P10.5 Vector problems

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested learners’ knowledge of vectors and proof of the properties of a triangle. It should be noted that in part (b) that, despite it being possible to gain full credit by finding the length of vectors and Pythagoras’ Theorem in a few lines, many learners chose to use more involved methods. This often led to them not gaining full marks due to errors in their working. Part (b) was well answered by most learners. Many responses lacked clarity when attempting to prove that the triangle was both isosceles and right-angled. In demonstrating that the triangle is isosceles, some calculated the magnitudes of the sides correctly but did not gain the accuracy mark as they just stated it was isosceles without reference two sides being equal. It seemed that learners found it harder to prove the triangle was right-angled. Most successful responses used Pythagoras’ Theorem to show that the triangle was right-angled and most of these responses gained full marks. When learners used the scalar dot product method, many of them did not show the calculation and therefore did not gain credit, whilst others were not very successful in showing the triangle was right-angled, with some learners incorrectly using it to show a perpendicular gradient of 1 - instead of 0. The least successful responses tended to involve SOHCAHTOA, mainly because this approach was more involved and thus required a fuller explanation, which was not always in evidence. As this was a ‘show’ question, a conclusion (either overall or separate for each property) and an assumption that the triangle was-right angled was required to score full marks. There were lots of incorrect attempts at the cosine rule, which learners either could not remember correctly, or did not rearrange correctly. A significant minority of responses attempted to show the right-angled property by splitting the triangle in half to create two right angled triangles and then used trigonometry. 15

Common mistakes

• magnitudes PQ = QR = √72 = 6√2 (state that two sides are equal) and PR = 12 with 72 + 72 = 144 (Pythagoras) or scalar product = 0 shown
• conclusion for both properties needed
• longer cosine-rule or trigonometry methods led to errors

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question about vectors proved to be a good differentiator between learners. Many were only able to make progress in part (a), whilst learners who had a good knowledge of the techniques required progressed through part (b) as well. Most learners made some progress, but also found this question to be challenging. Throughout, responses saw varied methods, and many made effective use of the given diagram to help answer the question. In general, those that wrote the route down using good vector notation fared better than those who did not in both parts of the question. In part (a), most successful responses attempted to use AD or DA , often finding an equation such as ( ) 4 15 5 2 3 k λ - + = - + + a b a b a b or ( ) 6 3 AD k = + - a b gaining the first mark. Some responses worked with the components of a and b from the start, rather than writing down expressions with vectors, but often produced correct work. A common mistake, however, was to incorrectly use AD AB DB = + and responses doing this made no progress in part (a). Those who had made a successful start were then often able to equate the a and b components of their expression to produce two simultaneous equations and used the a component to establish a value of their parameter λ, and the b component to find k. Others simply compared ( ) 6 3 k + - a b and 15a – 5b by inspection and were able to find the correct scale factor between them using it to find k with sufficient working. Regardless of the method used, some learners with incorrect work often wrote down the given answer, k = 5, without realising that their work did not lead to this value.

Common mistakes

• AD = AB − DB = 6a + (3 − k)b (not AB + DB)
• AD parallel to BC: compare 6a + (3 − k)b with 15a − 5b (scale factor 2.5)
• writing k = 5 from wrong work

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question about vectors proved to be a good differentiator between learners. Many were only able to make progress in part (a), whilst learners who had a good knowledge of the techniques required progressed through part (b) as well. Most learners made some progress, but also found this question to be challenging. Throughout, responses saw varied methods, and many made effective use of the given diagram to help answer the question. In general, those that wrote the route down using good vector notation fared better than those who did not in both parts of the question. Learners found part (b) of the question significantly harder. Many responses earned the first mark for identifying that 1 , 5 BN BC = and this was often the only mark scored in this question. A common error, however, was to misinterpret the ratio BN : NC, and to incorrectly use 1 4 BN BC = . Most learners who made progress found two expressions for one of the vectors BX ± , AX ± , NX ± or DX ± , one using a multiple of AN and one using a multiple of DB . Where responses found a second vector with a different parameter, they were typically successful in solving simultaneous equations and identifying the correct ratio. Common errors included: only finding one expression or using the same vector ( AN or DB ) in both of their expressions or using the same parameter in both expressions. There were other strategies employed, but invariably they had the same issues as the strategy covered in the main scheme. A neat approach was to establish that triangles AXD and NXB were similar and since AD = 2BN it followed immediately that BX : XD = 1 : 2. There were some responses that attempted to use a similar strategy, but often the work was not detailed enough, simply stating that AD and BN were parallel without further justification that they understood the triangles were similar.

