(ii) x = 0or 4. (ix) x = or 3. (xii) x = or. (vi) x = 4. (xiv) x = Exercise (i) ( x 3) 8. (iii) ( 3q. (iii) 3( x + 2) 16.

Transkript

1 Cambridge International A and AS Level Mathematics Pure Mathematics Practice Book Universit of Cambridge International Eaminations bears no responsibilit for the eample answers to questions taken from its past question papers which are contained in this publication. Algebra Eercise. i a 9b cd cd c d iii a + ab iv v vi vii viii f g 9 + i mn + n p p + iii q q + iv t u s + p i km bd i e v bc p n k m + w fu iii v u f d p iv e w Eercise. i or or iii or iv or v or vi or vii or viii or i or i ± ii or iii iv or i ± or or iii iv or v ± vi Eercise. i + iii iv + + i ± or iii or i + + iii + iv v + vi 9

2 P i a 9, b a, b a, b, c i iii iv Verte, Verte, Verte, Verte, v vi vii viii Verte, Verte, Verte.,. Verte,

3 i + iii k > or k < iv k < P iii + + iv + Eercise. i.9 or.9 s.f.. or. s.f. iii. or. s.f. iv. or. i iii a. Using Δ to be the value of the discriminant: i No solutions No solutions i m, n p Eercise. i,, iii, iv, Eercise. i 9 < iii > or < i < or > iii Two solutions iv One solution v No solutions vi Two solutions i k k or iii k < < iv k or i > k k < k < or k > iii k < or k > iv k > k > i < k < < k k >

4 P iii < or > vi < or > iv v < < 9 9 vii viii 9 9 < or >

5 i < or > ii < or > P < < Stretch and challenge i The equation could be thought of in the form a b + c. Since the verte is at approimatel 9, the equation becomes a 9 +. Substituting the point, we get: a 9 + a a i Equation is 9 will var depending on verte chosen. Man answers are possible. One approach is to determine the co-ordinates of the centre of the hoop and ensure these co-ordinates fit the equation. b i a + b + c + + a c a Given the roots we can write the equation as: α β α β + αβ α + β + αβ Equating, we get b α + β a b α + β a c αβ a

6 P + k + + iii The roots are α and α. k + α + α and α α Solving the second equation, α α 9 α ± Substituting into the first equation we get: k + + k k + or + k α + β and αβ α + β + α β αβ So the equation is: and αβ + or + iv α + β and αβ α + β α + α β + αβ + β α + β + αβ α + β Rearranging, we get α + β α + β αβ α + β αβ + 9 Let t t t t + t 9 t or 9 or 9 k + Let t kt has no solutions so 9, the equation can then be written as kt + t kt + t t + For a single solution b ac k k t t + t t Eam focus a, b, c a and b or i Verte is, The equation is + + < <

7 Co-ordinate geometr Eercise. i m m iii m i Length. i Eercise. A: + or + P Mid-point, B: Gradient Length. C: D: Mid-point, E: Gradient. F: iii Length.9 Mid-point, Gradient Gradient of line Gradient of perpendicular iv Length. m n i Mid-point, Gradient iii i m. m. m.9 d.p.. i iii.. Eercise. i Gradient:, intercept: Gradient:, intercept: iii Gradient:, intercept: iv Gradient:, intercept: i Equation: +, Gradient:, intercept: Equation:, Gradient:, intercept: iii Equation: 9 +, Gradient:, intercept: 9 iv Equation:, Gradient:, intercept: i + + iii i + +, +

8 P iii 9 Eercise. Area i d +. s.f. + a : b : i b + or + iii or + + iv D is,. Stretch and challenge The diagram shows the relationship between the tangent and normal at the point A. B A We need to find the co-ordinates of both A and B. To find where the line meets the curve, solve simultaneousl: So A is,. The gradient of the tangent to the curve is so the gradient of the normal is. The equation of the normal is: Eercise. i, or,, or, iii, iv, or, v, or, vi, k k > or k < k > 9 k or k or i < k < Point is, Area of triangle ABC. m + c + c c To find B, solve simultaneousl: or Substitute 9 into the equation of the curve to find the co-ordinate: 9 9 So B is,. Finall, use Pthagoras theorem to find the length of AB. 9 AB +.9 s.f.

