Unit 3C Specialist Mathematics

Transcription

1 Solutions to Unit C Specialist Mathematics by A.J. Sadler Prepared by: Glen Prideaux

2 ii Solutions to A.J. Sadler s

3 Unit C Specialist Mathematics Preface The answers in the Sadler text book sometimes are not enough. For those times when your really need to see a fully worked solution, look here. It is essential that you use this sparingly! You should not look here until you have given your best effort to a problem. Understand the problem here, then go away and do it on your own. Errors If you encounter any discrepancies between the work here and the solutions given in the Sadler text book, it is very likely that the error is mine. I have yet to find any errors in Sadler s solutions. Mine, however, have not been proofread as thoroughly and it is likely that there are errors in this work. Caveat discipulus! iii

4 Contents Chapter Exercise A Exercise B Miscellaneous Exercise Chapter Exercise A Exercise B Exercise C Exercise D Miscellaneous Exercise Chapter Exercise A Exercise B Miscellaneous Exercise Chapter Exercise A Exercise B Exercise C Exercise D Miscellaneous Exercise Chapter Exercise A Miscellaneous Exercise Chapter Exercise A Exercise B Miscellaneous Exercise Chapter Exercise 7A Exercise 7B Exercise 7C Exercise 7D Exercise 7E Exercise 7F Exercise 7G Miscellaneous Exercise Chapter Exercise 8A Exercise 8B Exercise 8C Exercise 8D Exercise 8E Miscellaneous Exercise iv

5 Unit C Specialist Mathematics CHAPTER Chapter Exercise A. a AB r + r r r cosθ θ + cos7 cos 9 If we had recognised before we began that A and B form a right angle we could have simplified this by using Pythagoras rather than the cosine rule. b AB r + r r r cosθ θ + cos cos c AB r + r r r cosθ θ cos 7 8 cos 8 + d AB r + r r r cosθ θ + cos cos. a If we recognise that θ θ is a right angle there is no need to use the cosine rule as Pythagoras will do. PQ r + r + b Here θ θ π π π which also represents a right angle, so we can use Pythagoras again. PQ r + r + c PQ r + r r r cosθ θ + cos π 7π cos 8π cos π + 7 d PQ r + r r r cosθ θ + cos π π cos π cosπ If we had recognised before we began that P and Q formed a straight angle we could simply have done PQ r + r + 7. AB ra + r B r Ar B cosθ B θ A + cos π π cos π 7 AC ra + r C r Ar C cosθ A θ C + cos π π cos π AC exceeds AB by 7.

6 Exercise B Solutions to A.J. Sadler s. a Cartesian coordinates of A: cos, sin Cartesian coordinates of B: cos, sin AB cos cos + sin sin cos cos cos + 9 cos + sin sin sin + 9 sin cos cos + sin sin + 9 cos cos + 7 b AB r + r r r cosθ θ + cos cos + 7. a Cartesian coordinates of P: cos π, sin π Cartesian coordinates of Q: cos 9π, sin 9π cos 9π 9π, sin cos π 9π cos PQ + sin π cos π 9π cos + sin π 9π + sin 8 cos π cos π +8 sin π + sin π sin 9π 8 cos π cos π + 9π cos π + 9π cos π cos 7π cos 9π sin 9π b PQ r + r r r cosθ θ + cos 9π + sin 9π 9π cos sin π 9π sin + + cos π 9π 8 + cos π + 9π cos π cos 7π Exercise B Questions to need no working. Refer to the solutions in Sadler.. a For the given point, π, r π θ π r θ k For point A θ π r θ π x π y Point A has Cartesian coordinates π,.

7 Unit C Specialist Mathematics Miscellaneous Exercise b For the given point π,, For point B r π θ π r θ k θ π r θ π x y π Point B has Cartesian coordinates, π. c For the given point,, r θ π r π/ θ π θ k π For point C θ π r π θ x y Point C has Cartesian coordinates,. d For the given point,, For point D r θ π r π θ k π θ π r π θ x y Point D has Cartesian coordinates,. Note regarding the extension work: these polar graphs are collectively called petal plots for obvious reasons. Miscellaneous Exercise No working required.. a a i j e cos θ a b a b b a b b i + j i + j i + j θ cos - 8 c di 9j a d d ± d a b +

8 Miscellaneous Exercise Solutions to A.J. Sadler s. p q ai + j i bj a ab b reject a since p is non zero. a + + b a + b + a a ± i bj ci + dj i j ci + dj i j ci dj i j c 7 d ei + fj ki j f e e + f c + d e c + d 7 + e e f rejecting e, f because s must be in the same direction as q which has a positive i component. 8. z z + i a + bi a bi + i a + bi + i a b 9. sin 7 sin + sin cos os sin + +. a + bi + i + i a + ai bi b + i a b + a bi + i This gives the following pair of equations by equating real and imaginary components: a b a b Solving simultaneously gives a b. a OP 7 + m z i 7. b i + j + 9 m c 7i j + i + j 8i + j m d 8i + j m OP OA + AB OA + OB OA OA + OB 7i + 7j + 8i + j + i + + j i + j. a + bi + i a b + abi + i a b ab b a a a a a a + a a a + 9 a ± b ± z ± + i Note that we disregard a + 9 as leading to any solutions because a must be real.

9 Unit C Specialist Mathematics Miscellaneous Exercise. a + bi + i a b + abi + i a b ab b a a a a a a a a + a 9 a ± b ± + i ± + i Note that the square root of a real number is defined uniquely to mean the positive square root. No such unique definition exists for the square root of complex numbers.. sin x + π sin x + π x + π π, π, 9π, π. y x when x x π, 7π, π, π y The equation of the tangent line is y y mx x y 9 x y x + 9 y x +. θ π θ π Polar coordinates: π, π. Assuming that we restrict r and θ to positive values as usual. The Cartesian coordinates of this point are π cos π, π sin π π, π 7. It may be simplest to first express r and s in terms of p and q: q q -p r -q p p s p then use these to work out the rest of the vectors: t r + s p + q + p q p q u t p + q v.r + s. p + q + p q.p +.q w.r + s. p + q + p q.p +.q x r +.s p + q +.p q p +.q y r s p + q p q p + q. y x x + + x x A t x x + + x 8x x + 8x x + when x y θ t The equation of the tangent line is y y mx x y x y x + y x sin θ t t θ sin - 8 B

