General Certificate of Education Advanced Level Examination January 2011

Mathematics

MFP4

Unit Further Pure 4 Friday 28 January 2011

9.00 am to 10.30 am

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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.

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Time allowed * 1 hour 30 minutes

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Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer the questions in the spaces provided. Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked.

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Information * The marks for questions are shown in brackets. * The maximum mark for this paper is 75. Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet.

P38407/Jan11/MFP4 6/6/6/

MFP4

2

   1 2 3     Let D ¼  x y z .   y þ z z þ x x þ y

1

(a)

Use a row operation to show that ðx þ y þ zÞ is a factor of D.

(2 marks)

(b)

Hence, or otherwise, express D as a product of linear factors.

(2 marks)

The non-zero vectors a and b have magnitudes a and b respectively.

2

Let c ¼ ja  bj and d ¼ ja . bj . By considering the definitions of the vector and scalar products, or otherwise, show that c2 þ d 2 ¼ a2 b2

(3 marks)

Find the values of t for which the system of equations

3 (a)

tx þ 2y þ

3z ¼ a

2x þ 3y 

tz ¼ b

3x þ 5y þ ðt þ 1Þz ¼ c does not have a unique solution.

(3 marks)

For the integer value of t found in part (a), find the relationship between a, b and c such that this system of equations is consistent. (3 marks)

(b)

2

3 1 4 The non-singular matrix X ¼ 2 4 4 2

4

(a) (i)

3 3 35. 1

Show that X2  X ¼ kI for some integer k.

(3 marks)

(ii) Hence show that X1 ¼ 1 ðX  IÞ . 20

(2 marks) 2

(b)

60 1 4 The 3  3 matrix Y has inverse Y ¼ 0 0

3 0 0 0 10 5 . 20 0

Without finding Y , determine the matrix ðXYÞ1 .

(3 marks)

P38407/Jan11/MFP4

3

2 3 2 3 6 10 The planes P1 and P2 have vector equations r . 4 2 5 ¼ 5 and r . 4 1 5 ¼ 4 9 11 respectively.

5

(a)

Write down cartesian equations for P1 and P2 .

(b)

Find a vector equation for the line of intersection of P1 and P2 .

(c)

The plane P3 has cartesian equation 5x þ 3y þ 11z ¼ 28 .

(1 mark) (5 marks)

Use your answer to part (b) to find the coordinates of the point of intersection of (4 marks) P1 , P2 and P3 . Determine a vector equation for the plane which passes through the point ð4, 1, 9Þ (3 marks) and which is perpendicular to both P1 and P2 .

(d)

2

3 12 The plane P has equation r . 4 15 5 ¼ 11 and the point Q has coordinates ð1, 1, 1Þ. 16

6 (a)

Show that Q is in P .

(b) (i)

Write down cartesian equations for the line l which passes through Q and is perpendicular to P . (2 marks)

(1 mark)

(ii) Deduce the direction cosines of l. (c)

The points M and N are on l, and each is 50 units from P . Find the coordinates of M and N .

(d)

(3 marks)

Given that the point Pð5, 1, 4Þ is in P , determine the area of triangle PMN . (3 marks) 2

3 4 Let Y ¼ 1 1

7

(2 marks)

1 3 1

3 1 15. 3

(a)

Show that 4 is a repeated eigenvalue of Y, and find the other eigenvalue of Y. (7 marks)

(b)

For each eigenvalue of Y, find a full set of eigenvectors.

(c)

The matrix Y represents the transformation T.

(5 marks)

Describe the geometrical significance of the eigenvectors of Y in relation to T. (3 marks) s

Turn over

P38407/Jan11/MFP4

4



3 The plane transformation T is represented by the matrix M ¼ 1

8

 8 . 3

The quadrilateral ABCD has image A 0 B 0 C 0 D 0 under T.

(a)

Evaluate det M and describe the geometrical significance of both its sign and its magnitude in relation to ABCD and A 0 B 0 C 0 D 0 . (3 marks) The line y ¼ px is a line of invariant points of T, and the line y ¼ qx is an invariant line of T.

(b)

Show that p ¼ 1 and determine the value of q.

(5 marks)

2

(c) (i)

Find the 2  2 matrix R which represents a reflection in the line y ¼ 1 x . 2

(2 marks)

(ii) Given that T is the composition of a shear, with matrix S , followed by a reflection in the line y ¼ 1 x , determine the matrix S and describe the shear as fully as possible. 2

(5 marks)

Copyright ª 2011 AQA and its licensors. All rights reserved.