9 multipling ever term b ab: a b b a ab b ab a b ab a b a b From this form the gradient of the line is b a, so b b a a The intercept is when so a a, so M is a,. b a The intercept is when so b, so N is, b. a b b B Pthagoras theorem, MN a + b a + b a + b a + b b + b b + b b + b b b 9 b a b k 9 k k k k k ± A B B i + are or. so the intercepts So the line must pass through, or,. k + k + k or k + k + k The intercepts are a or b. So the line must pass through a, or b,. k + k a + k a k + k b + k b iii The intercepts are a or b. So the line must pass through a, or b,. c m + c k a + c k a c m + c k b + c k b k k + + k k + + k + P Mid-point of A, and Bk, is + k, + k,. Gradient of AB is. k k Gradient of perpendicular is k Equation of perpendicular bisector is + k k m + c + c k k c Since the intercept is 9, + c + k. For two points of intersection, b ac > k > k + k > k k > k k Eam focus + > k < or k > The co-ordinates of D are,. i The equation of CD is +. The equation of BD is. iii The co-ordinates of D are,. k 9

10 P Sequences and series Eercise. i 9 $99 s.f. i or. iii iv. i iii 9 iv. m + 9n $ i minutes s.f. t. minutes s.f. i For the A.P., a u a + d + d u a + d + d First three terms of the G.P. are, + d, + d + d + d So r + d + d + d + d + 9d + 9d 9d + d 9 dd + d or 9 i Solving simultaneousl, d., a Since d, d When d, n 9 Eercise. i + d r iii n 9 + iii 9 iv 9 s.f. i. iii s.f. iv Parts and iv have a sum to infinit. ± u. S. 9.. S r u < < r u Eercise. i i iii a a b + a b... iv The term is The term is The term independent of is 9 i The coefficient is. The coefficient is. k ± b

11 9 i 9 + u + u +... The coefficient of the term is The coefficient of the term is i + The coefficient of the term is i The coefficient of the term is k a. Stretch and challenge u a + d [ ] S a + d a + 9d a + d [ ] Solving simultaneousl, d and a a + 9d So Anna used Facebook for minutes in week of her plan. General term for the first epansion is k r r r Term independent of is when r k 9 k 9 k General term for the second epansion is r r m k r Term independent of is when r m k m k k m So k m k 9 k m a + ar a + r ar + ar 9 ar + r 9 ar + r 9 a + r r r a + a a u u a + d a + d a + d a + d a + d a + d. S a + 9d [ ] a + d a + d 9 Solving simultaneousl: a + d a + d 9 a., d. Value of stamp :,, A.P. with a, d Value of stamp : G.P. with r.9 For Stamp, u + For stamp : a 9. a $ 9. So the second stamp was bought for $ i The general term is The epansion is: r r r a 9 + a + a + a + a + a + a + a + 9 a + a+ a Comparing the coefficients of the terms either side of the term: a > a a > a > a a < < a < The terms we are interested in are:... a + a + a We want a > a a > and a > a a < < a <... P

12 P Eam focus i The coefficient of the term is 9. The coefficient of the term is. 9 a a u n 9. a

13 Functions Eercise. i f + vi gf vii ff viii gg P f + iii f + iv f + All are functions. Is not one-to-one the rest are one-to-one. i Is a function. Is not one-to-one. Is not a function. Is not one-to-one. iii Is a one-to-one function. iv Is a function. Is not one-to-one. i f 9 f + i Is not a function. Is not one-to-one. Is a function. Is not one-to-one. a, b or a, b i fg + gf g + iii 9 iv k > a, b i or iii Is a one-to-one function. iv Is a function. Is not one-to-one. i Domain: Range: f Domain: Range: f iii Domain: Range: f iv Domain: Range: f Range is f Domain: Range: f 9 Domain: Range: f Given that, the range is f Eercise. i f, f f f Eercise..,. i f g iii fg iv gf v fg

14 P iii f + f f f, Domain f : Range f : f Domain f : Range f : f iv f + + f iii f, i f Domain f : Range f : f Domain f : Range f : f Domain f : Range f : f Domain f : Range f : f iv f Domain f : Range f : f Domain f : Range f : f

15 The largest possible value of k is. g j f + g Domain g : Range g : g Domain g : Range g k : g i Range is f f f ,,. Stretch and challenge i a b+ c + d f Constant Effect on graph a a > Vertical epansion < a < Vertical compression a < Reflects the graph in ais b b > Horizontal compression < b < Horizontal epansion c Shift left c > or right c < d Shift d units up d > or down d < The translation of f units right and one up is given b f P iii 9 i ff f g f g fg f g fg f g fg f We can see the pattern here, ever number is one n less than a power of so g n h g h gh g h gh g

16 P f + + f af + + a i a b a i k. g + fn an + b an a + b f n a n + b an + b f n an + b an + b an + b is common to all three epression so the three epressions can be simplified to a, and b. Since is one of the terms, the possible consecutive terms are i,,,, iii,, i a, b f n n a, b f n n a, b f n n + iii a, b f n n + i or g iii Since b iv v ac the equation has no real solutions. + f g + Point on f Point on f,,,,,, f Eam focus The domain of f is The range of The domain of The range of f is f is f is

