Vectors - WordPress.com€¦ · Higher Mathematics PSfrag replacements O x y Vectors Paper 1...

Higher Mathematics

PSfrag replacementsOxy

Vectors

Paper 1 Section A

Each correct answer in this section is worth two marks.

1. A vector v is given by

−326

.

What is the length, in units, of 3v ?

A. 7

B. 15

C. 21

D. 49

Key Outcome Grade Facility Disc. Calculator Content SourceC 3.1 C 0.47 0.56 NC G16 HSN 135


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Questions marked ‘[SQA]’ c© SQAAll others c© Higher Still Notes

Higher Mathematics


2. The point A has coordinates (9, 7, 2) and B(5, 5,−1) .

What is the value of |−→AB|?A.

√3

B. 3

C.√

29

D.√

21

Key Outcome Grade Facility Disc. Calculator Content SourceC 3.1 C 0.88 0.36 NC G16 HSN 041


3. What is the distance between the points (3,−1,−1) and (2, 7,−4)?

A.√

86

B.√

74

C. 2

D. −√

62

Key Outcome Grade Facility Disc. Calculator Content SourceB 3.1 C 0.61 0.55 NC G16, G1 HSN 040


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Higher Mathematics


4. The point B has coordinates (−3, 10,−5) and −→AB =

39−5

.

What are the coordinates of point A?

A. (0, 1,−10)

B. (0,−1,−10)

C. (−6, 1, 0)

D. (6,−1, 0)

Key Outcome Grade Facility Disc. Calculator Content SourceC 3.1 C 0.73 0.36 CN G17 HSN 12


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Higher Mathematics


5. Vectors p and q are defined by p = 2i − k and q = i + j + 3k .

Find 2 p − q in component form.

A.

3−1−5

B.

3−1−4

C.

1−1−4

D.

3−3−3

Key Outcome Grade Facility Disc. Calculator Content SourceA 3.1 C 0.76 0.27 NC G17, G18 HSN 035


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Higher Mathematics


6. The vector u is given by

k2k2k

where k > 0 is a constant.

Given that u is a unit vector, what is the value of k?

A. 19

B. 15

C. 1√5

D. 13

Key Outcome Grade Facility Disc. Calculator Content SourceD 3.1 C 0.33 0.36 NC G18 HSN 119


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Higher Mathematics


7. ABCDE is a square-based pyramid, and X is the centre of the base.


A

B C

D

E

X

Given that −→AC =

440

and −→CE =

−2−25

, find −→XE.

A.

−4−45

B.

005

C.

225

D.

445



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Higher Mathematics


8. Parallelogram OABC is shown below.

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O

xy

A B

C

D

a

c

The point D divides −→AB in the ratio 3 : 1.

Find −→CD in terms of a and c .

A. a − 14 c

B. a − 13 c

C. 14 c − a

D. 13 c − a

Key Outcome Grade Facility Disc. Calculator Content SourceA 3.1 C 0.52 0.5 CN G21 HSN 116


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Higher Mathematics


9. The point P has coordinates (4,−3, 7) and Q(7,−9, 4) . The point R divides −→PQ inthe ratio 1 : 2.

Find the components of −→PR.

A.

32−3−3

2

B.

1−2−1

C.

2−4−2

D.

−121

Key Outcome Grade Facility Disc. Calculator Content SourceB 3.1 C 0.54 1 NC G25 HSN 015


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Higher Mathematics


10. The vectors

3−17

and

k2−1

are perpendicular.

What is the value of k?

A. −3

B. 3

C. 103

D. 83



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Higher Mathematics


11. The vectors

a1b

and

0−2a3b

are perpendicular.

Find an expression for a in terms of b .

A. a = 3b2

B. a =32 b2

C. a =32 b2 − 1

2

D. a = 3b2 − 1



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Higher Mathematics


12. For two vectors u and v , |u | = 4, |v | = 7 and u .v = 3.

What is the value of u .(u + v)?