P38407/Jan11/MFP4

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011

Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp

mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) decimal place(s)

No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.

3

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011

MFP4 Q 1(a)

1 Δ=

x x+ y+z

Solution 2

y y+z+x 1

Marks

Total

3 z z+x+ y

e.g. R3′ = R3 + R2

M1

2 3

= ( x + y + z) x y z 1 1 1

A1

(b) Expanding remaining det. Δ = (x + y + z)(x – 2y + z) Total 2 c = | a × b | = ab sinθ d = | a • b | = ab |cosθ | c2 + d2 = a2b2(cos2θ + sin2θ ) = a2b2 Total 3(a) t 2 3

M1 A1 B1 B1 B1

2

2 4

3 3

M1

2 3 −t = 8t – 7t – 1 = 0 3 5 t +1

M1

t = 1, − 18

A1

x + 2 y + 3z = a t = 1 ⇒ 2x + 3y − z = b 3x + 5 y + 2 z = c

E.g. ➀ + ➁ – ➂ ⇒ a + b = c Total 4(a)

⎡ 23 1 −3⎤ (i) X = ⎢⎢ 2 24 3 ⎥⎥ ⎢⎣ −4 2 19 ⎥⎦ X2 – X = 20I i.e. k = 20

X

3 6

A1

(ii) Multg. X2 – X = 20I by X – 1 Re-arranging X – I = 20 X – 1 1 ( X − I) to get X – 1 = 20

–1

FT any integer value found

M1 A1

2

(b)

3

B1 M1 A1

⎡ 2 1 −3⎤ 1 ⎢2 3 3 ⎥⎥ = 20 ⎢ ⎣⎢ −4 2 −2 ⎦⎥

(XY) – 1 = Y – 1 X – 1 ⎡ 6 3 −9 ⎤ ⎢ 2 −1 1 ⎥ ⎢ ⎥ ⎢⎣ 2 3 3 ⎥⎦

Condone lack of | - | Legitimately shown Attempt at det. of coefft. mtx. (or equivalent) Equating to zero and solving a quadratic eqn. in t

2

(b)

Comments

≥ 5 correct for the M All 9 correct for the A 3

Shown legitimately

2

Legitimately

M1 A1

B1

Noted or used

M1

Incl. attempt at the multn.

A1 Total

3 8

4

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011

MFP4(cont) Q Solution 5(a) ∏1 : 6x + 2y + 9z = 5 ∏2 : 10x – y – 11z = 4 (b) Method 1 E.g. ∏1 + 2 ∏2 ⇒ 13(2x – z = 1) x z +1 = =λ 1 2 Using 1st two to find 3rd in terms of λ: x = λ, z = 2λ – 1 ⇒ y = 7 – 12λ ⎡0⎤ ⎡ 1 ⎤ ⎢ ⎥ r = ⎢ 7 ⎥ + λ ⎢⎢ −12 ⎥⎥ ⎢⎣ −1⎥⎦ ⎢⎣ 2 ⎥⎦ Method 2 (0, 7, –1) or

( 127 ,

0,

1 6

) or ( 12 ,

1, 0

Total

B1

1

M1 M1 A1

Comments

Both

Parametrisation attempt

M1 A1

)

⎡ 6 ⎤ ⎡ 10 ⎤ ⎡ 1 ⎤ ⎢ 2 ⎥ × ⎢ −1 ⎥ = (±)13 ⎢ −12 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 9 ⎥⎦ ⎢⎣ −11⎥⎦ ⎢⎣ 2 ⎥⎦ ⎡0⎤ ⎡ 1 ⎤ ⎢ ⎥ r = ⎢ 7 ⎥ + λ ⎢⎢ −12 ⎥⎥ ⎢⎣ −1⎥⎦ ⎢⎣ 2 ⎥⎦ (c) Substg. x = λ, y = 7 – 12λ, z = 2λ – 1 into 5x + 3y + 11z = 28 Solving a linear eqn. in λ : λ=–2 ( – 2, 31, – 5) (d)

Marks

⎡ 1 ⎤ n = ⎢⎢ −12 ⎥⎥ ⎢⎣ 2 ⎥⎦ ⎡4⎤ ⎡ 1 ⎤ d = ⎢⎢1 ⎥⎥ • ⎢⎢ −12 ⎥⎥ = 10 ⎢⎣ 9 ⎥⎦ ⎢⎣ 2 ⎥⎦

5

(M1A1)

Any correct vector eqn.form

Finding a pt. on the line

(M1) (A1)

Finding a d.v.