Common mistakes

• BN = ⅕BC = 3a − b (not ¼BC)
• two expressions for BX with different parameters (λ along BD, μ along AN) then compare coefficients
• BX : XD = 1 : 2
• similar triangles AXD and NXB (AD = 2BN) need justifying

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall, this question proved fairly tricky with lots of candidates losing marks throughout. Part (c) proved quite tricky for a lot of candidates many of whom did not seem to have fully understood the shape with which they were working. Those who were successful usually had a good diagram identifying it as a trapezium and were able to find the area. There were a number of attempts at splitting the shape up into a rectangle and triangle or 2 triangles, but often there was a misunderstanding about which lengths were which and how they should be combined to find the area. The most common error in the rectangle and triangle approach was that the candidates failed to spot that the triangle was a right-angled triangle so could use QP as the ‘height.’ The two triangles approach would also often have the same issue. There were quite a few attempts that treated the shape either as just a rectangle or just a triangle. It was pleasing however to see that whilst a large number of candidates were unsure of how to proceed to find the area many did at least acknowledge the need to attempt the lengths of various vectors, which generally resulted in two marks being awarded.

Common mistakes

• PQRS is a trapezium (PS parallel to QR) with height PQ = 3√5
• QR = 9√5, PS = 27√5
• area = ½(9√5 + 27√5)(3√5) = 270
• a diagram helps
• treating the shape as a rectangle or parallelogram
• not spotting the right angle

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be a differentiator between candidates. Some were unable to make any meaningful progress while others demonstrated a very intuitive grasp of the topic and were able to produce answers with very little working. In part (b), many candidates did not appreciate the scaffolding in the question and did not use their answer to part (a) to help them with a strategy for part (b). Relatively few candidates used the most straightforward method to find both two possible positions of P, e.g., 2 OP OA AB = + and 2 3 OP OA AB = + . Some candidates were able to find one of the positions (usually 8i+15j+11k) but not the other, with a common misconception being that the second point was on the opposite side of A, attempting 2 OP OA AB = - . There was also some confusion with the ratio with some candidates using 1 3 OP OA AB = + . Many candidates failed to appreciate that there were two points satisfying 2 AP BP = . Relatively few candidates drew a diagram or used the given diagram on the question paper to give them some idea of where P could be. It was much more common for candidates to take an unnecessarily complicated algebraic approach. Many responses calculated |𝐴𝐵⃗⃗⃗⃗⃗ |(= 3 sqrt 11) and progressed no further. Many responses attempted to express 2 AP BP = in terms of variables representing components of an unknown vector, showing a good understanding of how to calculate the magnitude, but struggled to formulate suitable equations. Some candidates used a fully correct method of considering the point P as the general point on the line OP OA AB = + and then setting up the equation 2 AP BP = . These candidates tended to arrive at two final answers for P, having found two log( ) log( ) log( ) ab a b = values for λ, but many had made processing errors and could only gain the method marks as a result. Other candidates set up a general point P in the form xi + yj + zk, then used 2 AP BP = , extracted and solved ( ) ( ) 2 2 2 4 5 , x x - = - ( ) ( ) 2 2 3 4 6 y y + = - and ( ) ( ) 2 2 5 4 8 z z - = - to generate 3 pairs of possible values for the x, y and z components. These candidates then tended to struggle to determine the possible positions of P from their 6 values. Vector notation was very variable with a wide variety of non-standard notations being encountered. It would be helpful for candidates to bear in mind that a four-mark question should not involve pages of work and that there are more efficient strategies. Whatever method was adopted, work was sometimes difficult to follow as there was little explanation and unclear notation used. In part (b), because of the volume of work they were producing, candidates often needed more space to complete their solution. Candidates are strongly advised that if they run out of space, they should use supplementary paper rather than complete their answer within the space of another question.