A. 7

B. 12

C. 19

D. 44

Key Outcome Grade Facility Disc. Calculator Content SourceC 3.1 C 0.52 0.09 NC G29, G16 HSN 132


[END OF PAPER 1 SECTION A]

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Higher Mathematics


Paper 1 Section B

13. Two vectors a and b are given by a =

−13

24

and b =

23792

.

A third vector c is defined by c = 2a − 3b .

Find the components of c and |c | . 3

Part Marks Level Calc. Content Answer U3 OC1

3 C CN G20, G16 c =

− 83

532

, |c| =53√

5 WCHS U3 Q1

•1 pd: start vector addition•2 pd: complete vector addition•3 pd: compute magnitude

•1

− 23 − 2

4 − 73

8 − 6

•2

− 83

532

•3√

649 +

259 + 4 =

53√

5

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Higher Mathematics


14. The cubic with equation y = x3 − 4x2 − 14x + 45and the line y = −3x + 15 are shown in thediagram.The line and curve intersect at the points P, Qand R.(a) Given that the x -coordinate of Q is 2, find

the coordinates of P, Q and R. 7

(b) Determine the ratio in which Q divides theline PR. 2

PSfrag replacements

O x

y

P Q R

Part Marks Level Calc. Content Answer U3 OC1(a) 7 C CN A23 P(−3, 24), Q(2, 9), R(5, 0) AT103(b) 2 A CN G24 5 : 3

•1 ss: know to equate•2 ss: use x = 2•3 ic: start to find quadratic factor•4 ic: complete quadratic factor•5 pd: factorise completely•6 ic: interpret points•7 pd: state coordinates

•8 ss: choose approach (e.g. vectors)•9 ic: interpret ratio

•1 x2 − 4x2 − 14x + 45 = −3x + 15•2 23 − 4 × 22 − 11 × 2 + 30 = 0•3 (x − 2)(x2 · · · )•4 (x − 2)(x2 − 2x − 15)•5 (x − 2)(x − 5)(x + 3)•6 xP = −3 and xR = 5•7 P(−3, 24), Q(2, 9), R(5, 0)

•8 −→PR =

(

5−15

)

and −→QR =

(

3−9

)

•9 5 : 3

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Higher Mathematics


15. The circles centred at A and B have equations x2 + y2 + 8x + 12y + 36 = 0 andx2 + y2 − 4x − 4y − 28 = 0 respectively.

(a) Write down the coordinates of A and B. 2

(b) Show that the circles touch externally. 4

(c) The circles touch at point C.(i) Find the ratio in which C divides AB.

(ii) Hence find the coordinates of C. 4

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y

A

B

C

Part Marks Level Calc. Content Answer U3 OC1(a) 2 C CN G9 A(−4,−6), B(2, 2) AT085(b) 4 B CN G14 proof(c) 4 C CN G24, G25, G20 (i) 2 : 3, (ii) C(− 8

5 ,− 145 )

•1 ic: interpret A•2 ic: interpret B

•3 pd: find distance between centres•4 ic: interpret radius•5 ic: interpret radius•6 ss: compare sum of radii to distance

•7 ic: state ratio•8 pd: find vector components•9 ss: use parallel vectors•10 pd: process vectors

•1 A(−4,−6)•2 B(2, 2)

•3 AB = 10•4 radA = 4•5 radB = 6•6 AB = radA + radB so touch

•7 2 : 3•8 −→AB =

(

68

)

•9 c = a + 25−→AB

•10 C(− 85 ,− 14

5 )

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Higher Mathematics


16.[SQA] The point Q divides the line joining P(−1,−1, 0) to R(5, 2,−3) in the ratio 2 : 1.