10 CHAPTER Solutions to A.J. Sadler s Chapter Exercise A. a z + b z + c z + d z + e z + f z +. Process: Determine which quadrant z lies in on the Argand diagram by examining the sign of the real and imaginary parts Use inverse tangent to determine the angle made with the real axis on an argand diagram. Combine these to obtain the principal argument. a st Quadrant: tan θ arg z π b th Quadrant: tan θ arg z π c nd Quadrant: tan θ d rd Quadrant: arg z π tan θ arg z π e nd Quadrant: tan θ arg z π f th Quadrant: tan θ arg z π. There seems little point in showing working for these problems. z, z and z : subtract π to obtain the principal argument. z : add π to obtain the principal argument. z to z : read r from the magnitude shown on the diagram, and determine θ as the directed angle measured anticlockwise from the positive real axis.. There seems little point in showing working for these problems. Determine r by Pythagoras as for question. Determine θ as for question.. There seems little point in showing working for these problems. You should be able to do them in a single step. Determine the real and imaginary components by evaluating the trigonometric expressions exactly and then multiplying by r. Exercise B. Read r from the magnitude of z and θ from the directed angle converted to radians measured anticlockwise from the positive real axis. 9 No working needed. You should be able to do these questions in a single step, converting to radians and adding or subtracting a multiple of π where necessary.. 7 cis π 7 cos π + i sin π 7 + i 7i

11 Unit C Specialist Mathematics. cis π cos π + i sin π i i. cis π cos π + i sin π + i Exercise C 8 Do these questions in the same way as question of exercise A i.e. using Pythagoras to find r and inverse tangent to determine θ, adding or subtracting π for z in the second or third quadrant respectively.. a The conjugate is a reflection in the real axis: Im w. cis π cos π + i sin π + i With a little practice you may find you can do questions like these by simply sketching or visualising an Argand diagram.. cis π cos π + i sin π + i. cis π cos π + i sin π w b z r cis α w r cis β. No working needed.. No working needed. z z Re + i. cis π cos π + i sin π 7. cis π i cis π cos π + i sin π + i. First subtract to get the principal argument, then multiply this by to obtain the conjugate.. First add to get the principal argument, then multiply this by to obtain the conjugate. 7. No working needed. 8. No working needed. 9. No working needed.. First subtract π to get the principal argument, then multiply this by to obtain the conjugate. Exercise C. zw + i i i + i + + i. zw cis + 9 cis 7 9 cis 9. zw + i + i + i i 7 + i. zw cis + cis 8. zw cis 8 9 cis π. zw cis + π cis 7π 7

12 Exercise C π 7. zw cis π 8 cis π Solutions to A.J. Sadler s. From z, zw has been rotated 8 and scaled by, so w cis 8. zw cos + + i sin + cos + i sin 9. zw cos i sin cos + i sin cos + i sin z w i i i + i i + i 8 + i 9i i. +.i z + i w + i i i. +.i z w 8 cis cis z w cis cis z w cis 8 cis cis + cis z w π cis π cis π z. w π cis π cis π cos z w. π π cos π + i sin π π + i sin π z w cos + i sin. cos + i sin 9. From z, zw has been rotated and scaled by, so w cis. From z, zw has been rotated 9 and scaled by, so w cis 9. From z, zw has been rotated 8 + and scaled by, so w cis. From z, zw has been rotated and scaled by, so w cis. From z, zw has been rotated and scaled by, so w cis. From z, z w has been rotated and scaled by, so w cis. From z, z w has been rotated 8 and scaled by, so w cis 8 7. From z, z w has been rotated and scaled by, so w cis 8. a z cis cis b w cis cis c zw cis + cis 7 d wz zw cis 7 e iz cis + 9 cis f iw cis + 9 cis g h w z cis cis z cis cis 9. a zw 8 cis π + π cis 7π cis 7π 8

13 Unit C Specialist Mathematics Exercise D b wz zw cis 7π w c z 8 cisπ π. cis π z d w 8 cisπ π cis π e z 8 cis π f w cis π g h z 8 cis. cis π π i w π cis π. cis π Exercise D - No working required.. First rewrite as z : z i 7. Think of this as points equidistant between 8+ i and + i. Plot those points on the Argand diagram, and the locus is the perpendicular bisector of the line segment between them. 8. Think of this as points equidistant between i and i. Plot those points on the Argand diagram, and the locus is the perpendicular bisector of the line segment between them. 9. Since x Rez, z : Rez is the vertical line x.. Since y Imz, z : Imz is the horizontal line y.. θ arg z tan θ y x tan π y x y x : x We need to specify x to exclude the th quadrant.. θ arg z tan θ y x tan π y x y x : x We need to specify x to exclude the nd quadrant.. Simply substitute x for Rez and y for Imz to give x + y.. z x + y x + y. This is a circular region centred at, having radius. You should be able to produce a Cartesian equation from this description, or you can do it algebraically: z i x + y i x + y x + y 9 You have to be careful with inequalities like this because multiplying both sides of the equation by a negative changes the direction of the inequality. It can be particularly tricky when you square both sides of an inequation as we have in the last step here: you have to make sure you know with certainty that both sides are positive. For example, a < is not the same as a <. We are safe here, however, since the square root on the left hand side is always positive and so is on the right hand side.. This is a circle radius centred at,. z + i x + y i x + y 9

