P(−3, −4)
DalkeithHighSchool–National5ExamStyleQuestions–Vectors
Practice Paper B: Higher Mathematics
1.
3.
12. If f(x) =
A
B
C
This is
setf ′(x)?
of drawers is being
x2 + 1 what
‘modelled’ on a computer
1
software design package as a
cuboid as shown. The edges
x
of the cuboid are parallel to
2
the x, y and z-axes. Three of
x +1
the vertices are P(−1,−1,−1),
x
3
S(−1,4,−1) and V(3,4,5)
x2 + 1
(
)
W
T
V(3, 4, 5)
U
S(−1, 4, −1)
P(−1, −1, −1)
R
Q
(a)
D Write
2x x2down
1 the lengths of PQ, QR and RV.
→
→
(b) Write down the components of VS and VP and hence calculate the size of
2.13. In the diagram ABCD represents
A
angle PVS.
a tetrahedron.
→
→
BC represents p, CD represents q,
→
→
B
DB represents r, BA represents s,
Page 21
→
→
CA represents t and DA represents u
LABK009_A.indd 21
p=−q+s−u
B
q=−p+s+u
C
r=−p−t+u
D
s=p+q+u
t
r
q
p
C
7/21/09 9:55:48 PM
3.14. P divides AB in the ratio 3:2 where A is the point (−3, 2, 6) and B is the point
y
4. OABC, DEFG is a cuboid.
z
(7, −3, 1). What is the y-coordinate of P?
G
F(5, 6, 2)
The vertex F is the point (5, 6, 2).
M
A −1
C
B
M is the midpoint of DG.
E
D
B
0
N divides AB in the ratio 1:2.
N
C
1
(a) Find the coordinates of M and N.
4.
6.
5.
0
A
x
D
3
→
→
(b) Write down the components of MB and MN .
2
−4
Two vectors are defined as = (
) and = (
).
3
(c) Find the size of angle BMN. −
(a)
Find the resultant vector + 3 .
The diagram below shows the graphs y = f(x) and y = g(x) where f(x) = m sin x
and g(x) = n cos x
(b)
H.MacLeod
Find | + 3 |.
1
Marks
2
2
5
1
2
y
√3
7
D
Practice Paper B: Higher Mathematics
u
s
One of these statements is false, which one?
A
1
DalkeithHighSchool
Page 31
DalkeithHighSchool–National5ExamStyleQuestions–Vectors
5.
3
2
5.
Vector 𝒖 has components (−2 ) and vector 𝒗 has components (−4 ).
−1
1
Calculate |4𝒖 − 2𝒗|.
2
6.
11.
Look at the cuboid shown on the coordinate diagram.
𝑧
The coordinates of point 𝐸 are (5,3,1)
6.
(a)
Factorise
2
𝑝 − 4𝑞 .
𝐴
𝐸
(a)
(b)
State the coordinates of 𝐺
(c)
What is the shortest distance between points 𝐷 and 𝐶?
1
𝐵
𝐹
𝐺
𝑝2 − 4𝑞 2
Hence simplify
.
3𝑝 + 6𝑞
State the coordinates of 𝐹
(b)
𝑦
𝐷
2
𝑂
𝑥
𝐶
2
4
National 5 Practice Paper B
7.
4.
Relative to coordinate axes, the point A
has coordinates (2, 4, 6).
12.
Last updated 27/01/15
At the carnival, the height, 𝐻 metres,
of a carriage
the big wheel
(a)
Find theoncoordinates
of Cabove
and D.
the ground is given by the formula
2
𝐻 = 10 + 5 sin 𝑡°,
𝑡 seconds
after
starting
to turn. of B.
(b)
Write
down
the coordinates
1
(a)
Find the height of the carriage above the ground after 10 seconds.
2
Find the two times during the first turn of the wheel when the
carriage is 12.5 metres above the ground.
4
(b)
5.