Find the coordinates of Q. 3

Part Marks Level Calc. Content Answer U3 OC13 C NC G25 (3, 1,−2) 2002 P1 Q2

•1 pd: find vector components•2 ss: use parallel vectors•3 pd: process vectors

•1 −→PR =

63−3

•2 −→PQ = 23−→PR

•3 Q = (3, 1,−2)

17. R is the point (3,−1, 7) and T(8, 14,−3) . The point S divides −→RT in the ratio 3 : 2.Find the coordinates of S. 3

Part Marks Level Calc. Content Answer U3 OC13 C CN G25 S(6, 8, 1) WCHS U3 Q11

•1 pd: find vector components•2 ss: use parallel vectors•3 pd: process vectors

•1 −→RT =

515−10

•2 −→RS = 35−→RT

•3 −→OS =−→OR +

−→RS leading to S(6, 8, 1)

18. A triangle has vertices at the origin,A(8, 6, 2) and B(10, 3,−1) as shown inthe diagram.AC is a median of triangle OAB, and thepoint M divides −→AC in the ratio 2 : 1.Find the coordinates of M. 4

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O

xy

A(8, 6, 2)

B(10, 3,−1)C

M

Part Marks Level Calc. Content Answer U3 OC14 C CN G25, G20 M(6, 3, 1

3 ) Ex 3-1-5

•1 ic: interpret ratio•2 ss: use/find a directed line segment•3 ss: use the fact that C is a midpoint•4 pd: process scalar multiplication

•1 m = c +−−→CM = c + 1

3−→CA

•2 m = c + 13 (a − c)

•3 m = 13 (a + b)

•4 13

1891

=

6313

so M(6, 3, 13 )

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Higher Mathematics


19.[SQA] VABCD is a pyramid with a rectangular base ABCD.

Relative to some appropriate axes,

−→VA represents −7i − 13 j − 11k−→AB represents 6i + 6 j − 6k−→AD represents 8i − 4 j + 4k .

K divides BC in the ratio 1 : 3.Find −→VK in component form. 3


A B

CD

V

K13

Part Marks Level Calc. Content Answer U3 OC1

3 C CN G25, G21, G20

1−8−16

2000 P1 Q7

•1 ss: recognise crucial aspect•2 ic: interpret ratio•3 pd: process components

•1 −→VK =−→VA +

−→AB +−→BK or

−→VK =−→VB +

−→BK

•2 −→BK =14−→BC or 1

4−→AD or

2−11

or

−1−7−17

•3 −→VK =

1−8−16

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Higher Mathematics


20. Vectors u and v are given by u =

k1−1

and v =

−1k3

k

where k is a constant.

(a) Given that u .v = 2, show that k3 − 3k − 2 = 0. 2

(b) Show that (k + 1) is a factor of k3 − 3k− 2 and hence fully factorise k3 − 3k− 2. 5

(c) Given that |u | = 3, find the value of k . 2

(d) Given that the angle between u and v is θ , find the exact value of cos θ . 2

Part Marks Level Calc. Content Answer U3 OC1(a) 2 C CN G26 proof AT102(b) 5 C CN A21 (k + 1)2(k − 2)

(c) 2 C CN G16 k = 2(d) 2 C CN G28 2

3√

65

•1 pd: find scalar product•2 ic: complete proof

•3 ss: know to use k = −1•4 pd: complete evaluation and

conclusion•5 ic: start to find quadratic factor•6 ic: complete quadratic factor•7 pd: factorise completely

•8 ss: find magnitude•9 ic: interpret k

•10 pd: find magnitude•11 ss: use formula

•1 u.v = k × (−1) + 1 × k3 + (−2) × k•2 k3 − 3k = 2 and complete

•3 know to use k = −1•4 (−1)3 − 3(−1) − 2 = 0•5 (k + 1)(k2 · · · )•6 (k + 1)(k2 − k − 2)•7 (k + 1)(k + 1)(k − 2)