(A1)

Any correct vector eqn.form

M1 M1 A1 B1

4

CAO FT their λ in their line eqn.

B1

FT when used as part of a plane eqn., not a line

M1

Attempted Any correct vector eqn. form

⎡ 1 ⎤ r • ⎢⎢ −12 ⎥⎥ = 10 ⎢⎣ 2 ⎥⎦ ⎡4⎤ ⎡ 6 ⎤ ⎡ 10 ⎤ ⎢ ⎥ Alt. r = ⎢ 1 ⎥ + λ ⎢⎢ 2 ⎥⎥ + ⎢⎢ −1 ⎥⎥ ⎢⎣ 9 ⎥⎦ ⎢⎣ 9 ⎥⎦ ⎢⎣ −11⎥⎦

A1

3

(M1) (A1) (A1) Total

Plane eqn. attempt Point + at least 1 d.v. All correct 13

5

FT

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011

MFP4(cont) Q Solution 6(a) ⎡ 1 ⎤ ⎡12 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ 1 ⎥ • ⎢15 ⎥ = 12 + 15 − 16 = 11 shown ⎢⎣ −1⎥⎦ ⎢⎣16 ⎥⎦ (b) (i)

eqn., or use, of line incorporating ⎡1⎤ ⎡12 ⎤ ⎢ ⎥ p.v. ⎢ 1 ⎥ and d.v. ⎢⎢15 ⎥⎥ ⎣⎢ −1⎦⎥ ⎣⎢16 ⎦⎥

1

Comments

2

B1

122 + 152 + 162 = 25

⎡1⎤ ⎡12 ⎤ ⎢ ⎥ Use of r = ⎢ 1 ⎥ + λ ⎢⎢15 ⎥⎥ with λ = ± 2 ⎢⎣ −1⎥⎦ ⎢⎣16 ⎥⎦ (25, 31, 31) and (– 23, – 29, – 33)

(d) Method 1 PQ = 5 Area ΔPMN = 12 × PQ × MN = 250

B1

FT 2

FT

M1 A1 A1

3

B1 M1 A1

Method 2 ⎡ −28⎤ ⎡ 20 ⎤ JJJJG ⎢ ⎥ JJJG ⎢ PM = ⎢ 30 ⎥ , PN = ⎢ −30 ⎥⎥ ⎢⎣ −29 ⎥⎦ ⎢⎣ 35 ⎥⎦ ⎡ 9 ⎤ JJJJG JJJG PM × PN = (±)20 ⎢⎢ −20 ⎥⎥ attempted ⎢⎣ 12 ⎥⎦ within an area formula Area ΔPMN = 250

B1

A1

12 15 16 , , 25 25 25 (c)

Total

M1

x −1 y −1 z +1 = = 12 15 16 (ii)

Marks

3

(Since Q = midpt.MN)

(M1)

Attempted

(M1)

||gm. or Δ

(A1) Total

CAO 11

6

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011

MFP4(cont) Q Solution 7(a) Attempt at Char.Eqn. λ3 – 9λ2 + 24λ – 16 = 0 Attempt at (at least) one linear factor (λ – 1)(λ – 4)2 = 0 (b)

(c)

λ = 1, 4, 4

2x − y + z = 0 ⎡1⎤ λ = 1 ⇒ − x + 2 y + z = 0 ⇒ α ⎢⎢ 1 ⎥⎥ ⎢⎣ −1⎥⎦ x + y + 2z = 0 λ=4 ⇒ x+y–z=0 Choosing any two independent vectors satisfying this ⎡1 ⎤ ⎡ 0 ⎤ ⎡ 1 ⎤ E.g.s: ⎢⎢0⎥⎥ , ⎢⎢1 ⎥⎥ , ⎢⎢1 ⎥⎥ ⎢⎣1 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ 2⎥⎦

⎡1⎤ α ⎢⎢ 1 ⎥⎥ ≡ a line of invariant points ⎢⎣ −1⎥⎦ ⎡1 ⎤ ⎡0⎤ ⎢ ⎥ e.g. α ⎢0 ⎥ + β ⎢⎢1 ⎥⎥ ≡ an invariant plane ⎢⎣1 ⎥⎦ ⎢⎣1 ⎥⎦

Marks M1 A3,2,1 M1 A1 A1

Total

Comments One each following coefft.