Common mistakes

• two positions: OA + 2AB = 8i + 15j + 11k and OA + (2/3)AB = 4i + 3j + 7k
• the second point is between A and B, not OA − 2AB
• a diagram helps
• finding the modulus of AB (3√11) alone gets nowhere
• long algebraic methods error-prone

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Given that candidates often find vector problems quite challenging, this question proved to be quite accessible, and most candidates made good progress through parts (a) and (b) of this question, with part (c) more mixed. Many candidates did not successfully reach p = 5. Typical errors in (c)(i) were assuming AB = BC , setting the sum of the position vectors of A, B and C equal to zero or assuming that ( ) 18 12 2 k p - + = - + i j i j . A significant number of candidates omitted (c)(ii) after an unsuccessful attempt at (c)(i), and when responses were seen they were often lengthy and involved unnecessary attempts at calculating the areas of the triangles AOC and AOB. A common incorrect answer was 4:9, reached after assuming incorrectly that the two triangles were similar. Some candidates did, however, successfully reach a ratio of 2:3 either after area calculations, or by, for example, examining the relative lengths of AB and AC. Part (ii) did not rely on an answer to part (i) and so a reasonable number of candidates scored the final mark having been unable to find p, with some doing so very concisely. It was notable that many of the concise solutions to both parts came from a simple, well-constructed and clear diagram.

Common mistakes

• (i) AB parallel to BC (or gradient/line AB) gives p = 5
• wrong assumptions: AB = BC, position vectors summing to zero
• (ii) triangles share height from O so ratio = AC : AB = 2 : 3 (4 : 9 from assuming similarity is wrong)
• a clear diagram helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was relatively well attempted. Most candidates attempted both parts and scored highly overall. In part (b) most candidates correctly identified the lengths of AB, BC, CD and AD, but many left out their calculations by not showing the application of Pythagoras’ Theorem. This is an important step that should be included to fully demonstrate where their answers came from. There was a misunderstanding by some students who thought they could add the 4 vectors of the sides together to calculate a total vector and work out the length of the resulting vector. A few instead found the lengths of the position vectors and added them. Some also assumed that there were two pairs of equal length lines and/or two pairs of parallel lines and just used the two lengths from part (a) then doubled them. Despite this, candidates often achieved the first two marks. The calculation of the average speed was also done well by many candidates. Although several candidates based their calculation on a single lap and arrived at half the correct answer, the majority completed it correctly and clearly showed division by a fraction equivalent to 5 60 (most commonly 1 12 ) or multiplication by its reciprocal. Candidates should be aware of the general instruction in the rubric to give non-exact answers to 3 significant figures; it was common to see 7 or 7.0 instead of the correct final answer.

Common mistakes

• lengths 20, 130, 28√17 ≈ 115.5, 26 by Pythagoras (show it)
• 2 laps = 583 m in 5 min
• speed 6.99 km/h
• one lap only gives half
• adding the vectors or using lengths of position vectors
• assuming opposite sides equal

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many candidates tackled this question well, scoring full marks. Only a few made either no attempt or scored 0 for their efforts. The use of diagrams was widespread and seemed to aid those candidates in parts (a) and (c) to clarify their work. Part (c) did provide problems for a number of candidates, although a pleasing number provided full solutions which were often concise. Some continued down the line of attempting to calculate magnitudes using the answer to (b) and magnitudes of the given vectors, whilst others failed to find the appropriate proportion of QR (usually using 2 3 or 1 3 ). The other main error seen was candidates failing to take proper account of the direction of RS . Others just found the appropriate proportion of QR but did not provide a full method and scored no marks as a result.