•8 |u|2 = k2 + 5 = 9•9 k = 2

•10 |v| =√

65•11 cos θ = 2

3√

65

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Higher Mathematics


21. (a)[SQA] Roadmakers look along the tops of a setof T-rods to ensure that straight sectionsof road are being created. Relative tosuitable axes the top left corners of theT-rods are the points A(−8,−10,−2) ,B(−2,−1, 1) and C(6, 11, 5) .Determine whether or not the section ofroad ABC has been built in a straight line. 3


A

B

C

(b) A further T-rod is placed such that D hascoordinates (1,−4, 4) .Show that DB is perpendicular to AB. 3PSfrag replacements

Oxy

A

B

C

D

Part Marks Level Calc. Content Answer U3 OC1(a) 3 C CN G23 the road ABC is straight 2001 P1 Q3(b) 3 C CN G27, G17 proof

•1 ic: interpret vector (e.g. −→AB)•2 ic: interpret multiple of vector•3 ic: complete proof

•4 ic: interpret vector (i.e. −→BD)•5 ss: state requirement for perpend.•6 ic: complete proof

•1 e.g. −→AB =

693

•2 e.g. −→BC =

8124

= 43−→AB or

−→AB = 3

231

and −→BC = 4

231

•3 a common direction exists and acommon point exists, so A, B, Ccollinear

•4 −→BD =

3−33

•5 −→AB.−→BD = 0•6 −→AB.−→BD = 18 − 27 + 9 = 0

or

•5 −→AB.−→BD = 18 − 27 + 9•6 −→AB.−→BD = 0 so AB is at right angles to

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Higher Mathematics


22. Find the value of c for which the vectors u =

−14

c − 2

and v =

5−5−1

are

perpendicular. 3

Part Marks Level Calc. Content Answer U3 OC13 C CN G27, G26 c = −23 OB 09-001

•1 ss: know u.v = 0•2 pd: compute u.v•3 pd: complete for c

•1 u.v = 0•2 −5 − 20 − (c − 2) = 0•3 c = −23

23. Two vectors u and v are such that |u | = 7, |v | = 4 and u .v = 14.

The vector w is defined by w = 2u + 12 v .

Evaluate w .w and hence state |w | . 4

Part Marks Level Calc. Content Answer U3 OC14 B CN G29, G16 w.w = 228, |w| =

√228 Ex 1-3-7

•1 ss: use distributive law•2 ic: use a.a = |a|2•3 pd: substitute and process•4 ic: state length

•1 w.w = 4u.u + 2u.v + 14 v.v

•2 4|u|2 + 2u.v + 14 |v|2

•3 4.72 + 2.14 + 14 .42 = 228

•4 √228 units

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Higher Mathematics


24.[SQA]PSfrag replacements

Oxy


[END OF PAPER 1 SECTION B]

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Paper 21.[SQA]




Oxy


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3.[SQA]




Oxy


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5.[SQA]



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6.[SQA]



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7.[SQA]



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8.[SQA]



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9.[SQA]



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10.[SQA] The vectors p , q and r are defined as follows:

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


11.[SQA]



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12.[SQA]



13.[SQA]




Oxy


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15.[SQA]



16.[SQA]



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17.[SQA]



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18.[SQA]PSfrag replacementsOxy


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19.[SQA] A cuboid measuring 11 cm by 5 cm by 7 cm is placed centrally on top of anothercuboid measuring 17 cm by 9 cm by 8 cm.

Coordinates axes are taken as shown.