14 Exercise D Solutions to A.J. Sadler s 7. This is a circle radius centred at,. z + i x + y + i x + y + 8. This is a line defined by the locus of points equidistant from, and,, i.e. the vertical line x. Showing this algebraically: z z x + yi x + yi x + y x + y x x + + y x x + + y x + x + 8x x 9. This is a line defined by the locus of points equidistant from, and, : z i z x + y i x + yi x + y x + y x + y y + x x + + y y + x + x y x y 8. This is a line defined by the locus of points equidistant from, and, : z + i z i x + y i x + y + i x + y x + y + x x + + y y + x 8x + + y + y + x y x y 9. This is a doughnut-shaped region centred at the origin, including all points on or inside the -unit circle and on or outside the -unit circle. For the Cartesian equation, substitute x + y for z.. This is the region in the first quadrant bounded below by the line z π and above by the line z π. The step taken in the last step multiplication by x is only valid because we restrict x. If x was negative, this step would change the direction of the inequalities. An alternative, possibly simpler, approach to this question would be to first observe that the specified region lies above or on the line y x and below or on the line y x, again using tan θ y x to obtain these equations. This is essentially doing the same thing as the above, but it s perhaps a more intuitive way of looking at it.. We have a circle centred at, having a radius of. a The minimum value of Imz is. b The maximum value of Rez is + and the minimum is so the maximum value of Rez is. c The distance between the origin and the centre of the circle is +. The point on the circle nearest the origin is units nearer, so the minimum z is. d The point on the circle furthest from the origin is units further away than the centre, so the maximum z is +. e The locus of z is the reflection in the real axis of the locus of z. The point on this image furthest from the origin is the same distance from the origin as the corresponding point on the original: +.. We have a circle centred at, having a radius of. a The minimum value of Imz is. b The maximum value of Rez is +. c The distance between the origin and the centre of the circle is +. The point on the circle furthest from the origin is units further, so the maximum z is + 7. d The point on the circle nearest the origin is units nearer than the centre, so the minimum z is. e Consider the geometry of the situation: Im π θ π tan π tan θ tan π x y x y x : x Note that the second line above where we take the tangent is only valid because the tangent function is strictly increasing in the first quadrant, so a > b implies tan a > tan b. This is only true for functions that have positive gradient in the given domain. O β α β θ C:, T Re

15 Unit C Specialist Mathematics Miscellaneous Exercise The minimum value of arg z corresponds to the angle for the lower tangent, T. tan α α. f The maximum value of arg z corresponds to the angle for the upper tangent. θ α + β. sin β β. θ α β. Miscellaneous Exercise. a z + w i i b z w + i 7i c zw i + i + 9i 8i i d z i e f 9 i 7 i z w i + i i i + i i 9i 8i + 9 7i 7 i w z + i i + i + i i + i + 8i + 9i i + 7 i. a AB c b AD AB c c DB AB c d DE DB + BE c + c c e OB OA + AB a f OD OA + AD a + c g CE CB + BE a + c h OE OA + AE a + c alternatively OE OC + CE c + a + c a + c. a r tan θ θ π third quadrant i cis π

16 Miscellaneous Exercise b 8 cis π 8 8 cos π i i. a cos π, sin π, + i sin π Solutions to A.J. Sadler s f pa + a pb b + qb qa p + + qa + q + pb p + q p + q p q q q p b cos π, sin π, c cos π π, sin,. z + cis tan - cis π w + cis tan - cis π zw cis cis π cis z w cis π π + π π π. Show that the variable part of each expression is a scalar multiple of that of the other, i.e. i 7j i + 7j. 7. Show that the variable parts are perpendicular vectors, using the scalar product, i.e. i j i + j. 8. a The only scalar multiples of non-parallel, non-zero vectors that equates them is zero, so p q. 9. i + j + λ7i j i j + µi j + 7λ µi µ + λj 7λ µ i λ µ 7j 7λ µ λ µ 7 λ + µ 8. a d x x b c d e d x x λ λ r i + j + 7i j 7i 9j d x 7x + x 7 d x+x+ d x +x+8 x+ You could use the product rule here if you preferred. d x +x x d x + x You could use the quotient rule here if you preferred, but this approach is probably simpler. b p so p, and q. c p + so p, and q so q. f d x + x + x + 8 d pa + b a qb pa a qb b p a q b p q. π θ θ π polar coordinates : π, π e pa + qa + pb qb a + b p + q a p + qb p + q p q p q x π cos π π y π sin π π Cartesian coordinates : π, π

17 Unit C Specialist Mathematics Miscellaneous Exercise. a pi + j + qi j 9i + j p + q 9 p q p q 8 q 9 q p + 9 p p 9i + j a b b pi + j + qi j i 8j p + q p q 8 p q 8 q q p + p p i 8j a + b c pi + j + qi j 7i + j p + q 7 p q p q q q p + 7 p p. 7i + j.a b d pi + j + qi j i + j p + q p q p q 8 q 9 q 7 p + 7 p p. i + j.a 7b. sin A cos A cos A cos A cos A cos A. sin θ cos θ sin θ cos θ cos θ sin θ cos θ sin θ sin θ sin θ sin θ. cos x sin x sin x sin x sin x sin x sin x sin x sin x or sin x sin x. x {, π, π} { π x, π } Solutions are, π, π, π, π.. From the first set of points given we can obtain the Argand diagram: Im - - +i +i Re From the second set we see that the line is the set of points equidistant from + i and a + bi, hence a + bi is the reflection of + i in the line, thus: Im i +i giving us a 8 and b. +i a + bi From the third set we see that the line is the set of points equidistant from + i and c + di, hence c + di is the reflection of + i in the line, thus: -+i Im +i +i c + di Re Re

18 Miscellaneous Exercise Solutions to A.J. Sadler s 7. giving us c and d. From the fourth set we see that the line is the set of points equidistant from +i and e fi, hence e fi is the reflection of + i in the line, thus: Im e fi 9 -+i 8 7 +i +i giving us e and f t 8 i components: 8t + t j components: t 8 t. + t Re Since the value of t. satisfies both i and j components, a collision occurs at :pm +. hours, i.e. at pm. The location of the collision is given by 8 r km 8. a A v B v A v B i + j i + 8j 8i j BA r A r B i + j 9i + j 7i j BP BA + t A v B 7i j + t8i j 7 + 8ti + tj BP Av B 7 + 8ti + tj 8i j t t + t + + t t t. BP i +.j i j km The minimum distance is. km and occurs at :pm. b The position of the ships at time t is: r A t i + j + ti + j + ti + + tj r B t 9i + j + ti + 8j 9 + ti + + 8tj The displacement AB at time t is given by ABt r B t r A t 9 + ti + + 8tj + ti + + tj 7 8ti + + tj d AB d 7 8t + + t d dt d 7 8t t + 8t + + 7t t At the turning point, d dt d t t. d km The minimum distance is. km and occurs at :pm.