Common mistakes

• PS = PQ + (3/5)QR (not 2/3 or 1/3 of QR)
• take care with the direction of RQ if going via R
• working with magnitudes is wrong
• 9i − 7j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This vector question was set within the context of a rhombus. Surprisingly, many candidates were unable to use (or remember) the basic geometric of a rhombus and over 30% did not manage to make any progress. A vast array of approaches were in evidence in part (b) to find the area of the rhombus. The easiest and most direct approach was to find 1 2 PR QS x . A more common method used involved finding the length of one diagonal, followed by the cosine rule to find an internal angle and then use of the formula 'ab sin C '. Unfortunately most attempts using this method only scored 3 marks as it was usual for answers to be given as 22.7 rather than the one required, Ö517. Other methods seen involved use of Heron’s formula, the “shoelace” determinant method, scalar product and vector product….. with many successfully used. Students who added a “narrative” – so explaining what they attempted to do – were in a better position to gain credit if they suffered a processing error early in their solution. This has to be considered good practice. The most common incorrect solution, seen many times over was in stating area = 29 29 29 PQ PS x = x =

Common mistakes

• diagonals PR = PQ + QR = 7i + 3j − 6k, QS = −PQ + QR = 3i − 3j + 2k
• area = ½ × PR × QS = √517 (exact)
• the cosine-rule/½ab sin C route often gave 22.7 not √517
• over 30% made no progress

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was a good differentiator, with the majority of candidates gaining some credit, but only the strongest achieving full marks. Overall, candidates often demonstrated good reasoning skills and were rewarded with good marks on this question but a significant number of candidates struggled with this topic. Throughout, methods of solution were many and varied, and often involved non vector approaches that were sometimes difficult to follow. In part (a) candidates using a vector method often attempted to subtract the given position vectors to form at least two of the vectors ±𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗, ±𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ or ± 𝐵𝐵𝐵𝐵 ሬሬሬሬሬ⃗ although errors in their components were common. Candidates unfortunately often misread 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗ as ൭ 4 0 6 ൱ or ൭ 4 6 0 ൱. This was as far as many candidates reached, but some were able to successfully complete this approach by establishing 𝐵𝐵𝐵𝐵 ሬሬሬሬሬ⃗= 4𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗, 𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗= 5𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ or 5𝐵𝐵𝐶𝐶 ሬሬሬሬሬሬሬሬ⃗= 4𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ and reached p = 32. Some candidates incorrectly used methods like 𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗= 𝐴𝐴𝐴𝐴 ሬሬሬሬሬ⃗ or assumed that the original position vectors were parallel. Candidates attempting to work out the magnitudes of vectors usually gained no marks. Others who made progress with vector methods, established the vector equation of the straight line through A, B and C and used this to find p. Candidates often used non vector methods as well. A common approach was to ignore one component and compare ‘gradients’ using the other two e.g. 4-(-3) 0-4 = 𝑝𝑝-4 -16-0 . This was judged to be an acceptable approach, although errors with the substitution of values were common. Some even established an ‘equation’ such as 𝑦𝑦= - 7 4 𝑥𝑥+ 4 and then substituted in 𝑥𝑥= -16 to find p, which, again, was acceptable, but it was a shame that so many candidates resorted to such methods and seemed to lack confidence in their ability to construct a solution using vectors. In some cases where the first method mark was awarded, candidates went on to an incorrect equation such as 𝑝𝑝= 4 x 7 and were much less likely to be awarded the second method mark, as it was often unclear which vector they were working with for this step. Many fully correct answers for (a) were seen, sometimes with minimal working. It should be noted here that candidates should try to make their intention clear, as there were many occasions where examiners struggled to follow the structure of candidates’ work and were unable to award marks as a result.