PSfrag replacements

O

xy

5 7

89

11

17

z

A

BC

(a) The point A has coordinates (0, 9, 8) and C has coordinates (17, 0, 8) .Write down the coordinates of B. 1

(b) Calculate the size of angle ABC. 6

Part Marks Level Calc. Content Answer U3 OC1(a) 1 C CN G22 B(3, 2, 15) 2000 P2 Q9(b) 6 C CR G28 92·5◦

•1 ic: interpret 3-d representation

•2 ss: know to use scalar product•3 pd: process vectors•4 pd: process vectors•5 pd: process lengths•6 pd: process scalar product•7 pd: evaluate scalar product

•1 B= (3, 2, 15) treat

32

15

as bad form

•2 cos ABC =−→BA.−→BC|−→BA||−→BC|

•3 −→BA =

−37−7

•4 −→BC =

14−2−7

•5 |−→BA| =√

107, |−→BC| =√

249•6 −→BA.−→BC = −7•7 ABC = 92·5◦

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Higher Mathematics


20. The square-based pyramid ABCDT is shown below.PSfrag replacementsOxy

A

B C

D

T

All of the edges of ABCDT have length 4 units.

(a) Find the exact value of cos TAC. 2

(b) Hence find the exact value of −→AT.(−→AB +−→AC) . 4

Part Marks Level Calc. Content Answer U3 OC1(a) 2 B CN G22

√2

2 WCHS U3 Q3(b) 4 B CN G29, G26 24

•1 ss: use Pythagoras•2 ic: complete

•3 ss: use distributive law•4 pd: expand scalar products•5 ic: complete “TAB” value•6 ic: complete, using cos TAC

•1 AC = 4√

2•2 let M = midpointAC so AM = 2

√2

cos TAC =AMAT =

√2

2 or equivalent

•3 −→AT.−→AB +−→AT.−→AC

•4 |−→AT||−→AB| cos TAB + |−→AT||−→AC| cos TAC•5 4 × 4 cos 60◦ = 8•6 +4 × 4

√2 ×

√2

2 , so 24

21.[SQA]



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Higher Mathematics


22.[SQA]



23.[SQA]



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24.[SQA]



25.[SQA]



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Higher Mathematics


26.[SQA] ABCD is a quadrilateral with vertices A(4,−1, 3) , B(8, 3,−1) , C(0, 4, 4) andD(−4, 0, 8) .

(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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Higher Mathematics


27. In the diagram, the circle centred at A hasequation x2 + y2 − 10x − 6y + 9 = 0 and the circlecentred at B has equation (x + 1)2 + (y + 5)2 = 4.(a) Find the coordinates of A and B. 2

(b) Find the shortest distance between the twocircles. 4

(c) Points A, B and C are collinear, and the circlecentred at C touches the other two circlesexternally.

(i) Given that −→BC = k−→BA, find the value ofk .

(ii) Hence find the coordinates of C.(iii) Write down the equation of the circle

centred at C. 5

PSfrag replacements

Ox

y

A

B

C

Part Marks Level Calc. Content Answer U3 OC1(a) 2 C CN G9 A(5, 3), B(−1,−5) Ex 2-6-4(b) 4 B CN G9, G1 3(ci) 1 B CN G24 k = 7

20(cii) 3 B CN G25 C( 11

10 ,− 115 )

(ciii) 1 B CN G10 (x− 1110 )2 +(y + 11

5 )2 = 2·25

•1 ic: interpret circle equation•2 ic: interpret circle equation

•3 ss: use distance between centres•4 ic: use radius formula•5 ic: interpret circle equation•6 pd: conclusion

•7 ic: interpret ratio•8 ss: use a vector pathway•9 pd: process•10 pd: complete•11 ic: state circle equation

•1 A(5, 3)•2 B(−1,−5)

•3 dAB = 10•4 radiusA =

√

(−5)2 + (−3)2 − 9 = 5•5 radiusB = 2•6 10 − 5 − 2 = 3

•7 k = 720

•8 c = b +−→BC

•9 −→BC = 720−→BA =

( 2110145

)

•10 C( 1110 ,− 11

5 )

•11 (x − 1110 )2 + (y + 11

5 )2 = 2·25

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28.[SQA]



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29.[SQA]



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30.[SQA]



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31.[SQA]



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32.[SQA]



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33.[SQA]



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34. Vectors u , v and w are given by u =

k−13k

, v =

−1−25

and w =

141

,

where k is a constant.