19 Unit C Specialist Mathematics CHAPTER Chapter Exercise A. For students with limited experience with hardware, the right hand screw rule explained in Sadler is not a very helpful mnemonic. An alternative method of remembering the relationships between axes in three dimensions is the Right Hand Rule. In this rule, the x, y and z axes correspond to thumb, index finger and middle finger respectively as illustrated below.. a a + b + i + + 8j + k i + j + k b a b i + 8j + k i j + k c a + b + i + + 8j + k 7i + j + k d a + b i + j + k i + 8j + k e a b f b a a b g a h a + b a c + d b c d + c c + d d c + d 7 8 e c d + + f d c c d g c + + h c + d a e f,,,, b e f,,,, c e + f +, +, +,, d e + f +, +, +,, e e f f e f e f 7 g e f e f h e f a AB OB OA i + j + k i + j + k i j + k b BC OC OB i j + k i + j + k 8i j

20 Exercise A c CA OA OC i + j k i j + k 7i + j k d AC AC 7i + j k 7i j + k +. a p + q b q + r c p + q q + r a u b v + + c u v d u v u v cos θ 8 7 cos θ θ cos OA OB 8. cos AOB OA OB AOB cos - 9. cos θ p q p q θ cos - 8 Solutions to A.J. Sadler s. cos θ s t s t θ cos - 7. a Let u be a scalar multiple of r having unit magnitude. r u 7 r i j + k 7 b Let v be a scalar multiple of r having the same magnitude as s. If it s a scalar multiple of r then it s also a scalar multiple of u, so s + v u i j + k 7 c Scale s by the ratio of the magnitudes of r and s. r s s 7 i + k d cos θ r s r s θ cos - 7. a Vectors are scalar multiples of each other the second is double the first so they are parallel. b Not scalar multiples, so not parallel. i + j k i j + k Dot product is not zero, so not perpendicular. c Not scalar multiples, so not parallel.,,,, + 9 Dot product is not zero, so not perpendicular.

21 Unit C Specialist Mathematics Exercise A d Not scalar multiples, so not parallel.,,,, + 9 Dot product is zero, so vectors are perpendicular. e Not scalar multiples, so not parallel. 7 Dot product is zero, so vectors are perpendicular. f Not scalar multiples, so not parallel Dot product is not zero, so vectors are not perpendicular. g Not scalar multiples, so not parallel. 8 + Dot product is zero, so vectors are perpendicular.. F + F + F + i j + + k i + 8j + 9k N F + F + F N. a + b + a b a 7 + a a + b a b b 7 b. b is parallel to a so they are scalar multiples, i.e. b ka. By examining the i and j components we can see that k so p. c is perpendicular to a so they have zero dot product, i.e. 7 + q + + q q d is perpendicular to b, but b is parallel to a so d is perpendicular to a. d a + r + r r Note that even though both are perpendicular to a, c is not parallel to d as they would be in two dimensions. 7. a i j + k + i + j k i + 9k m b i j + k + i + j k j + 7k m. BA OA OB i + j + k i + j k OB OB i + j k i + j + k i 8k The position vector of B is i 8k. c i j + k + i + j k i + 8j + k m i + 8j + k m d i j + k + t i + j k + ti + + tj + tk + t + + t + t + t + + t + t t 9t + t.s ignoring the negative root 7

22 Exercise A 8. A, B and C are collinear if AB k AC. AB i + j k 7i + j i j k AC i k 7i + j i j k AB AC A, B and C are collinear. 9. Let P be the point that divides AB in the ratio :. OP OA + AB AB i j + k i + j k i j + k OP OB + BP OB + AB i j + k + i j + k i j + k AB 9i + j 9k i j + k i + 8j k OP OA + AB i j + k + i + 8j k i j + k + i + j 9k 8i + j k AB k i + j + k i j + k AC i j + k i + j + k i j BC i j + k k i j k AB AC AB BC Solutions to A.J. Sadler s AB BC and ABC is right angled at B.. Let α be the angle a makes with the x axis. cos α a i a i i + j k i + + α cos Let β be the angle a makes with the y axis. cos β a j a j i + j k j β cos -.7 Let γ be the angle a makes with the z axis. cos γ a k a k i + j k k γ cos -. We want the acute angle, so we need the supplementary angle: Vector d: d λa + µb + ηc 7i j + k λ i j + k + µ i + j k + η i j + k Equating i, j and k components gives three equations in three unknowns: λ +µ +η 7 λ +µ η λ µ +η Solving this using the Classpad. On the D tab, tap on to expand to three lines. Tap the same icon a second time. Fill in the 8

23 Unit C Specialist Mathematics three equations. Use x, y, z in place of the Greek letters: d a b Vector e: e λa + µb + ηc i j + 8k λ i j + k + µ i + j k + η i j + k Equating i, j and k components gives three equations in three unknowns:. a AO i + j i j AO + + AE 8k i + j i j + 8k AE AO AE cos OAE AO AE Exercise A λ +µ +η λ +µ η λ µ +η 8 Solving this using the Classpad gives λ, µ, η. e a b Vector f: f λa + µb + ηc j k λ i j + k + µ i + j k + η i j + k Equating i, j and k components gives three equations in three unknowns: λ +µ +η λ +µ η λ µ +η Solving this using the Classpad gives λ, µ, η. f a b. a DC i cm DB i + k cm DI j + k cm DI DB b cos IDB DI DB j + k i + k j + k i + k IDB 8 9 b OAE.8 DB i + j i j 8i + j DB AE DB cos θ AE DB θ a AB BC AC b AB + + BC + + AC + + AB AC so ABC is isosceles. c AC AC + + Alternatively, AC AC AC

24 Exercise B AB AC d cos A AB AC + + A. A + B + C 8 C B isosceles A + B 8 B 8.. B C 7 Bv A v B v A i + j k v B i + j + k Solutions to A.J. Sadler s v B i + j k + i + j + k i + j m/s v B + + m/s 9. If the aircraft are following the same path, their velocity vectors must be scalar multiples. B is behind A but is closing the gap at m/s, so it must be flying m/s faster than A. v A + + m/s v B + 7 m/s v B 7 v A v A 8. Let v A be the velocity of the first bird, and v B that of the second. The apparent velocity of the second from the point of view of the first is i j + k 7i j + k m/s Exercise B. a r i + j k + λ i j + k x + λ b y λ z + λ. Note that there are many possible correct answers to questions like these. This answer is different to Sadler s. Convince yourself that they are both correct. What are some other possible correct answers? a AB i + j + k i + j + k b i j k r i + j + k + λ AB i + j + k + λ i j k x λ y λ z λ. r i j + k i j + k i j + k r i j + k + + r i j + k 9. r r + 9 r. No working required. See Sadler for solution.. No working required. See Sadler for solution. 7. Substitute the given point as r: ai + 7j + k i + bj k + λ i + j + k k components: i components: + λ λ a λ 7