Common mistakes

• form two of AB, AC, BC by subtraction (OB = 4j + 6k, no i component)
• parallel vectors are multiples: AC = 5AB gives p + 3 = 35, p = 32
• magnitudes are no help

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was a good differentiator, with the majority of candidates gaining some credit, but only the strongest achieving full marks. Overall, candidates often demonstrated good reasoning skills and were rewarded with good marks on this question but a significant number of candidates struggled with this topic. Throughout, methods of solution were many and varied, and often involved non vector approaches that were sometimes difficult to follow. Part (b) was found to be challenging by most candidates, with few gaining any marks and with many not attempting it. Many candidates again worked with magnitudes of vectors and made no progress. Those candidates who did gain credit usually used the fact that 𝑘𝑘𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗-𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗= ൭ 16 4𝑘𝑘-𝑝𝑝 6𝑘𝑘-10 ൱ was parallel to 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗ to find 𝑘𝑘= 5 giving 𝑂𝑂𝑂𝑂 ሬሬሬሬሬሬ⃗= ൭ 0 20 30 ൱. Many candidates failed to realise that 𝑂𝑂𝑂𝑂 ሬሬሬሬሬሬ⃗= 𝑘𝑘൭ 0 4 6 ൱, and made no progress, whilst others incorrectly assumed that ൭ 16 4𝑘𝑘-𝑝𝑝 6𝑘𝑘-10 ൱= 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗. Candidates often drew diagrams to visualise the problem but full marks in this part were rarely awarded. A significant number of candidates who had found the correct vector, failed to find its magnitude.

Common mistakes

• OD = λOB and CD = OD − OC parallel to OA gives λ = 5, OD = 20j + 30k
• modulus 10√13 (fully simplified)
• CD is not equal to OA
• many stopped at the vector

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question using vectors was generally well answered with most candidates scoring more than half of the marks available. In part (a) candidates seemed unsure how to go about proving that the stone passed through the origin. The most successful attempts typically involved finding the equation of the line for the stone which was 5 12 y x = and showed that the origin was on this line or stated that this linear equation would pass through the origin. Other attempts which candidates proceeded with involved either comparing the position vectors and showing that one was a multiple of the other or showing that the gradients of the position vectors were the same. Plotting the points on a graph was not an acceptable method to prove.

Common mistakes

• show AO and OB are parallel, e.g. −24i − 10j = −2(12i + 5j), and conclude that as the stone moves in a straight line it passes through O
• or show the line y = (5/12)x passes through the origin
• plotting points is not a proof

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question using vectors was generally well answered with most candidates scoring more than half of the marks available. In part (b) most candidates were able to find the vector AB and proceeded to find the magnitude of their vector. Some forgot to then find the speed by dividing by 9, whilst others omitted or gave the wrong unit, or just gave the generic units per second as their unit for speed.

Common mistakes

• AB = √(36² + 15²) = 39 m
• divide by the 4 seconds: speed 9.75 m s⁻¹
• units required (not "units per second")
• forgetting to divide by the time

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates found the first part of this question challenging, but they were more successful with the second part, with many fully correct responses. There were a number of common mistakes in candidates’ approaches to part (a). Firstly, many candidates tried to find the resultant vector by adding the two given vectors, rather than subtracting them. Secondly, candidates seemed unclear on exactly what angle the question was looking for, with lots of time being spent on finding various different (but unfortunately not relevant) angles in their diagrams. Finally, some candidates took a rather long-winded 4 approach to find the correct angles (through use of the cosine rule to find one part of the required answer, then additional work using tan) as they had not spotted the more straightforward use of 7 arctan 3 . Even those who found a relevant angle were often unable to proceed to find a bearing.

Common mistakes

• displacement is (−3i − 5j) − (4i − 2j) = −7i − 3j (subtract, not add)
• find the right angle (arctan 7/3) and convert to a bearing 180° + 66.8° = 246.8°

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates found the first part of this question challenging, but they were more successful with the second part, with many fully correct responses. Part (b) was generally answered successfully, with a few candidates mistakenly stating that 2 hours and 45 minutes was equal to 2.45 hours, rather than 2.75 hours.