7 Substg. back and solving

M1 A1 B1 M1

A1

5

M1 A1

B1

Total

Invariant line LoIPs

3 15

7

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011

MFP4(cont) Q Solution 8(a) Det(M) = – 1 Magnitude = 1 ⇒ area invariant – ve sign ⇒ cyclic order of vertices is reversed OR “reflection” involved (b) Method 1 Char. Eqn.: λ2 – 1 = 0 ⇒ λ = ± 1 Substg. back: λ = 1 ⇒ y = 12 x and λ = –1 ⇒ y =

1 4

x

4⎤ 5⎥ ⎥ −3⎥ 5⎦

=

Finding and solving attempt 5

(M1)

Attempted

(M1) (M1A1)

From (4m – 1)(2m – 1) = 0

⎡ −3 8⎤ ⎢ −1 3⎥ = ⎣ ⎦

M1

A1

⎡3 4 ⎤ ⎡ −3 8 ⎤ (ii) Use ⎢ 5 5 ⎥ S = ⎢ ⎥ ⎢ 4 −3 ⎥ ⎣ −1 3 ⎦ ⎣5 5⎦ S found using inverse matrix 4⎤ 5⎥ ⎥ −3⎥ 5⎦

FT area s.f.

(A1)

⇒ cos2θ = 53 and sin2θ = 54

⎡3 ⎢5 ⎢4 ⎢⎣ 5

3

A1

(i) p = 12 = tanθ ⎡3 ⎢5 ⎢4 ⎢⎣ 5

B1

Comments

M1 A1

p = 12 and q = 14

R=

Total

M1 A1

Method 2 ⎡ −3 8⎤ ⎡ x ⎤ ⎡(8m − 3) x ⎤ ⎢ −1 3⎥ ⎢ mx ⎥ = ⎢ (3m − 1) x ⎥ ⎦ ⎣ ⎦⎣ ⎦ ⎣ Use of y′ = mx′: 3m – 1 = 8m2 – 3m Solving a quadratic eqn. in m = 14 , 12

(c)

Marks B1 B1

1 ⎡ −13 5 ⎢ −9

⎣

36 ⎤ 23⎥⎦

Shear, parallel to y = 12 x mapping (e.g.) (1, 1) → (4.6, 2.8) Total TOTAL

For these attempted and used in a reflection matrix 2

M1

FT their R

M1

Or equivalent method

A1 B1 B1

8

CAO 5 15 75

FT any pt. and its image

AQA – Further pure 4 – Jan 2011 – Answers Question 1:

∆

Exam report

1 2 3 = x y z y+z

z+x

x+ y

1 x

2 y

3 z

x+ y+z

x+ y+z

x+ y+z

' Combination : R= R2 + R3 3

1 ∆ = ( x + y + z) x

2 3 y z =

1 ( x + y + z) x

1 1 1

1 y−x

2 z−x

0

0

1

Combinations : C2' = C2 − C1 and C3' = C3 − C1

∆= ( x + y + z ) ( ( z − x) − 2( y − x) ) = ( x + y + z )( x − 2 y + z )

Question 2:

c = a × b = abSinθ and d = a.b = abCosθ + d 2 a 2b 2 Sin 2θ + a 2b 2Cos 2θ c 2=

= c 2 + d 2 a 2b 2 (Cos 2θ += Sin 2θ ) a 2b 2

This was a straightforward starter to the paper, and was generally found to be so by candidates. However, almost half of all candidates failed to score all 4 marks, due largely to arithmetical slips and carelessness with signs. This was mostly due to unhelpfully lengthy work on row- and/or column-operations after the (x + y + z) factor had been extracted, when the simplest approach was to just go ahead and expand the remaining determinant. This highlights the view that many of these candidates had had insufficient practice with handling determinants algebraically in order to arrive at the point when they might have a “feel” for what is a good approach to take at different stages. Moreover, even in this example, it was about 50-50 whether candidates added R3 into R2 or vice versa; the latter approach clearly producing a simpler determinant to continue to work with afterwards, if some thought had been given to it.