25 Unit C Specialist Mathematics Exercise B j components: 7 b + λ 7 b + b x 8. Substitute r y : z x y z x + y z You should be able to do this by observation in a single step. 9. No working required. Use the inverse of the process used for the previous question.. The line is parallel to i j + k. i j + k i + j k the line is parallel to i + j k i + j k is perpendicular to the plane the line is perpendicular to the plane.. Equate the two expressions for r and simplify: + λ 8 + µ λ µ. Equate corresponding components to obtain three equations: λ + µ λ + µ λ µ Solve the first two of these simultaneously to obtain λ, µ. Substitute these values into the third equation. If it is consistent, then the two lines intersect. is consistent, so the lines intersect. Determine the point of intersection by substituting either λ or µ into its original equation: r + λ λ µ 8 + µ Equating corresponding components: λ + µ λ λ µ 8 Solve the first two of these simultaneously to obtain λ, µ is consistent, so the lines intersect. The position vector of the point of intersection is:. a r i j + k + i + j + k i + j + k λ + λ 8 µ 8 Equating corresponding components: λ µ λ µ λ + 8µ + µ 8 Solve the first two of these simultaneously to obtain λ 8, µ is inconsistent, so the lines do not intersect. b + λ + β 8 8 λ β Equating corresponding components: λ β 8 λ β λ + β Solve the first two of these simultaneously to obtain λ, β. + is consistent, so the lines intersect. The position vector of the point of intersection is: r i + j + 8k i j + k i + j 7k The angle between the lines is given by i j + k i + j k cos θ i j + k i + j k i j + k i + j k The angle between the lines is 9.

26 Exercise B Solutions to A.J. Sadler s. a Choose one component say, i and solve for λ, then confirm that the same value of λ satisfies both of the other components. i components: + λ λ j components: + k components: 7 The same value of λ satisfies all three components, so point A is on line L. b i components: + λ λ 9 Check that is not parallel to r + λ + µ Parametric equations: x +λ+µ y λ z λ+µ. j components: + 9 The same value of λ does not satisfy both i and j components, so point B is not on line L. c Point A: i j + 7k i + j + k + Point B: i + j + k i + j + k d If line L lies on plane Π then every point on L must satisfy the defining equation for Π. + λ + λ + + λ λ which is true for all λ. AB t A v B 8 t 7 8 t t 7 s r r A + tv A + 7 m. c Eliminate λ from and : { x+y 8+µ + y z 7 µ Now eliminate µ and simplify: x+y z x+y z In scalar product form: x y z r + 7. i + j + k + λ i + j k i j k + λ + λ r i + j + k 7 i + j k 9i 8j + k λ 7 8. Let D represent the position of the debris and S the position of the spacecraft, then for a collision to occur DS t debris v spacecraft 7 7 t 9 t 7 7 i : t. j : t. k : t 7 7. The spacecraft and debris will collide at time t. hours.

27 Unit C Specialist Mathematics 9. Position of the fighter at the time of interception is r i + 7j + k + i + 8j i + j + k km Call this point P, and call the initial position of the fighter point A, then AP i + j + k 8i + j + k i + j k km AP tv A i + j k v A v A i + j k 7i + j k km/h. Let P be the point of minimum separation. P B Av B AB r B r A Av B v A v B BP AB + t A v B t t 8 A BP Av B t t t s. BP BP + + 7m Exercise B is normal to plane Π. is normal to plane Π.. is normal to plane Π. Π Π Another approach: proof by contradiction. Suppose Π and Π are not parallel. If that is the case, there exists a line of intersection between the planes, i.e. a set of points r that simultaneously satisfies and Starting with : r r r r r r

28 Exercise B Solutions to A.J. Sadler s substituting which means that our original supposition leads to a contradiction, hence Π and Π are parallel. To find the distance the planes are apart, consider the line r λ We know that this line is perpendicular to both planes, so the distance along this line between the point where it intercepts Π and where it intercepts Π will represent the perpendicular and hence minimum distance between the planes. Call these points P and P. λ λ λ + + λ P λ λ P P P P P The planes are 8 units apart.. a r i j + k + 8i + j + k i j + 8k b When it is due west, the j component will be zero. + t t s i.e. at minute and seconds after pm. c When it is due north, the i component will be zero. + 8t t s i.e. at minutes and seconds after pm. The altitude at that time is given by the k component: + t m d Five minutes is seconds: r i j + k + 8i + j + k i + j + k distance m km e Let P be the position nearest O. OP 8 + t 8 OP + t t t t 8s OP OP m.km

29 Unit C Specialist Mathematics Miscellaneous Exercise Miscellaneous Exercise. a No working needed. Refer to the answer in Sadler.. b No working needed. Refer to the answer in Sadler. c cos θ p q p q θ d cos θ p i p i 7 θ e cos θ q j q j 9 θ + λ λ µ λ + µ λ + µ λ µ λ λ µ + µ + subst. µ 7 We would normally need to confirm that these values work in the third equation i.e. but in this case and are redundant i.e. they can be rearranged to be identical equations so any solution of must also be a solution of. This also means that we didn t really need to find µ. The point of intersection is given by:. The resultant is r i + k + λ i + j k i + k + i + j k i +.j k i + j k + i j + k i + j k This has magnitude + + so a unit vector parallel to the resultant is. fx x + x i + j k lim fx + h fx h h x + h + x + h x + x lim h x + xh + h + x + h x x lim h h xh + h + h lim h h lim x + h + h x +. a z + i + i + i + i i i b r 9 i i i i + i i + θ is in the th quadrant positive real component, negative imaginary component and tan θ θ π z cis. cos θ 7 θ 97. π h