17 Differentiation Eercise. Eercise. i Stationar points are, and,. P i d d d d iii iv v d d d d d d vi d d vii d d viii + + d + d i f f iii f iv f f Point is,. Points are, or,. Eercise. Equation of the tangent is: i PR + S is, i + + Q is, k, is a local minimum, is a local minimum Points are, and,.... > f, is alocal minimum f <, is alocal maimum iii Point is, is a local minimum

18 P < < Since the discriminant is <, there are no solutions when + i So + is alwas < h i h + h h h Perimeter g g + A + i k iii Stationar value when. m s.f. da d iii The co-ordinate of the other stationar point is. Eercise. f + > isaminimum f + < is amaimum Maimum value of P is 9 9 Dimensions are m b. m. iv. m is a maimum i V r h h r S r + rh r + r r r + r Stationar value when r 9.cm ds dr iii r 9. is a minimum value

19 i b + r b r b r r.m s.f. b... m Let the length of the end be Let the length be + so V dv d Maimum volume when cm or cm Since, cm cm Volume is cm dv d The largest area is. units... cm s.f. 9 Dimensions are: length. m, width. m, height. m vi vii viii + + d + d + d d + + d + d + Equation of the tangent is m + c + c c Point is, g 9 + Since g > for all values of, the curve is alwas increasing so there are no stationar points. Hence the function is one-to-one so it has an inverse. The graph of g for > and its inverse are shown on the net page. g P Eercise. d i d d + + d + 9 d iii 9 d iv + + d + d + v 9 9 d 9 9 d 9 i k < or > i d d d d + maimum point A is,, minimum point B is, + 9

20 P.cm/s.m/min s.f..cm/min s.f. 9. units/s i. cm/s s.f.. rad/s cm /s s.f. i Using similar triangles h a + b h h + + h h + h +. cm/s h Stretch and challenge i g k so if the gradient of the normal is, the gradient of the tangent is. cm h cm cm cm + cm h + h + h+ h+ Area of trapezium cross-section h + + h Volume h + h h + hcm h.cm or.cm h + h h + h Since h cannot be negative, h.cm k k k k The value when is g. Equation of the normal is m + c + c c so To find the other point, solve simultaneousl + or So the co-ordinate of the other point is. g The co-ordinates of the point P are,. g iii i.9cm/min h P B In the right-angled triangle h h h tan h Length of CD is + + Volume area of trapezium length i SP + so cosα PW + cos β

21 Time T f P + + T [cosα cos β] T when [cosα cos β] cosα cos β AE + AE km km B km i V r h R r h V dv dh dv dh h + r R h R h h Rh h R h when R h r R R r R V V clinder sphere h R R h R h r h R r h R R R R R R R R R So ratio is : or : R R R R A D km E km In triangle BDE, BD + BD + 9 Total length of road, L, is: L AD + BD + CD C dl d Minimum length is when km L km 9 km s.f.

22 P Eam focus The points are, and,. The equation of the tangent is + or + The slope of the normal is m + c + c c The equation of the normal is + or + The stationar points are,,, and,. d d d At, so, is a maimum. d i Volume r h + r r h + r r h r r h r r r r r r S rh + r + r r S r + r + r r r + r r + r r + r r d At, so, is a d minimum. d At, so, is a d minimum. The function is increasing for < <. mm /minute. units/second iii Stationar point is when + r r + r r r.cm s.f. iv r. cm is a minimum value.

23 Integration Eercise. i + + c + + c iii + c iv v vi + c + + c + + c Area. 9,, P i 9 + c Area iii + c iv + c + + c + c + + f + The equation is + Eercise. i A A + d Area between curve and the ais: Area between curve and the ais: 9 9 k Eercise. i i iii iv c + c + c + c + c + c The area is. Area between the curve and the ais: Area between the curve and the ais: iii iv c + + c + c + + Area Area Total area + v + c vi + + c

24 P Eercise. 9 Eercise. i. s.f.. s.f..9 s.f. V solid V solid 9. s.f. Points of intersection: or V solid. d. + d. + d a since a > i A is,, B is, M is, Final volume units Stretch and challenge + k + d + k + + d + k d + k + k + + k + k k k h Gradient of the sloping line is m R r Equation is h hr m + c r + c c R r R r h R r hr R r hr h + R r R r R r + h r h V d h R r r R + h h R r h R r h R r h r + r r R r + + r h d r R r + h h r R r h + h hr r + hr R r + rh + r + rh hr Rr + r + hr R r + rh h R Rr + r + rr r + r h R + Rr + r h R + rr + r k + Area d k k + k k k k k k + k h