Exam report This turned out to be much more troublesome for candidates than had been anticipated, with almost two-thirds of the candidature failing to score a single mark on the question, despite the injunction to consider the definitions of the two products. Even amongst those who did proceed as planned, a large number of them insisted on using modulus signs and/or a unit vector throughout, indicating a lack of grasp as to what should, and what should not, have been involved. Some of these misuses of both notation and understanding of vector and scalar matters were condoned.

Question 3:

Exam report

The system does not have a unique solution when the determinant of the associated matrix is 0 t

2 3 3 −t 2 −t 2 3 −t −2 +3 t det= 2 3 = 5 t +1 3 t +1 3 5 3 5 t +1 det= t (3t + 3 + 5t ) − 2(2t + 2 + 3t ) + 3(10 − 9) det = 8t 2 + 3t − 10t − 4 + 3 = 8t 2 − 7t − 1 det= 0 ⇔ 8t 2 − 7t − 1 =0 (8t + 1)(t − 1) = 0 1 8 b) for t = 1, the system becomes t = 1 or t = −

a l1 a  x + 2 y + 3z =  x + 2 y + 3z =   y + 7 z = 2a − b (2l1 − l2 )  2 x + 3 y − z = b l2 ⇔  3 x + 5 y + 2 z = c l  y + 7 z = 3a − c (3l1 − l3 ) 3   The system is consistent if and only if 2a − b = 3a − c a+b−c = 0

This was generally well done, with marks relatively high. It was surprising to find that those who failed to get any integer value for t generally didn’t appear to go back and check their working. Weaker candidates often didn’t seem to know how to go about establishing consistency.

Question 4:

Exam report

 23 1 −3   3 1 −3      a= ) i ) X 2 − X  2 24 3  −  2= 4 3  −4 2 19   −4 2 −1      X2 − X = 20I

 20 0 0     0 20 0   0 0 20   

ii ) By multiplying both sides by X −1 , we have X −1 ( X 2 − X) 20X −1I ⇔ =

X= − I 20 X −1

 2 1 −3  1 1   X (X − I) =  2 3 3  = 20  20   −4 2 −2  1 b) ( XY) −1 = Y −1X −1 = Y −1 × ( X − I ) = 20 0   2 1 −3   6 3 −9   60 0 1      =  0 0 −10   2 3 3  =  2 −1 1  20       0 20 0   −4 2 −2   2 3 3  −1

This was another popular question, in that responses were generally at least partially successful. Even so, around half of all candidates failed to realise that the “hence” in part (a)(ii) meant that they were supposed to work algebraically with the result of part (a)(i) rather than go through the lengthy –1 approach of finding X directly. About the same proportion –1 –1 –1 incorrectly opted for (XY) = X Y . In addition to all these errors, a lot of marks were lost due to carelessness; mistakes 2 made in calculating X were often not corrected, even when the candidates clearly failed to arrive at a multiple of I in part (a)(i). –1 Then, despite being told that X could be found by

1 ( X − I ) in part (b), many made a slip somewhere in the 20 –1

working, and the correct (XY) appeared far less often than should have been the case, given the information given in the question.

Question 5:

Exam report

 x  6    5 ⇔ 6x + 2 y + 9z = a ) Π1 :  y  .  2  = 5  z  9     x  10     4 Π 2 :  y  .  −1  =4 ⇔ 10 x − y − 11z =  z   −11     6  10      b) A direction vector of the line of intersection is u =  2  ×  −1       9   −11 1   −13      u = 156  , the vector  −12  is also a direction vector. 2   −26      5 6 x + 2 y + 9 z = A point belonging to both plane satisfies the equations  4 10 x − y − 11z = = z 13 (2l2 + l1 ) 1 26 x − 13 2 x − z = ⇔ ⇔ 5 5 6 x + 2 y + 9 z = 6 x + 2 y + 9 z = For example, the point P(0, 7, − 1) belongs to both planes.  0  1      An equation of the line of intersection is r =  7  + t  −12   −1  2      c) The point of intersection of the three planes is the point of intersection between the line and Π 3 Q(t , 7 − 12t , − 1 + 2t ) belongs to the plane with equation 5 x + 3 y + 11z =28 so 5t + 3(7 − 12t ) + 11(−1 + 2t ) =28 − 9t + 10 = 28 t= −2 and the coordinates of Q are ( − 2,31, −5) d ) Let's call Π 4 the plane perpendicular to Π1 and Π 2 going through (4,1,9). A normal vector to Π 4 is i − 12j+2k ( the direction vector of the line of intersection)   4   1  1         An equation of the plane Π 4 is :  r − 1   .  −12  =0 ⇔ r.  −12  = 10 2   9  2        

