P10.4 Position vectors & distance

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

Examiner's comment

In part (a), a good proportion of candidates arrived at the correct value for OB as 40 or 2 10 , either in part (a) or (b). Some candidates mistakenly found OB as OA AB - rather than OA AB + whilst others found OB but not its magnitude, although they were allowed recovery if they subsequently found the magnitude in part (b). A few just found AB in this part and others slipped up on one of the components of OB .

Common mistakes

• OB = OA + AB = −2i − 6j (not OA − AB)
• exact magnitude √40 = 2√10
• stopping at the vector OB or finding the magnitude of AB

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

Examiner's comment

In part (b), many drew a diagram to help and there were many correct answers. Those who did not draw a diagram found this more challenging. The vast majority achieved at least one mark for finding the length of OA or OB . Most understood the need to use the cosine rule, and this was often the rearranged form with cos(OAB) as the subject. A surprising number of candidates found the wrong angle of the triangle by substituting their vector lengths into the wrong places in the cosine rule formula, especially those without a diagram. Others made slips in the rearranging. Some candidates incorrectly assumed that the triangle was right-angled and attempted to find angle OAB using trigonometry in a right-angled triangle. For those with a good diagram, the 1 tan- approach worked well, and many successful strategies were seen. A few students used the scalar product approach, but again those without a diagram often used incorrect vectors and found the wrong angle.

Common mistakes

• OA = √73, AB = √29, OB = √40
• cosine rule with √40 opposite angle A gives 47.6°
• lengths in the wrong places find the wrong angle
• do not assume a right angle (sin⁻¹(√40/√73) = 47.75° is wrong)
• draw a diagram

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. In part (a), most candidates knew they needed to subtract the vectors, there were a few signs slips when doing this and quite a few candidates subtracted the wrong way round so found QP instead. There were a small but significant number who added the vectors.

Common mistakes

• PQ = q − p = −6i + 3j
• subtracting the wrong way round (QP) or adding the vectors

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. In part (b), quite a few candidates seemed unclear as to how to show 90 PQR = ; some made no attempt whilst others used the given fact within their proof often finding two angles in a triangle and showing that they were complementary, but they had assumed a right angle in order to utilise trigonometry which achieved no marks. The candidates who tended to be most successful at this were the ones who attempted to show the required property using gradients. There were a small number of signs slips and incorrect attempts at gradients but overall, this approach was usually successful. Quite a lot attempted to use Pythagoras’ theorem but there were a significant number who would only find PQ and QR and either did not attempt to find PR or had a poor attempt at finding it. It is of note that for those who successfully used Pythagoras’ theorem to complete the proof, diagrams proved useful. There were a small number who attempted the dot product or cosine rule and usually these attempts were good enough to achieve the marks. It was disappointing that the most common reasons for the M1A0 score were due to incorrect/incomplete solutions or simply a lack of conclusion.

Common mistakes

• gradients −1/2 and 2 multiply to −1
• or PQ² + QR² = PR² with PR found
• must conclude angle PQR = 90°
• assuming the right angle to use trigonometry scores 0

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

Examiner's comment

This question required problem solving skills and focused on coordinate geometry and circles. Part (b) required candidates to find the equation of the circle. Most candidates found both values of k but ignored the information in the question that k was a positive constant. Those who did appreciate this and used 5 k = went on with good success to find the coordinates of P, giving them the centre of the circle and Q which allowed them to calculate the radius and complete the question. This part was more often successfully answered than the “show” part (a), which showed good accessibility, as well as good exam technique by many candidates to still attempt a later part, even if they were unable to successfully answer the earlier part.

Common mistakes

• k = 5 only (k positive, reject −2)
• centre (−1, 13)
• r² = PQ² = 4² + 2² = 20
• (x + 1)² + (y − 13)² = 20

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. 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.

Common mistakes

• AB = b − a = 3i + 9j + 3k (subtract, not add)
• a vector, not coordinates (3, 9, 3)

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. In part (a), most candidates successfully reached 18 12 - + i j although a few subtracted the vectors the wrong way round or added the position vectors of A and B instead.

Common mistakes

• AB = OB − OA = −18i + 12j
• subtracting the wrong way round or adding

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 (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.

Common mistakes

• AD = d − a = 10i + 24j, BC = 50i + 120j (subtract, not add, position vectors)
• BC = 5AD so parallel, with a conclusion
• equal magnitudes do not show parallel

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. In part (a), errors were either adding the given vectors, subtracting the incorrect way round, or arithmetic errors involving negative numbers. Many candidates chose to use PQ PR - + rather than PR PQ - and seemed to have more success with the j component.

Common mistakes

• QR = PR − PQ (not PQ + PR)
• sign slips with the j component
• 10i − 20j

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 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 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 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.

Common mistakes

• AB = b − a (subtract the right way round) = i − 8j + 2k
• coordinates (1, −8, 2) not accepted

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

Examiner's comment

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.

Common mistakes

• PQ = q − p (subtract, do not add, position vectors)
• use PR = 3PQ or q = p + ⅓(r − p)
• careful vector notation (Q is a point, not a vector)

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

Examiner's comment

This question was very accessible to the vast majority of students with many scoring full marks. Errors seen in part (a) included finding the sum of the two vectors instead of the difference, or, finding the difference ‘the wrong way around’. Correct vector notation was seen in all but a few cases with incorrect notation such as (-9i,3j) being penalised.

Common mistakes

• AB = OB − OA = −9i + 3j
• common errors: adding the vectors or OA − OB
• incorrect notation such as (−9i, 3j) penalised

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

Examiner's comment

students scoring the 3 marks. Part (b) required the students to find the equation of the circle C using the information given. Again most students attempted this with ease, finding firstly the coordinates of point P, and then the radius PA, before writing down the equation of the circle. Errors witnessed tended to be arithmetic, although some students complicated the question by attempting to solve 2 2 simultaneously in an attempt to 2 and 7 5 x y r 2 1 y x find r. There were numerous ways to find the value of k in part (c). One of the best involved

Common mistakes

• solve y = 2x + 1 with 2y + x = 17 to get P(3, 7)
• r² = PA² = 20
• equation (x − 7)² + (y − 5)² = 20, not (√20)²

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. 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

Common mistakes

• AC = OC − OA
• (a − 2)² + 2² + 2² = 16 is already completed-square form, (a − 2)² = 8
• reject 2 + 2√2 because a < 0, answer a = 2 − 2√2

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

Examiner's comment

The vast majority of students recognised the requirement to integrate to find the position vector and subsequently to substitute the given values of t. Occasionally, these were substituted straight into the velocity vector but such instances were rare and much correct working was seen. Most proceeded to find the magnitude of the vector joining the two points to give the exact distance as a surd as required; only a very small minority went straight to a rounded decimal value.

Common mistakes

• integrate v to get r = 2t^½ i − 2t² j (+ C)
• positions at t = 1 and t = 4, not substituting into v
• exact distance √904 = 2√226, not a rounded decimal