DalkeithHighSchool–National5ExamStyleQuestions–Vectors

H.MacLeod DalkeithHighSchool

1.

2.

3.

4.

Page 21

Practice Paper A: Higher Mathematics

1. The diagram shows a cubic curve with equation

y = x2 − 13

3x .

A tangent PQ to the curve has point of contactM(3, 0).

(a) Find the equation of PQ

A circle has equation x2 + y2 + 4x − 26y + 163 = 0

(b) Show that PQ is also a tangent to this circleand fi nd the coordinates of the point ofcontact N

(c) Find the ratio in which the y-axis cuts the line MN

2. Triangle PQR has coordinates P(−3, −4), Q(−3, 4)and R(5, 12)

(a) Find the equation of the median MR

(b) Find the equation of the attitude NQ

(c) Median MR and altitude NQ intersect atpoint S. Find the coordinates of S.

(d) The point T(2, 9) lies on QR. Show that STis parallel to PR

3. This set of drawers is being‘modelled’ on a computersoftware design package as acuboid as shown. The edgesof the cuboid are parallel tothe x, y and z-axes. Three ofthe vertices are P(−1,−1,−1),S(−1,4,−1) and V(3,4,5)

(a) Write down the lengths of PQ, QR and RV.

(b) Write down the components of→VS and

→VP and hence calculate the size of

angle PVS.

4

6

3

3

3

3

2

1

7

Marksy

x0M(3, 0)

P

Q

Q(−3, 4)

R(5, 12)

P(−3, −4)

M

N

S(−1, 4, −1)

P(−1, −1, −1)

V(3, 4, 5)

WT

U

QR

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Page 31

Practice Paper B: Higher Mathematics

12. If f(x) = x2 1+ what is f ′(x)?

A 1

B x

x2 1+

C

x

x23

1+( ) D 2 1x x2

13. In the diagram ABCD represents a tetrahedron.

→BC represents p,

→CD represents q,

→

DB represents r, →BA represents s,

→

CA represents t and →

DA represents u

One of these statements is false, which one?

A p = − q + s − u

B q = − p + s + u

C r = − p − t + u

D s = p + q + u

14. P divides AB in the ratio 3:2 where A is the point (−3, 2, 6) and B is the point (7, −3, 1). What is the y-coordinate of P?

A −1

B 0

C 1

D 3

A

D

C

B

su

q

p

t

r

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Page 38

Practice Paper B: Higher Mathematics

4. OABC, DEFG is a cuboid.

The vertex F is the point (5, 6, 2).

M is the midpoint of DG.

N divides AB in the ratio 1:2.

(a) Find the coordinates of M and N.

(b) Write down the components of →

MB and →

MN .

(c) Find the size of angle BMN.

5. 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

1

y

0ππ–2

3π—22π

√3

−1

−√3

x

(a) Write down the values of m and n

(b) Write f(x) − g(x) in the form k sin (x − a) where k > 0 and 0 < a <π2

(c) Hence fi nd, in the interval 0 ≤ x ≤ π the x-coordinate of the point on the curve y = f(x) − g(x) where the gradient is 2.

2

5

4

2

2

2

Marks

x

z y

0

G

M

DN

B

A

E

F(5, 6, 2)

C

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National 5 Practice Paper A Last updated 18/03/14

6. Two vectors are defined as = ( 2− ) and = (−4 3 ).

(a) Find the resultant vector + 3 . 1

(b) Find | + 3 |. 2

7.

Part of the graph of = cos is shown in the diagram.

State the value of . 1

8. Find the point of intersection of the straight lines with equations

2 + = and − 3 = . 4

9. A parabola has equation = 2 − 3 + .

Using the discriminant, determine the nature of its roots. 3

DalkeithHighSchool–National5ExamStyleQuestions–Vectors

H.MacLeod DalkeithHighSchool

5.

6.

7.

National 5 Practice Paper B Last updated 27/01/15

1. Evaluate

7.18 − 2.1 × 3. 2

2. Evaluate

2

3. Solve the inequality 5 − 𝑥 > 2(𝑥 + 1) 3

4. Given that 𝑓(𝑥) = 𝑥2 + 5𝑥 , evaluate 𝑓(−3). 2

5. Vector 𝒖 has components ( 3−2−1

) and vector 𝒗 has components ( 2−4 1

).

Calculate |4𝒖 − 2𝒗|. 2

6. (a) Factorise 𝑝2 − 4𝑞2. 1

(b) Hence simplify .

2

118

÷34

𝑝2 − 4𝑞2

3𝑝 + 6𝑞

National 5 Practice Paper B Last updated 27/01/15

11. Look at the cuboid shown on the coordinate diagram.

The coordinates of point 𝐸 are (5,3,1)

(a) State the coordinates of 𝐹

(b) State the coordinates of 𝐺

(c) What is the shortest distance between points 𝐷 and 𝐶? 4

12. At the carnival, the height, 𝐻 metres, of a carriage on the big wheel above the ground is given by the formula

𝐻 = 10 + 5 sin 𝑡°,

𝑡 seconds after starting to turn.

(a) Find the height of the carriage above the ground after 10 seconds. 2

(b) Find the two times during the first turn of the wheel when the carriage is 12.5 metres above the ground. 4

[End of question paper]

𝑧

𝑥

𝑦

𝐴 𝐵

𝐶

𝐷 𝐸

𝐹

𝑂 𝐺

National 5 Practice Paper E Last updated 08/04/14

4. Relative to coordinate axes, the point A has coordinates (2, 4, 6).

(a) Find the coordinates of C and D. 2

(b) Write down the coordinates of B. 1

5. Shampoo is available in travel size and salon size bottles.

The bottles are mathematically similar.

The travel size contains 200 millilitres and is 12 centimetres in height.