Common mistakes

• distance √58 km
• time 2 h 45 min = 2.75 h, not 2.45 h
• speed ≈ 2.77 km/h

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In (b) those who gained both marks were able to compare the two vectors AB and OC to show that they were parallel and conclude that OABC is a trapezium. Usually candidates made a sketch of a trapezium so they could visualize which vectors or sides they would compare. This is a very good idea and should be encouraged by centres. Marks were lost when • candidates did not give a minimal conclusion after finding OC and AB were parallel. • candidates incorrectly stating that 2 AB OC = x • candidates finding the length of each side OABC rather than their directions

Common mistakes

• OC = 2AB so OC is parallel to AB, hence a trapezium
• a reason and minimal conclusion needed
• finding side lengths instead of directions scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q10 discriminated well between the medium and higher ability candidates, with lower ability candidates struggling to gain any creditable access. Q10 increased in difficulty as the candidates progressed through the question, and only a small proportion of candidates scored full marks. In Q10(a), most candidates followed a correct method to achieve the correct answer 3 1 CM Of the candidates who identified a suitable route from C to M, the . a b 2 2 1 . CM CA AM CA AB most common approach was Less efficient routes such 2 as CM CO OB BM or even CM CO OA AM were also observed. Some candidates struggled to apply the vectors a and b to the routes they had chosen. Incorrect methods included applying CA a (instead of CA a ) or applying AB AB ). b a a b (instead of There were a few candidates who incorrectly deduced the position of C to be in the opposite direction to OA and they were mostly 1 , CM CO OA AM CO OA AB able to apply the approach, to gain some 2 credit. Q10(b) proved to be demanding, with some candidates struggling to understand what the question required them to do, and many seemed to be confused by the introduction of the scalar multiple . Only a minority deduced that CN CM and applied this . ON OC CN deduction to the route Most of these candidates progressed well to the given answer unless they had found an incorrect CM in Q10(a). Many candidates used ON OC CM MB BN or ON OA AB BN and made no creditable progress. These candidates usually produced solutions which led to the elimination of from the a component of the resulting vector for . ON A few candidates used ON OM MN and proceeded to obtain a correct expression 1 1 3 1 ON a b a b (or ), in terms but only a few deduced that this 2 2 2 2 (or ). 1 could lead to the stated solution by setting 1 Q10(c) proved to be challenging with many candidates failing to make any creditable is parallel to OB progress. The crucial step was to recognise that ON and hence is a multiple of b with no a component. Only a few candidates set the a component ON 3 4 , 2 2 0. to zero to give the equation Almost all of these candidates found 3 2 ON b with many finding and giving a correct explanation of the given ratio. 3

Common mistakes

• CM = CA + ½AB with CA = −a (not a) and AB = b − a: −3/2a + ½b, simplified

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q10 discriminated well between the medium and higher ability candidates, with lower ability candidates struggling to gain any creditable access. Q10 increased in difficulty as the candidates progressed through the question, and only a small proportion of candidates scored full marks. In Q10(a), most candidates followed a correct method to achieve the correct answer 3 1 CM Of the candidates who identified a suitable route from C to M, the . a b 2 2 1 . CM CA AM CA AB most common approach was Less efficient routes such 2 as CM CO OB BM or even CM CO OA AM were also observed. Some candidates struggled to apply the vectors a and b to the routes they had chosen. Incorrect methods included applying CA a (instead of CA a ) or applying AB AB ). b a a b (instead of There were a few candidates who incorrectly deduced the position of C to be in the opposite direction to OA and they were mostly 1 , CM CO OA AM CO OA AB able to apply the approach, to gain some 2 credit. Q10(b) proved to be demanding, with some candidates struggling to understand what the question required them to do, and many seemed to be confused by the introduction of the scalar multiple . Only a minority deduced that CN CM and applied this . ON OC CN deduction to the route Most of these candidates progressed well to the given answer unless they had found an incorrect CM in Q10(a). Many candidates used ON OC CM MB BN or ON OA AB BN and made no creditable progress. These candidates usually produced solutions which led to the elimination of from the a component of the resulting vector for . ON A few candidates used ON OM MN and proceeded to obtain a correct expression 1 1 3 1 ON a b a b (or ), in terms but only a few deduced that this 2 2 2 2 (or ). 1 could lead to the stated solution by setting 1 Q10(c) proved to be challenging with many candidates failing to make any creditable is parallel to OB progress. The crucial step was to recognise that ON and hence is a multiple of b with no a component. Only a few candidates set the a component ON 3 4 , 2 2 0. to zero to give the equation Almost all of these candidates found 3 2 ON b with many finding and giving a correct explanation of the given ratio. 3