30 Miscellaneous Exercise 7. cis π cis π π cos + sin π i i i + i i + i + i + i i + + i i Alternatively: cis π cis π cis π cos π i sin π i i 8. No working required. If you re having trouble understanding this question, think about what operation a 9 rotation in the Argand plane represents. 9. a The first line is parallel to i j + k and the second parallel to i j + k. Since these are scalar multiples, the lines are parallel. b First observe that the lines are not parallel. Next equate the lines and rearrange to make three equations from the three components: λ µ 7 λ µ λ µ Solve the first two simultaneously λ µ Does this solution satisfy the third equation? No: they are skew lines. c First observe that the lines are not parallel. Next equate the lines and rearrange to make three equations from the three components: λ µ λ λ µ 7 Solutions to A.J. Sadler s Solve the first two simultaneously λ µ Does this solution satisfy the third equation? 7 Yes: the lines intersect. d The first line is parallel to i + j k and the second parallel to i j + k. Since these are scalar multiples, the lines are parallel.. The height corresponds to the k component, so λ 8 λ and the initial position is r i + j + k i + j + 8k m The displacement from there to the touchdown point at i + j + k m is giving a velocity of s i + j + 8k i j 8k m v i j 8k i j k m/s The distance travelled is d i j 8k + + 7m π. a zw cis + π b cis π z w π cis π cis π c w cis π cis π

31 Unit C Specialist Mathematics Miscellaneous Exercise d z cis π 8 cis π e w 9 9 cis 9 π cis π cis π f z 9 9 cis 9 π cis 9π cis π. The first part of this question is really a D question, since passes directly over means we are not concerned with depth, so ignore the k component until it comes to finding the depth of the submarine. i + 87j + t i j i + 97j + t i j t i j i j i + 97j i + 87j t i + j 9i + j t satisfies both components, so the tanker passes over the submarine at t seconds. The depth of the submarine at that time is d m below the surface.. The plane r is perpendicular to the line r λ. The plane r is perpendicular to the line r λ. Two planes perpendicular to the same line are be parallel to each other. Therefore the two planes are parallel. The perpendicular line r λ intersects plane r at point A: λ λ λ λ 7 r A 7 The same line intersects plane r at point B: λ λ λ λ 7 r B 7 AB AB The distance between the planes is 8 units. We can generalise this result. Suppose we have a pair of parallel planes expressed as and r n a r n b These planes are perpendicular to the line r λn The points of intersection between the planes and this line are given by λ A n n a λ A a n n r A λ A n a n n n a n n 7

32 Miscellaneous Exercise similarly r B b n n AB r B r A b a n n AB b a n n b a n. Let P be the point on the line nearest the origin. OP is perpendicular to the line. + λ P + + λ P λ P λ P. OP... OP Solutions to A.J. Sadler s. Let A be the point. Let P be the point on the line nearest to A. AP is perpendicular to the line. OP AP 7 + λ P λ P λ P 7 + λ P 8 + λ P λ P AP 7 + AP + + units λ P 8

33 Unit C Specialist Mathematics CHAPTER Exercise A Chapter Proofs can be particularly challenging because it is often not clear what path will lead to success. Even the best mathematician will take several attempts at some problems before arriving at a successful proof. Indeed, some conjectures have remained unproven for centuries before a proof is finally found. Because of this you should be particularly slow to reach for these worked solutions when you find the work difficult. Make sure you try several different approaches before taking even a peek here. The more you can do on your own or together with other students, the more satisfying it will be for you.. a e + f 8 angles in a st. line f + g 8 angles in a st. line e + f f + g e g. A a e g b B h c e c g Alternate angles D d a e + g f c C b e + f 8 angles in a st. line c f + h c e Alternate angles d + e + f 8 angles in b + g + h 8 angles in. a c + f 8 cointerior angles e + f 8 angles in a st. line c + f e + f c e d + e + f +b + g + h d + e + g +b + f + h d + a + b b d + e 8 cointerior angles a + d 8 angles in a st. line a + d d + e a e. x a y z. d A a e a + y + z 8 angles in a + x 8 angles in a st. line a + x a + y + z x y + z C c b B z d + a + e 8 angles in a st. line d c alternate angles. a c + a + e 8 e b alternate angles x y c + a + b 8 a x + y ext. angle of Now that this is proved, it can be used as a theorem in other proofs. a + z 8 angle in a st. line x + y + z 8 Fermat s Last Theorem states that there are no positive integers a, b and c such that a n + b n c n for n an integer greater than two. Originally proposed in 7 it was finally proven in 99 by British mathematician Andrew Wiles although it had been proven for a number of special cases by various different people previously, including Fermat himself. The proof runs to well over typewritten pages. 9

34 Exercise A Solutions to A.J. Sadler s 7. There are often many possible counter examples; citing any one will do to disprove a statement. The following are representative. a is both prime and even. b A rectangle with length cm and width cm has four right angles but is not a square. c Rhombus ABCD has angles A and C and angles B and D and all four sides equal but is not a square. d If n then n n + is a multiple of and not prime. e can not be expressed as the sum of two other square numbers. The only possible candidates to add to would be and and none of +, + or + yield. f x + i.e. a, b, c has no real solutions. 8. Any odd number can be represented as n+ for a suitably chosen integer n. Then. Any odd number can be represented as n for a suitably chosen integer n. The next consecutive odd number is Then n + n + n n + n. Given n an even number square of an an even no. T n n then the next consecutive term is then T n+ n + n + + n + n + + n + n + n + n + an integer a multiple of 9. Any odd number can be represented as n+ for a suitably chosen integer n. Then + n + + n + + n + n + + n + n + + 8n + n + n + 8n + n + n + n + n + n + an integer a multiple of T n + T n+ n + n + n + n + n + n + an integer + an even number + an odd number. F n+ F n+ + F n+ F n+ + F n+ + F n+ F n+ + F n+ F n+ + F n+ + F n+ F n+ + F n+ F n+ + F n + F n+ F n+ + F n