(a) Given that u + v is perpendicular to w , find the value of k . 3

(b) Hence calculate the smallest angle between u and v . 5

Part Marks Level Calc. Content Answer U3 OC1(a) 3 C CN G26, G27 k = 2 New 3.1 G6(b) 5 C CR G28, G26, G16 31·2◦

•1 ss: know to set scalar product tozero

•2 pd: expand scalar product•3 pd: process

•4 pd: find length of vector•5 pd: find length of vector•6 pd: find scalar product•7 ss: use scalar product•8 pd: evaluate angle

•1 since perp., (u + v).w = 0

•2

k − 1−3

3k + 5

.

141

= k− 1− 12 + 3k + 5

•3 k = 2

•4 |u| =√

22 + (−1)2 + 62 =√

41•5 |v| =

√

(−1)2 + (−2)2 + 52 =√

30•6 u.v = 2.(−1) + (−1).(−2) + 6.5 = 30•7 cos θ = u.v

|u||v| = 30√41

√30

•8θ = 31·2◦ (1 d.p.) or 0·544 rads (3d.p.)

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Higher Mathematics


35.[SQA] A box in the shape of a cuboidis designed with circles of differentsizes on each face.The diagram shows three of thecircles, where the origin representsone of the corners of the cuboid. Thecentres of the circles are A(6, 0, 7) ,B(0, 5, 6) and C(4, 5, 0) .Find the size of angle ABC. 7

PSfrag replacements

O

x

y

z

A

B

C

Part Marks Level Calc. Content Answer U3 OC15 C CR G17, G16, G22 2001 P2 Q42 A/B CR G26, G28 71·5◦

•1 ss: use−→BA.−→BC|−→BA||−→BC|

•2 ic: state vector e.g. −→BA•3 ic: state a consistent vector e.g. −→BC•4 pd: process |−→BA|•5 pd: process |−→BC|•6 pd: process scalar product•7 pd: find angle

•1 use−→BA.−→BC|−→BA||−→BC|

stated or implied by •7

•2 −→BA =

6−51

•3 −→BC =

40−6

•4 |−→BA| =√

62•5 |−→BC| =

√52

•6 −→BA.−→BC = 18•7 ABC = 71·5◦

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Higher Mathematics


36. The parallelogram PQRS is shown in the diagram below.PSfrag replacements

Oxy

P

Q

R

S Tu

v

The vectors u and v represent line segments −→QP and −→QR respectively, and aresuch that |u | = 5, |v | = 2 and u .v = 5. The point T divides −→QP in the ratio 2 : 3.

(a) Express −→RP and −→RT in terms of u and v . 3

(b) Hence evaluate −→RP.−→RT. 3

Part Marks Level Calc. Content Answer U3 OC1(a) 3 C CN G21, G25 −→RP = u − v, −→RT = 2

5 u − v OB 09-003(b) 3 B CN G26, G29 −→RP.−→RT = 7

•1 ss: use pathways•2 ss: use pathways•3 ic: interpret ratio

•4 ss: substitute and expand•5 pd: use magnitudes, and u.v•6 pd: complete

•1 −→RP = u − v•2 −→RT =

−→RQ +−→QT

•3 −→RT = 25 u − v

•4 −→RP.−→RT = (u − v).( 25 u − v)

•5 |u|2 − 75 u·v + |v|2

= 25 52 − 7

5 5 + 22

•6 7

37.[SQA] For what value of t are the vectors u =

t−23

and v =

210t

perpendicular? 2

Part Marks Level Calc. Content Answer U3 OC12 C CN G27 t = 4 2000 P2 Q7

•1 ss: know to use scalar product•2 ic: interpret scalar product

•1 u.v = 2t − 20 + 3t•2 u.v = 0 ⇒ t = 4

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Higher Mathematics


38.[SQA] A(4, 4, 10) , B(−2,−4, 12) and C(−8, 0, 10) are the vertices of a right-angledtriangle.