Common mistakes

• ON = OC + λCM (CN is a multiple of CM)
• routes via M, B or A eliminate λ and get nowhere

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q10 discriminated well between the medium and higher ability candidates, with lower ability candidates struggling to gain any creditable access. Q10 increased in difficulty as the candidates progressed through the question, and only a small proportion of candidates scored full marks. In Q10(a), most candidates followed a correct method to achieve the correct answer 3 1 CM Of the candidates who identified a suitable route from C to M, the . a b 2 2 1 . CM CA AM CA AB most common approach was Less efficient routes such 2 as CM CO OB BM or even CM CO OA AM were also observed. Some candidates struggled to apply the vectors a and b to the routes they had chosen. Incorrect methods included applying CA a (instead of CA a ) or applying AB AB ). b a a b (instead of There were a few candidates who incorrectly deduced the position of C to be in the opposite direction to OA and they were mostly 1 , CM CO OA AM CO OA AB able to apply the approach, to gain some 2 credit. Q10(b) proved to be demanding, with some candidates struggling to understand what the question required them to do, and many seemed to be confused by the introduction of the scalar multiple . Only a minority deduced that CN CM and applied this . ON OC CN deduction to the route Most of these candidates progressed well to the given answer unless they had found an incorrect CM in Q10(a). Many candidates used ON OC CM MB BN or ON OA AB BN and made no creditable progress. These candidates usually produced solutions which led to the elimination of from the a component of the resulting vector for . ON A few candidates used ON OM MN and proceeded to obtain a correct expression 1 1 3 1 ON a b a b (or ), in terms but only a few deduced that this 2 2 2 2 (or ). 1 could lead to the stated solution by setting 1 Q10(c) proved to be challenging with many candidates failing to make any creditable is parallel to OB progress. The crucial step was to recognise that ON and hence is a multiple of b with no a component. Only a few candidates set the a component ON 3 4 , 2 2 0. to zero to give the equation Almost all of these candidates found 3 2 ON b with many finding and giving a correct explanation of the given ratio. 3

Common mistakes

• ON is parallel to OB so the a-component is 0: λ = 4/3, ON = ⅔b, NB = ⅓b

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

find r. There were numerous ways to find the value of k in part (c). One of the best involved using vector geometry to find the coordinates of the point N on C, where meets C. 2 y x k Using the fact that PA enabled students to find N = (11,3) and then k by substituting AN this point into Again simultaneous equation methods were common, some more 2 . y x k successful than other. The most common correct attempt involved solving with 2 y x k 2 2 and then finding where the resulting equation had one root. Care was 7 5 20 x y required with the algebra, especially the squaring of terms, as well as the application of the 2 discriminant This proved to be too great a challenge for many. 4 0. b ac

Common mistakes

• best route: other tangent point N = A + PA vector = (11, 3), then k = −19
• discriminant route 5x² + (4k − 34)x + k² − 10k + 54 = 0 with b² − 4ac = 0 has algebra/squaring errors
• reject k = 1

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a well-answered question with most students scoring at least 4 of the 5 marks available. Part (a) was often completed successfully. Most students found AB and used the given result AB BD to deduce and apply . OD OB AB A few students applied 2 . OD OA AB There were some sign, manipulation and bracketing errors seen in this part and students who used column vectors were generally more successful than those who used the , , i j k notation.

Common mistakes

• OD = OB + BD = OB + AB = 2OB − OA = 6i − 7j + 10k
• sign, bracketing errors with i j k notation (column vectors safer)