 3), B(8, 3, −1), C(0, 4, 4) and
D(−4, 0, 8).
[SQA]
(a) Find the coordinates of M, the midpoint of AB.
1
(b) Find the coordinates of the point T, which divides CM in the ratio 2 : 1.
3
(c) Show that B, T and D are collinear and find the ratio in which T divides BD.
4
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31.
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32.
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33.
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34.
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Higher Mathematics
35. A box in the shape of a cuboid
is designed with circles of different
sizes on each face.
[SQA]
The diagram shows three of the
circles, where the origin represents
one of the corners of the cuboid. The
centres of the circles are A(6, 0, 7),
B(0, 5, 6) and C(4, 5, 0).
z
B
Find the size of angle ABC.
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7
A
O
y
C
x
Part
Marks
5
2
•1 ss: use
•2
•3
•4
•5
•6
•7
ic:
ic:
pd:
pd:
pd:
pd:
Level
C
A/B
Calc.
CR
CR
Content
G17, G16, G22
G26, G28
−→ −→
BA.BC
−→ −→
|BA||BC|
Answer
2001 P2 Q4
71·5◦
•1 use
−→
state vector e.g. BA
−→
state a consistent vector e.g. BC
−→
process |BA|
−→
process |BC|
process scalar product
find angle
U3 OC1
•2
•3
•4
•5
•6
•7
−→ −→
BA.BC
−→ −→
|BA||BC|
stated or implied by •7


6
−→  
BA = −5
1
 
4
−→
BC =  0 
−6
√
−→
|BA| = 62
√
−→
|BC| = 52
−→ −→
BA.BC = 18
b = 71·5◦
ABC


 
t
2



36. For what value of t are the vectors u = −2 and v = 10 perpendicular?
3
t
[SQA]
Part
replacements
O
x
y
Marks
2
Level
C
Calc.
CN
Content
G27
•1 ss: know to use scalar product
•2 ic: interpret scalar product
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Answer
t=4
U3 OC1
2000 P2 Q7
•1 u.v = 2t − 20 + 3t
•2 u.v = 0 ⇒ t = 4
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Higher Mathematics
37. A(4, 4, 10), B(−2, −4, 12) and C(−8, 0, 10) are the vertices of a right-angled
triangle.
[SQA]
Determine which angle of the triangle is the right angle.
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39. The diagram shows a square-based
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pyramid of height 8 units.
[SQA]
D(3, 3, 8)
Square OABC has a side length of 6 units.
The coordinates of A and D are (6, 0, 0)
and (3, 3, 8).
y
C
C lies on the y-axis.
(a) Write down the coordinates of B.
−→
(b) Determine the components of DA
−→
and DB.
B
O
A(6, 0, 0)
x
2
4
(c) Calculate the size of angle ADB.
Part
(a)
Marks
1
(b)
2
Level
C
C
Calc.
CN
CN
Content
G22
Answer
(6, 6, 0)
G17
−→
DA
U3 OC1


3
−3,
−8
=

(c)
4
C
CR

3
−→
DB =  3 
−8
38·7◦
G28
•1 ic:
interpret diagram
•4
•5
•6
•7
use e.g. scalar product formula
process lengths
process scalar product
process angle
•2 ic: write down components of a
vector
3
• ic: write down components of a
vector
ss:
pd:
pd:
pd:
2002 P2 Q2
•1 B = (6, 6, 0)
 
3
−
→
•2 DA = −3
−8
 
3
−→  
3
• DB =
3
−8
.DB
b = −DA
•4 cos ADB
→ −→
|DA||DB|
√
√
−→
−→
•5 |DA| = 82, |DB| = 82
−→ −→
•6 DA.DB = 64
b = 38·7◦
•7 ADB
−→ −→
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40.
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41.
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42.
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44.
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46.
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47. PQRS is a parallelogram with vertices P(1, 3, 3), Q(4, −2, −2) and R(3, 1, 1).
[SQA]
3
Find the coordinates of S.
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[END OF PAPER 2]
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