Determine which angle of the triangle is the right angle. 3


39.[SQA]PSfrag replacementsOxy


40. The vectors u and v have components

k−2−4

and

k−k−1

respectively.

Show that there is no value of k for which u and v are perpendicular. 4

Part Marks Level Calc. Content Answer U3 OC14 C CN G27, G26, A17 proof AT042

•1 ss: know to use scalar product•2 pd: compute scalar product•3 ss: know to use discriminant•4 ic: conclusion

•1 u ⊥ v iff u.v = 0•2 u.v = k2 + 2k + 4•3

∆ = −12•4 since ∆ < 0, u.v 6= 0

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Higher Mathematics


41.[SQA] The diagram shows a square-basedpyramid of height 8 units.Square OABC has a side length of 6 units.The coordinates of A and D are (6, 0, 0)and (3, 3, 8) .C lies on the y-axis.(a) Write down the coordinates of B. 1

(b) Determine the components of −→DAand −→DB. 2

(c) Calculate the size of angle ADB. 4

PSfrag replacements

Ox

y

A(6, 0, 0)

BC

D(3, 3, 8)z

Part Marks Level Calc. Content Answer U3 OC1(a) 1 C CN G22 (6, 6, 0) 2002 P2 Q2

(b) 2 C CN G17 −→DA =

3−3−8

,

−→DB =

33−8

(c) 4 C CR G28 38·7◦

•1 ic: interpret diagram

•2 ic: write down components of avector

•3 ic: write down components of avector

•4 ss: use e.g. scalar product formula•5 pd: process lengths•6 pd: process scalar product•7 pd: process angle

•1 B = (6, 6, 0)

•2 −→DA =

3−3−8

•3 −→DB =

33−8

•4 cos ADB =−→DA.−→DB|−→DA||−→DB|

•5 |−→DA| =√

82, |−→DB| =√

82•6 −→DA.−→DB = 64•7 ADB = 38·7◦

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Higher Mathematics


42.[SQA]



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Higher Mathematics


43. Calculate the acute angle between the vectors u =

6√3

3

and v =

−12

4−1

. 4

Part Marks Level Calc. Content Answer U3 OC14 B CR G28, G16 θ = 88·15◦ WCHS U3 Q8

•1 ss: use scalar product formula•2 pd: process lengths•3 pd: process scalar product•4 pd: process angle

•1 cos θ =

u.v|u||v|

•2 |u| =

√48, |v| =

√

17 14

•3 u.v = 4√

3 − 6•4 cos θ ≈ 0·0323, so θ = 88·15◦ (to 2

d.p.)

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Higher Mathematics


44.[SQA]



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Higher Mathematics


45. Calculate the acute angle between the two vectors p = 2i + 4 j − kand q = i − 3 j + 2k . 5

Part Marks Level Calc. Content Answer U3 OC15 C CR G28, G26, G16 134·4◦ Ex 3-1-3

•1 pd: find length of vector•2 pd: find length of vector•3 pd: find scalar product•4 ss: use scalar product•5 pd: evaluate angle

•1 |p| =√

22 + 42 + (−1)2 =√

21•2 |q| =

√

12 + (−3)2 + 22 =√

14•3 p.q = 2.1 + 4.(−3) + (−1).2 = −12•4 cos θ =

p.q|p||q| = − 12√

21√

14•5

θ = 134·4◦ (1 d.p.) or 2·346 rads (3d.p.)

46.[SQA]



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Higher Mathematics


47.[SQA]



48.[SQA]



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Higher Mathematics


49.[SQA]



50.[SQA]



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Higher Mathematics


51.[SQA] PQRS is a parallelogram with vertices P(1, 3, 3) , Q(4,−2,−2) and R(3, 1, 1) .

Find the coordinates of S. 3


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Higher Mathematics


52.[SQA]



[END OF PAPER 2]

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