35 Unit C Specialist Mathematics Exercise B Exercise B... AB AO + OB OA + OB CD CO + OD OC + OD h OA + h OB h OA + OB h AB Since CD is a scalar multiple of AB, CD is parallel to AB. PQ. OA +. AB.a +. OA + OB.a.a +.b.b SR. OC +. CB.c +. OC + OB.c.c +.b.b SR PQ This is sufficient to prove that PQRS is a parallelogram, since it shows that SR and PQ are both parallel and equal in length. However, since the question also asks us to find expressions for QR and PS... QR. AB +. BC. OA + OB +. OB + OC.a +.b.b +.c.c.a PS. AO +. OC.a +.c.c.a PS QR O A a M c b C CM CB. OB a.b CA CB OB + OA a b + a a b CA CM B. Hence M lies on CA and is the midpoint of CA. OM OA + AM k OB OA + h AC OB OA + OC AC OA + OC k OA + OC OA + h OA + OC k OA + k OC OA h OA + h OC h + k OA h k OC h k h + k h k h h k.. Refer to the diagram provided for question. AB AM + MB OC OM + MC OM MB since M bisects OB AM MC since M bisects AC OC MB + AM Similarly AB OA OM + MA CB CM + MB OA MB + CM CB OABC is a parallelogram.. a i. AB AO + OB a + b ii. AC AO +. OB a +.b iii. AD. AB.a +.b iv. OD OA + AD a.a +.b.a +.b v. OM OD a + b vi. AM AO + OM a + a + b a + b b M lies on AC such that AM:MC: if and only if AM MC. To prove: AM MC Proof:

36 Exercise B R.H.S.: MC AC AM a + b a + b a + b a + b AM L.H.S. c M lies on BE such that BM:ME: if and only if BM ME. To prove: BM ME Proof: R.H.S.: ME BE BM BE BO + OE a b BM BO + OM b + a + b a b ME a b a b a b 8. CE ka b Solutions to A.J. Sadler s ka kb DE AD + AC + CE ha + b + ka kb k ha + kb But DE is parallel to AC, so it must be a scalar multiple of b, so the component that is not parallel to b must be zero: k h k h CE h CE OA + AM OM a + k AC h OD OD OA +. AB a +. OA + OB a +.b a.a +.b AC OA +. OB.b a a + k.b a h.a +.b ka + k b h a + h b ka h a h b k b k h a b BM L.H.S. and h k h k k k B k 7. A D E C h k BE OB +. OA.a b BM OB + OD AB a AC c AD h AB ha Let CE k CE CB AC + AB a b b +.a +.b b + a.a b BE the medians intersect at a point two thirds of the way along their length measured from the vertex.

37 Unit C Specialist Mathematics 9. O To prove: R.H.S. A X Y C OA + BA + OC + BC XY XY XY + XY + XY + XY. BO + OA +. AC B. BO + BA +. AC +. BO + OC. AC. BO + BC. AC OA + BA + OC + BC L.H.S. This proof depends on expressing XY in four different ways. It may not be obvious at first that this is going to work, but after you ve done enough of these you sometimes get a hunch about an approach that will pay off. Practice is the key.. a a c because they are perpendicular. b AC a OB a c To prove: AC OB First consider the L.H.S.: AC AC AC AC a a a a a c c a a c a a c Now consider the R.H.S.: OB OB OB OB a a a a + a c c a a + c a a c L.H.S.. a AB a + b b To prove: AB OA + OB. O OA a a OB b b AB a + b a + b a a a c c a a + c a a c Exercise B OA + OB A C Let OA a and OC c OB OA + AB a CA OA OC a c a c a a c c B a a c OB CA If the perpendiculars are parallel, then OB CA a c. a AC a BD a + AC a c a + a a b AC BD a a a a c a + c but a c because the triangle is isosceles, so AC BD c + c AC and BD are perpendicular and BDA is a right angle.

38 Exercise C Solutions to A.J. Sadler s. a CB CO + OB c + b AO OB b AC AO + OC b b AC CB b b c b b c c b c but b c because they are both radii of the circle, so AC CB c c AC and CB are perpendicular and ACB is a right angle.. BC b a BC a b a b + a c a c a b AC a b AC b a a b + b c b c a b AB a + b c AB c a + b a c + b c a b + a b CF is perpendicular to AB. Exercise C. If the number is even, we can represent it as n, for a suitably chosen integer value n. n n which is a multiple of, and hence is even. If the number is odd, we can represent it as n+ for a suitably chosen integer value n. n + n + n + n + n + which is one more than a multiple of, and hence is odd.. There are five possible cases. The number is a multiple of ; one more than a multiple of ; two more than a multiple of ; three more than a multiple of ; or four more than a multiple of. Considering each of these exhaustively: If the number is a multiple of : n n n The square is a multiple of. If the number is one more than a multiple of : n + n + n + n + n + The square is more than a multiple of. If the number is two more than a multiple of : n + n + n + n + n + The square is more than a multiple of. If the number is three more than a multiple of : n + n + n + 9 n + n + + The square is more than a multiple of. If the number is four more than a multiple of : n + n + 8n + n + n + + The square is more than a multiple of.

39 Unit C Specialist Mathematics Exercise C. The the number is a multiple of ; one more than a multiple of ; or two more than a multiple of. Considering each of these exhaustively: If the number is a multiple of : n 7n 9n The cube is a multiple of 9. If the number is one more than a multiple of 9: n + 9n + n + n + 7n + 9n + 8n + n + n + 7n + 7n + 9n + 9n + n + n + The cube is more than a multiple of 9. If the number is two more than a multiple of : n + 9n + n + n + 7n + 8n + n + n + n + 8 7n + n + n + 9 9n + n + n + The cube is less than a multiple of 9.. Suppose T n is even, i.e. T n x for some integer x, then T n+ T n + x + x + Hence T n+ is also even. Suppose T n is odd, i.e. T n x+ for some integer x, then T n+ T n + x + + x + Hence T n+ is also odd. x + + T n+ has the same parity as T n.. Consider x x xx x + x + If x n then x x has x n as a factor, so it is a multiple of. If x n + then x x has x n + n as a factor, so it is a multiple of. If x n + then x + n + + n + n + n + n + Hence as x x has x + n +n+ as a factor, it is a multiple of. If x n + then x + n + + n + n + n + n + Hence as x x has x + n +n+ as a factor, it is a multiple of. If x n + then x x has x + n + + n + n + as a factor, so it is a multiple of. Hence, x x for x > is always a multiple of. As one or other of x and x is even, x x always has as a factor. Since it has both and as factors, it is always a multiple of. If x is odd, both x and x + are even, so x x is a multiple of. If x is even, x and x + are both odd. For x + : x + n + n + which is also odd, so x x has only one factor of, and so is not a multiple of. We can check this with an example. If x, x x which is not a multiple of.. If x 7n then x 7 x has x 7n as a factor, so it is a multiple of 7. If x 7n + then x 7 x has x 7n + 7n as a factor, so it is a multiple of 7. If x 7n + then x + x + 7n + + 7n + + 9n + 8n + + 7n + + 9n + n n + n + Hence as x 7 x has x + x + 77n + n + as a factor, it is a multiple of 7.