Around a third of the candidature attempted little beyond part (a), which was very surprising. Of the majority who did proceed well into the question, part(b) was usually done well, although there were many sign errors that arose in the vector product of the two normals. Another surprise was the lack of appreciation that there was a factor of 13 which could be ignored here. Of those who realised in part(c) that they simply had to substitute part(b)’s answer into Π 3 ’s equation, a slight majority had incorrect components to work with, and this made follow through marking difficult. In part(d), only a very few spotted that they could answer this without any need to rely on previous answers, as a plane equation in the form r = a +λd 1 + μd 2 could be written straight down using the two normals given in the question. The most popular form was, of course, r . n = d, although a lot of candidates opted for a line equation here instead.

Question 6:

Exam report

1  12     a ) Let's work out 1  . 15  = 12 + 15 − 16 = 11  −1 16     Q belongs to the plane Π. b) i ) A direction vector of the line is n =12i +15j+16k x −1 y −1 z +1 and cartesian equations of the line are: = = 12 15 16 ii ) n = 122 + 152 + 162 = 25 12 15 3 16 , = , 25 25 5 25 1  12      c) A vector equation of l is : r = 1  + t 15   −1 16      The direction cosines are :

A point P on the line has coordinates P(1+12t,1+15t, − 1+16t)  The distance PQ is 50 i.e PQ = 50 12t      PQ = 15t  and PQ = (12t ) 2 + (15t ) 2 + (16t ) 2 = 50 16t    2 t2 4 = 625t 2 50 = t = 2 or t = −2

Around a quarter of all candidates made attempts only at part (a), for just the one mark, and around half the candidates failed to proceed beyond part (b). Even amongst these, there was a large number of candidates who apparently didn’t know the difference between vector and cartesian equations, with many others not happy to work with the given normal vector. Of that half of the candidature that did proceed beyond part (b), most of them decided that “a distance of 50 units” meant that λ was ± 50, rather than ± 2. They had failed to realise that they had just worked out that the direction vector 12i + 15j + 16k was 25 units long. This was very disappointing. Even in the final part of the question, despite being given the coordinates of P and Q and the information that the line PQ was the height of a right-angled triangle with base 100, only about five candidates (of 336) appreciated that they could work out the area of the triangle just as it was. Even more disappointing than this was that so few could employ alternative vector methods to get the correct answer, mostly due to not having had much success with parts (a) to (c).

This gives M ( 25,31,31) and N ( −23, −29, −33) d ) In the triangle PMN, the base MN =100 and the height is PQ  PQ = −4i + 3k and PQ= 16 + 9= 5 1 1 The area of the triangle is NM × PQ = ×100 × 5 = 250 2 2

Question 7:

Exam report

a ) Let's work out det(Y-λ I )

−1 3 − λ −1 1 4−λ det(Y-λ I ) = −1 3 − λ 1 = −4 + λ 3 − λ 1 1 3−λ 0 1

0 4−λ 4−λ

We did : C1' = C1 − C2 and C3' = C2 + C3 −1 0 1 1 −1 0 det(Y-λ I ) =(4 − λ ) 2 −1 3 − λ 1 =(4 − λ ) 2 0 2 − λ 1 0

1

1

0

1

1

det(Y-λ I ) = (4 − λ ) (2 − λ − 1) = (4 − λ ) (1 − λ ) det(Y-λ I )= 0 ⇔ λ = 4 (repeated value) or λ = 1 2

2

x   b) To find the eigenvectors we need to solve (Y-λ I )  y  = 0 z    − y+z 0 x − 3y 0  2 −1 1   x  2 x = 3=      For λ = 1,  −1 2 1   y  = 0 ⇔ − x + 2 y + z = 0 ⇔ − x + 2 y + z = 0  x + y=  x + y=  1 1 2 z  + 2z 0 + 2z 0      x = y i +j − k ⇔ An eigenvector is u= 1 z = −x  −1 −1 1   x  − x − y + z =0     For λ = 4,  −1 −1 1   y  = 0 ⇔ − x − y + z = 0 ⇔ { x + y − z = 0 x + y − z =  1 1 −1  z  0     i +j − 2k and u3 = j + k (two independent vectors) Eigenvectors are u 2 = c) The line going through O with direction vector u1 is a line of invariant points. The plane going through O with direction vectors u 2 and u3 is an invariant plane.