Shampoo is available in travel size and salon size bottles.
The bottles are mathematically similar.
H.MacLeod
DalkeithHighSchool
DalkeithHighSchool–National5ExamStyleQuestions–Vectors
8.
6.
The diagram shows a square based pyramid
.
A line of best fit is drawn as shown above.
(a)
Find the equation of this straight line.
3
(b)
Use this equation to estimate the sport score for a team with
a film score of 20.
1
Express ⃗⃗⃗⃗⃗ in terms of
and
.
3
9.
5.
Given that ⃗⃗⃗⃗⃗
(
) calculate |⃗⃗⃗⃗⃗ |.
Give
yourWatson
answer Fast
as a Foods
surd inuse
its asimplest
form.
William
logo based
on parts of three identical parabolas.
7.
3
The logo is represented on the diagram below.
National 5 Practice Paper F
Last updated 24/04/14
The first parabola has turning point P and equation
=
+2
2
−1
(a)
State the coordinates of P.
2
(b)
If R is the point (2,0), find the coordinates of Q, the minimum turning
point of the second parabola.
1
(c)
H.MacLeod
Find the equation of the parabola with turning point S.
DalkeithHighSchool
National 5 Practice Paper D
2
Last updated 27/01/15
0
−1
180
360 x
S
A
x
T C
So x = 180
DalkeithHighSchool–National5ExamStyleQuestions–Vectors
for cos x° negative.
Solutions are:
Solutions:
Interpretation
143·1, 180, 216·9
✓
• The diffi
culty p37
is that the axes are
3. (a)
HMRN:
1. (to 1PQ
not
shown.
Look
for a single change
decimal
place).
= 4 units
in coordinates: P(−1,−1,−1) to S
QR = 5 units
5 marks
(−1,4,−1): This is 3 units parallel to the
RV = 6 units ✓
y-axis since only the y-coordinate has
changed.
Point M
1 mark
• M is the midpoint. Half-way along DG
2. B
HMRN: p 42
which is 6 units long (y-coordinate) is
4. (a)
3 units
M(0,3,2)
✓
3.
Point N
N(5,2,0) ✓
1
1
• N is 3 of the way along AB so 3 of
2 marks
6 = 2 units is the y-coordinate.
HMRN: p 42
4. (b)
M(0,3,2) and B(5,6,0)
→
MB = b − m
Page 73
Components
• The basic result used is:
⎛ 5⎞ ⎛ 0⎞ ⎛ 5 ⎞
= ⎜ 6⎟ − ⎜ 3⎟ = ⎜ 3 ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜⎝ 0⎟⎠ ⎜⎝ 2⎟⎠ ⎜⎝ −2⎟⎠
73
✓
⎛
⎞
⎛
⎞
⎛
⎞
x
x
x −x
→
⎜ 2 ⎟ ⎜ 1⎟
⎜ 2 1⎟
⎜
⎟
⎜
⎟
⎜
so PQ = q − p = ⎜ y2 ⎟ − ⎜ y1⎟ = ⎜ y2 − y1⎟⎟
also M (0,3,2) and N (5,2,0)
→
MN = n − m
⎛ 5⎞ ⎛ 0⎞ ⎛ 5 ⎞
= ⎜ 2⎟ − ⎜ 3⎟ = ⎜ −1⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜⎝ 0⎟⎠ ⎜⎝ 2⎟⎠ ⎜⎝ −2⎟⎠
7/21/09 9:59:47 PM
P(x1,y1,z1) and Q (x2,y2,z2)
⎜ ⎟
⎜⎝ z ⎟⎠
2
⎜ ⎟
⎜⎝ z ⎟⎠
1
⎜
⎟
⎜⎝ z − z ⎟⎠
2 1
HMRN: p 43–44
✓
2 marks
Page 95
swer.indd 95
H.MacLeod
7/21/09 10:00:48 PM
DalkeithHighSchool