40 Exercise D Solutions to A.J. Sadler s If x 7n + then x x + 7n + 7n + + 9n + n + 9 7n + 9n + n n + n + Hence as x 7 x has x x + 77n + n + as a factor, it is a multiple of 7. If x 7n + then x + x + 7n + + 7n + + 9n + n + + 7n + + 9n + n + 77n + 8n + Hence as x 7 x has x + x + 77n + 8n + as a factor, it is a multiple of 7. If x 7n + then x x + 7n + 7n + + 9n + 7n + 7n + 9n + n + 77n + 9n + Hence as x 7 x has x x + 77n + 9n + as a factor, it is a multiple of 7. If x 7n + then x 7 x has x + 7n + + 7n + 7 7n + as a factor, so it is a multiple of 7. Hence, x 7 x for x > is always a multiple of 7. Exercise D. Suppose the triangle is right angled. Then the hypotenuse is the longest side, cm. By Pythagoras s theorem, the length of the hypotenuse is given by h but : a contradiction. Therefore a triangle with sides 8cm, 9cm and cm is not right angled.. Suppose that the lines intersect at point P. P i + j k + λ i j + k and P 8i + µ i + j k Equating corresponding components: + λ 8 + µ λ µ + λ µ + λ 8 + µ λ µ µ µ λ λ. From µ we get P 8i + i + j k 8i From λ. we get P i + j k +. i j + k 8i +.k resulting in a contradiction. Therefore the lines do not intersect.. Suppose that there exist integers p and q such that p + q. p + q p + q is even. But is odd: a contradiction. Therefore no such integer p and q exist.. Suppose that for some positive real a and b a b + b a < a + b < ab a + b < ab a ab + b < a b < But this requires the square of a real number to be negative: a contradiction.

41 Unit C Specialist Mathematics Miscellaneous Exercise Therefore no such positive real a and b exist. Note that the step taken in the third line above is only valid because we know ab is positive. If ab was negative then the inequality would change direction.. Suppose the triangle is right angled. Then x + x + x + 9x + x + x + x + x + x + x + x + x + Therefore the triangle can not be right angled for any value of x.. Suppose that log is rational. Then log a b for integer a and b with no common factors. a b a b a is an integer power of, and hence is even. But b is an integer power of an odd number, and hence is odd: a contradiction. Therefore log is irrational. 7. Suppose there exists some smallest positive rational number a b where a and b are positive integers. Because this is the smallest rational number, every other positive rational is greater than it. Then a b + > a b ab > ab + ab > ab + a > a But this is a contradiction, since we specified a and b both positive. Therefore there is no smallest rational number greater than zero. 8. Suppose that there are a finite number of primes. Since every finite set of numbers must have a largest member, there is a largest prime, a, a very large number, since it is larger than all known primes. Now consider the number b obtained from the product of all the primes: b 7... a Now consider the number b +. This is clearly larger than a. Because b has every prime as a factor, b + will have a remainder of when divided by every prime number. This means that the only factors of b+ will be and itself. Therefore b is a prime number larger than a: a contradiction. Therefore there is an infinite number of primes. Miscellaneous Exercise. cos x cos x cos cos x + sin sin x cos x + sin x cos x π. cos x sin π cos x cos π sin x. a z cis π cos x sin x cos x cos π + i sin π + i 7 b w i + cis tan - cis π rd quadrant c zw cis π cis π π cis π cis π cos π i + i sin π

42 Miscellaneous Exercise d. a z w cis π cis π π cis + π cis π cos π + i sin π + i OB a OB OA a a a + a c OB OC a c c + a c b Since OAOC then a c and it follows from the above that OB OA OB OC OBOA cos α OBOC cos β cos α cos β α β since both α and β are between and 8. L.H.S.: + sin A cos Asin A cos A sin A cos A + sin A cos A sin A cos A sin A cos A + cos A cos A sin A sin A sin A cos A os A cos A sin A + sin A cos A + sin A sin A cos A R.H.S.. L.H.S.: cos θ sin θ cos θ + sin θcos θ sin θ 7. cos θ sin θ sin θ sin θ sin θ R.H.S. lim x + h + x + h x + x h h x + x h + xh + h + x + h x x lim h h x h + xh + h + h lim h h lim x + xh + h + h x + 8. There are infinite possible correct solutions. First find any vector perpendicular to the one given Solutions to A.J. Sadler s for example by setting one component to zero and then swapping the other two, changing the sign of one such as: then scale this vector so that it has unit magnitude: The set describes the locus of a circle radius centred + i on the Argand plane. a The minimum possible value of Imz is. b The maximum possible value of Rez is + 7. c The distance between the origin and the circle s centre is + i. Since this is greater than, the minimum possible value of z is. d Similarly the maximum possible value of z is +. e Since z is just the reflection of z in the real axis, the maximum value of z is equal to the maximum value of z, i.e. +.. r F r A + r B i + j k i + j k The radius of the sphere is AF: AF r F r A i + j k i + j 8k i j + k CF, DF and EF are all also equal to the radius of the sphere, i.e. 7. CF 7 r F r C 7 i + j k 7i k 7 i + j + ck c 9 c c ± c rejecting the solution c because we are given that c is non-negative. 8

43 Unit C Specialist Mathematics Miscellaneous Exercise. DF 7 r F r D 7 i + j k i + dj k 7 i + dj + k 7 + d + 9 d d ± d d 9 rejecting the negative solution again. DE 7 r F r E 7 i + j k i + j + ek 7 i + j + ek e 9 e e ± e e rejecting the negative solution again. AB r B r A Av B v A v B For the missiles to intercept, AB t A v B t 8 i components: k components: t t 8t t Because AB is not a scalar multiple of A v B the missiles will not intercept. To find how much it misses by i.e. the minimum distance, let P be the point of closest approach. BP BA + AP AB + t A v B + t 8 BP Av B + t BP BP +t t + t m The missile misses by about m. After seconds the positions are now r A r B AB r B r A 7 7 t