This question produced many more at least partially successful attempts. Nonetheless, sign errors and arithmetical slips abounded and there was, yet again, very little evidence of tracking back to check obviously incorrect working and answers. It was only a very small minority who appreciated that a repeated eigenvalue should lead to an invariant plane, which needs two representative eigenvectors. Uncertainty in parts (b) and (c) often led to marks not exceeding 10 out of the 15 available, even amongst the attempts of the better candidates.

Question 8:

Exam report

−3 8 =−9 + 8 =−1 a ) det(M ) = −1 3

In part(a), many candidates realised the significance of both the sign and magnitude of the determinant of the The area is invariant through the tranformation transformation matrix, although there are still far too but the cyclic order of the vertices is reversed. many who try to describe area scale factors without using the key word “area”.  x   x  In part(b), almost all attempts managed to find or verify b) • Any point P( x, px) is invariant, this mean that M   =   the given value of p, but most who did go on to try and  px   px  find a value for q as well simply did what amounted to the same working again. The more successful bids found the  −3 8   x   (−3 + 8 p ) x   x  ⇔ = eigenvalues first and then deduced p and q.    =     −1 3   px   (−1 + 3 p ) x   px  Attempts at part(c)(i) fell almost equally into the three o o We need to have − 3 += 8 p 1 AND − 1 += 3p p categories of correct, incorrect – usually a 45 or 90 rotation matrix – or nothing at all. In part(c)(ii), however, 1 1 = p = AND p (the system is consistent) candidates’ lack of exam-readiness was once again 2 2 markedly to the fore. The order of the two matrices was 1 the right way round only about 50% of the time. More The line of invariant points has equation y = x 2 significantly the great majority decided that the best way • The line y = qx is an invariant line.This means that for all x: to find a matrix  a b  such that   c d   x   x'

M   = =  with y ' qx '.  qx   y '   x   (−3 + 8q ) x   x '  M   = =     qx   (−1 + 3q ) x   y ' 

 0.6 0.8   a b   −3 8    =  (either way round on  0.8 −0.6   c d   −1 3 

the LHS) was to multiply out and solve four simultaneous equations in the four unknowns a, b, c, d rather than premutliply by an inverse matrix. Needless to say, most such attempts went wrong somewhere with the accompanying arithmetic and/or algebra, often due to pre-multiplying on one side of the matrix equation whilst post-multiplying on the other. Even more disappointing was the almost total lack of success of those attempts which did try to employ the inverse of a reflection matrix – candidates should have realised that such a matrix is self-inverse. In describing the shear in detail few candidates were very clear as to what was wanted. Even amongst those who reached the end of the question still with the opportunity to offer an answer, most didn’t. Rather bizarrely, many offerings from candidates who did state something opted for a shear either parallel to one of the coordinate axes or parallel to (their) y = qx, despite the rather obvious flagging of y = 1 x at several stages of the question. 2

y ' = qx ' gives (−1 + 3q ) x = q (−3 + 8q ) x 3q − 1 =−3q + 8q 2 ⇔ 8q 2 − 6q + 1 =0 1 ⇔ (4q − 1)(2q + 1) = 0 ⇔ q = or q = 4 1 c) i )= y = x Tanθ x 2 2 = = −1 θ −1 Cos 2θ 2Cos 2= 1 + Tan 2θ 4 Sin 2θ =− 1 Cos 2θ = 5 The matrix R, reflection in the line y = 3  cos 2θ sin 2θ   5 R =    sin 2θ − cos 2θ   4  5 = ii ) M RS and because R −1

1 = p 2

3 5

1 x, is 2

4  5   3 −  5 R (reflection)

1  3 4   −3 8  1  −13 36  then RM = S ⇔ S =   =   5  4 −3   −1 3  5  −9 23  1 S represents a shear parallel to the line y = x 2 13 9 which maps (1,0) onto ( − , − ) 5 5

Grade Mark

Max 75

A* 63

Grade boundaries A 55

B 47

C 40

D 33

E 26