Vectors questions

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• Edexcel 9MA0 · June 2024 · Paper 2 · Q7 · 5 marks

  5 marks

  7

  
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  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. Most candidates were successful in part (a) although there were a small number of incorrect responses which had arithmetic errors, used addition of vectors rather than subtraction or had poor notation. 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√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.

  Text version

  A straight line l passes through points A and B. Relative to a fixed origin O, A has position vector 2i - 3j + 5k and B has position vector 5i + 6j + 8k.

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• Edexcel 9MA0 · June 2023 · Paper 1 · Q3 · 3 marks

  3 marks

  3

  
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  This was another short question on two-dimensional vectors with many candidates able to score 2 out of the 3 marks. Part (a) was generally answered accurately by all candidates. Sometimes the mark was lost because the candidate omitted the square root and wrote 25 9 4 OA = + + . Sometimes the notation was not made clear or i, j and k appeared. In part (b), the most common approach was to proceed to 2 18 a  or 2 18 a = , and the majority of candidates were able to achieve this. However, a significant number of candidates did not conclude that 5 a = , with some claiming that e.g. 18 is an integer or missing the requirement that “a” had to be an integer. Some other incorrect answers included 19 a = following 18 a  and the answer 5 a  was not condoned, which lots the final mark for a number of candidates. A common processing error was the simplification of 2 20 a + to 20 a + . Very few candidates used the substitution method and some of those did not reach any conclusion in the end.

  Text version

  Relative to a fixed origin O, point A has position vector 5i + 3j + 2k and point B has position vector 2i + 4j + ak, where a is a positive integer.

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• Edexcel 9MA0 · June 2023 · Paper 2 · Q6 · 6 marks

  6 marks

  6

  
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  This question was relatively well attempted. Most candidates attempted both parts and scored highly overall. In part (a) the majority of candidates correctly identified 10i + 24j, and 50i + 120j. Those who didn’t, often mistakenly added the position vectors instead of subtracting them. Most then went on to show that one was a multiple of the other (or that they were both different multiples of the same vector). However, some wrote the vectors the wrong way round for their multiple, which lost the second mark. Some candidates used other successful methods involving ratios or “gradients”. A common error here was candidates attempting to conclude the vectors were parallel by calculating and using their magnitudes. The most common reason for the loss of a mark in part (a) was with candidates correctly showing that the two vectors were multiples of each other but then not drawing a conclusion afterwards. It is important that candidates state what is shown by their working and provide a conclusion. 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.

  Text version

  Relative to a fixed origin O, A has position vector 12i, B has position vector 16j, C has position vector 50i + 136j, and D has position vector 22i + 24j.

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• Edexcel 9MA0 · June 2022 · Paper 1 · Q9 · 6 marks

  6 marks

  9

  
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  (Mean mark 1.9 out of 6) 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. Part (a) required candidates to show that parallelogram PQRS was a rhombus. Although the requirement to 'show' is becoming more familiar to candidates, a significant number in this case didn't recognise the level of detail expected. So a lack of conclusion, or stating the sides were of length Ö29 without evidence of calculation, led to loss of credit. It was also necessary to conclude that the sides were equal, and not simply find the lengths and leave it to the reader to decide. 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 × . 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 × = × =

  Text version

  PQRS is a parallelogram. The adjacent side vectors are vector PQ=2i+3j-4k and vector QR=5i-2k.

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• Edexcel 9MA0 · June 2022 · Paper 2 · Q13 · 6 marks

  6 marks

  13

  
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  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 × 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. 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.

  Text version

  Relative to a fixed origin O, point A has position vector 4i - 3j + 5k, point B has position vector 4j + 6k, and point C has position vector -16i + pj + 10k, where p is a constant. Points A, B and C lie on a straight line. The line segment OB is extended to a point D so that vector CD is parallel to vector OA.

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• Edexcel 9MA0 · October 2021 · Paper 1 · Q6 · 5 marks

  5 marks

  6

  
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  This was another question where the modal mark was full marks, this time achieved by over 40% of the candidates Part(a) A very significant majority of candidates obtained vector AC correctly. Only a few subtracted rather than adding the given vectors AB and BC. Part (b) The vast majority of candidates made some progress in this part, realising that the lengths of the vectors would be required to solve this problem. A small minority then simply stopped, having done no more than finding the lengths of the sides of the triangle ABC, with some assuming that the triangle was right angled. However most then went on to use the cosine rule to try to show the required result. In almost all cases a correct cosine rule was stated and used, so that following on from a previous accuracy error, the method marks could be awarded. Fully correct calculations were necessary in order to gain the final accuracy mark and very many candidates managed to do this and so gained full marks for this question. A few candidates, presumably those who had studied further maths, used the scalar product. This method would be perfectly acceptable, as long as the vectors are used the correct way around. Many using this method showed that cos ABC = − 9 10 but failed to explain why the answer should be positive.