The salon size contains 1600 millilitres.

Calculate the height of the salon size bottle. 3

DalkeithHighSchool–National5ExamStyleQuestions–Vectors

H.MacLeod DalkeithHighSchool

8.

9.

National 5 Practice Paper D Last updated 27/01/15

6. The diagram shows a square based pyramid .

Express ⃗⃗⃗⃗ ⃗ in terms of and . 3

7. William Watson Fast Foods use a logo based on parts of three identical parabolas.

The logo is represented on the diagram below.

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) Find the equation of the parabola with turning point S. 2

National 5 Practice Paper F Last updated 24/04/14

4. Teams in a quiz answer questions on film and sport.

This scatter graph shows the scores of some of the teams.

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

5. Given that ⃗⃗⃗⃗ ⃗ ( ) calculate | ⃗⃗⃗⃗ ⃗|.

Give your answer as a surd in its simplest form. 3

DalkeithHighSchool–National5ExamStyleQuestions–Vectors

H.MacLeod DalkeithHighSchool

Solutions:

1.

2. B

3.

Solutions to Practice Paper A: Higher Mathematics

Page 73

2. (c)To fi nd the point of intersection:Solve 2 5

2 3 9

yy x3 =x3

⎫⎬⎫⎫⎬⎬⎫⎫⎫⎫

⎭⎬⎬⎭⎭⎬⎬⎬⎬ ✓

Subtract: 4x = −4 ⇒ x = − 1 ✓Substitute x = −1 in 2y + x = 5

⇒ 2y − 1 = 5 ⇒ 2y = 6⇒ y = 3

The point of intersection isS(−1,3) ✓

3 marks

Strategy• Simultaneous equations: if your method

is clear you will gain this strategy mark.

Calculation• Correct calculation of the other

variable gains you this last mark.

HMRN: p 6

Calculation• Correct calculation of either x or y

gains the 2nd mark.

2. (d)from (b) mPR = 2For S(−1,3) and T(2,9)

mST = = =9 3−2 −

63

2( )1− ✓

So mST = mPR = 2and so ST is parallel to PR ✓

2 marks

Gradient• You must fi nd the gradient of this new

line ST if you are to compare it’s slopewith that of line PR

Parallel lines• If the gradients of two lines are equal

then the lines are parallel.

• Clear statements are needed here sothat it is obvious you understand theresult:

equal gradients ⇔ parallel lines.

HMRN: p 4

3. (a) PQ = 4 units QR = 5 units RV = 6 units ✓

1 mark

Interpretation• The diffi culty is that the axes are

not shown. Look for a single changein coordinates: P(−1,−1,−1) to S(−1,4,−1): This is 3 units parallel to they-axis since only the y-coordinate haschanged.

HMRN: p 42

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Page 95

Solutions to Practice Paper B: Higher Mathematics

5 marks

3. Continued.For cosx° = − 1

y

x0 180 360−1

So x = 180

Solutions are:143·1, 180, 216·9 ✓(to 1 decimal place).

Solutions• A lot of knowledge and work needed

for this fi nal processing mark!

• The quadrant diagram is used:

A

T

S

C

y

x

for cos x° negative.

HMRN: p37

Point M• M is the midpoint. Half-way along DG

which is 6 units long (y-coordinate) is3 units

Point N

• N is13 of the way along AB so

13 of

6 = 2 units is the y-coordinate.

HMRN: p 42

4. (a) M(0,3,2) ✓ N(5,2,0) ✓

2 marks

4. (b)M(0,3,2) and B(5,6,0)

MB→

=

=⎛

⎝

⎜⎛⎛

⎜⎜⎜⎜⎝⎝⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟⎟⎠⎠⎟⎟ −

⎛

⎝

⎜⎛⎛

⎜⎜⎜⎜⎝⎝⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟⎟⎠⎠⎟⎟ =

−

⎛

⎝

⎜⎛⎛

b m−5

6

0

0

3

2

5

3

2⎜⎜⎜⎜⎜⎜⎜⎝⎝⎜⎜⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟⎟⎠⎠⎟⎟ ✓

also M (0,3,2) and N (5,2,0)

MN→

=

=⎛

⎝

⎜⎛⎛

⎜⎜⎜⎜⎝⎝⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟⎟⎠⎠⎟⎟ −

⎛

⎝

⎜⎛⎛

⎜⎜⎜⎜⎝⎝⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟⎟⎠⎠⎟⎟ = −

−

⎛

⎝

n m−5

2

0

0

3

2

5

1

2

⎜⎜⎛⎛⎛⎛

⎜⎜⎝⎝⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟⎟⎠⎠⎟⎟ ✓

2 marks

Components• The basic result used is:

P(x1,y1,z1) and Q (x2,y2,z2)

so PQ→

= = −

⎛

⎝

⎜⎛⎛

⎜⎜⎜

⎜⎜⎜⎜⎜⎝⎝⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟

⎟⎟⎟⎟⎟⎠⎠⎟⎟

⎛

⎝

⎜⎛⎛

⎜⎜⎜

⎜⎜⎜

⎜⎜⎜⎜⎝⎝⎜⎜

⎞

q p−x

y

z

x

y

z

2

2

2

1

1

1⎠⎠

⎟⎞⎞

⎟⎟⎟

⎟⎟⎟

⎟⎟⎟⎟⎠⎠⎠⎠⎟⎟

⎛

⎝

⎜⎛⎛

⎜⎜⎜

⎜⎜⎜

⎜⎜⎜⎜⎝⎝⎜⎜

⎞

⎠

⎟⎞⎞

⎟⎟⎟

⎟⎟⎟

⎟⎟⎟⎟⎠⎠⎟⎟

=x x−

y y−

z z−

2 1x

2 1y

2 1z

HMRN: p 43–44

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