  Text version

  Triangle ABC has vectors AB=-3i-4j-5k and BC=i+j+4k.

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• Edexcel 9MA0 · October 2020 · Paper 1 · Q3 · 4 marks

  4 marks

  3

  
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  In part (a) most candidates were able to successfully subtract vectors the correct way round gaining both marks. Very few candidates added the two vectors together. 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 = ×   • candidates finding the length of each side OABC rather than their directions

  Text version

  Relative to a fixed origin O, points A, B and C have position vectors 2i+5j-6k, 3i-3j-4k and 2i-16j+4k respectively.

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• Edexcel 9MA0 · October 2020 · Paper 2 · Q2 · 3 marks

  3 marks

  2

  
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  This proved to be a demanding question for many candidates, with false starts and seemingly random statements. Those who were able to make some progress generally gained the initial method mark for expressing one of the line segments in terms of the difference of two position vectors, but many candidates were unable to make further progress. Unfortunately, the addition of two position vectors was a common error. Diagrams helped with some responses, but they frequently depicted OP, OQ and OR as collinear, which tended to cause more confusion. The most common approach amongst those who were able to make further progress were either 𝑃𝑃𝑃𝑃 ሬሬሬሬሬ⃗= 1 3 𝑃𝑃𝑃𝑃 ሬሬሬሬሬ⃗ or 𝑂𝑂𝑂𝑂 ሬሬሬሬሬሬ⃗= 𝑂𝑂𝑂𝑂 ሬሬሬሬሬ⃗+ 1 3 𝑃𝑃𝑃𝑃 ሬሬሬሬሬ⃗. Vector notation was often poor, for example Q being used to denote a vector rather than a point, leading to such confused statements as Q = 1/3 PR.

  Text version

  The points P, Q and R have position vectors p, q and r respectively. The points are collinear and Q lies one-third of the way from P to R.

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• Edexcel 9MA0 · June 2019 · Paper 2 · Q10 · 6 marks

  6 marks

  10

  
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  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 . 2 2 CM   a b Of the candidates who identified a suitable route from C to M, the most common approach was 1 . 2 CM CA AM CA AB     Less efficient routes such 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   a b (instead of ). AB   b a There were a few candidates who incorrectly deduced the position of C to be in the opposite direction to OAand they were mostly able to apply the approach, 1 , 2 CM CO OA AM CO OA AB       to gain some 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 deduction to the route . ON OC CN   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 2 2 2 2 ON            a b a b in terms (or ),   but only a few deduced that this could lead to the stated solution by setting 1     (or 1    ). Q10(c) proved to be challenging with many candidates failing to make any creditable progress. The crucial step was to recognise that ON is parallel to OB and hence is a multiple of b with no a component. Only a few candidates set the a component ON to zero to give the equation 3 2 0. 2    Almost all of these candidates found 4 , 3  with many finding 2 3 ON  b and giving a correct explanation of the given ratio.

  Text version

  In triangle OAB, point C lies on the ray OA beyond A and OC = 2OA. Point M is the midpoint of AB. The straight line through C and M meets OB at N. The position vectors OA = a and OB = b.

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• Edexcel 9MA0 · June 2018 · Paper 2 · Q2 · 5 marks

  5 marks

  2

  
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  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. In part (b), most students found the difference between OC and , OA and applied Pythagoras’ Theorem to 4 AC  to give the correct 2 2 2 2 ( 2) (2) (2) (4) . a     The resulting quadratic equation was generally solved correctly but not always efficiently. Too often, students expanded to form a 3-term quadratic rather than spotting that their quadratic was already in a ‘completed square’ form. Despite the question stating that 0, a  some students gave both solutions 2 2 2 a   as their final answer, while others mistakenly rejected 2 2 2 a   to leave 2 2 2. a  

  Text version

  Relative to a fixed origin O, A has position vector 2i + 3j - 4k, B has position vector 4i - 2j + 3k, and C has position vector ai + 5j - 2k, where a < 0. Point D is such that vector AB = vector BD.

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• Edexcel 9MA0 · sample · Paper 1 · Q7 · 5 marks

  5 marks

  7

  
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  Triangle ABC has vector AB=2i+3j+k and vector BC=i-9j+3k.

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• Edexcel 9MA0 · specimen · Paper 1 · Q4 · 5 marks

  5 marks

  4

  
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  Relative to a fixed origin O, A has position vector i + 7j - 2k, B has position vector 4i + 3j + 3k, and C has position vector 2i + 10j + 9k. ABCD is a parallelogram. The vector AX has the same direction as AB and |AX| = 10sqrt(2).

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