25 k + k + k + k 9 k + k + 9 k + k + 9 k + k + 9 k + k Since >, k k + k k + k k or + + c Area of DFGE area of OBGE area of ABGF area of OAFD Shaded Area P + R R w w Also + r R r R w / V d V clinder w / w / d rw w / R d r w w / R rw w R w rw Rw w rw w Rw rw w w Rw R w w Rw Rw+ w w Since the volume does not depend on R, the claim is true. Volume units i Area under line Shaded area Area under curve area of Δ 9 i a Volume of curve rotated around the ais: Finalvolume or b Final volume or or Final volume units. units to s.f. Volume or Finalvolume units Eam focus The curve is + A,

26 P Trigonometr Eercise. i a..cm s.f. b..9 m s.f. iii c. cm iv θ. i d iii.cm β.9 θ.9..9cm iv θ i sintan cos cos sin LHS sin cos sin cos cos cos cos cos cos cos cos RHS tan sin sin cos LHS tan cos sin cos cos sincos RHS Using rounded answers from question i Area Area.cm cm iii Area 9.cm iv Area i.cm cm Eercise. Angle.9 m Other angle that has the same value sin sin sin sin cos cos tan tan Eact value cos cos tan tan iii iv i + cos sin sin cos sin + cos RHS cos + cos sin + cos cos sin + cos sin + cos sin LHS tan sin tan sin sin LHS sin cos sin sin cos cos sin cos cos sin sin cos sin sin cos tan sin RHS sin cos sin cos LHS sin cos sin + cos sin cos sin cos RHS

27 cos tan + cos sin sin LHS + cos cos sin + cos sin + cos cos cos sin + cos sin sin + cos sin + sin cos sin RHS iii iv + cos + cos sin + cos + cos LHS cos + cos cos sin RHS + tan tan cos sin + LHS cos sin cos cos + sin cos cos sin cos cos cos sin cos cos cos cos sin cos sin cos cos cos RHS Eercise. i sin sinθ sinθ sinθ sinθ sinθ sinθ RHS cos cos cos + cos tan cos cos LHS cos + cos cos + cos cos cos cos + cos cos + cos cos + cos cos cos sin sin cos tan RHS tan i cos sin i or or or or iii. or. d.p. iv. or. or. d.p. P i cosθ tanθ + sinθ sinθ cosθ LHS sinθ + sinθ cosθ cos θ sinθ + sinθ sin θ sinθ + sinθ sinθ + sinθ sinθ + sinθ sinθ i.. or. d.p.

28 P i 9,,,, iii. i cos + sin sin + sin sin + sin. or. sin + sin + sin sin sin + sin iii sin orsin Since sin is never greater than, sin θ.,.. or. Eercise. Degrees Radians iv i. or. + sinθ + tanθ cosθ sinθ LHS sinθ + cosθ cosθ + sinθ cosθ + sinθ cos θ + sinθ sin θ + sinθ sinθ RHS + sinθ sinθ + sinθ Degrees Radians θ. or.

29 i Angle sin cos tan cos sin tan Other angle that has the same value sin cos Eact value tan cos sin tan tan or tan,,,.,. i Perimeter. cm s.f. Area of shaded region.cm s.f. i θ sinθ θ sinθ θ sinθ θ sinθ θ sinθ + θ. Total length is: 9.cm s.f. AC i tan θ AC tan θ Shaded area Area of triangle OAC area of sector OAB tan θ θ tanθ θ tan θ θ P sin or sin, iii sin. or.. or. iv cos orcos,,, v tan ortan.9,.,.,. vi sin orsin, Perimeter + +.cm s.f. i Since triangle OCD is equilateral, angle OCD So angle ACD Perimeter AD + BD + AB cm iii Shaded area Eercise. i Eercise. 9 i P.cm A 9.cm P.cm s.f. A cm s.f. θ Perimeter:.cm s.f. Area:.cm 9

30 P iii 9 T t iv iii From the graph, the highest points occur at t.,.,,.,. and d.p. t. pm,. pm, pm,. pm,. pm, pm v 9 i vi 9 The line we need to draw is: i f iii f f A, B, C A, B, C A, B, C A, B, C A, B, C A, B iv f tan i b. a. i period

31 i cos k cos k θ,,,,,,, 9, i H m H sin t + Amplitude Period : t hours H sin t + P iii sin k Period hours iv cos v tan + k i tan k tan k k k iii cos + k k iv sin + + k v tan + k i or iii am. 9 cos t cos t. cos t t cos t, t, t, or or iii tan or iv.9 or. or.or. s.f. Since t is hours after 9. am, refloating can occur between and hours after 9. am, i.e. between. pm and. pm. i Period seconds Sun Moon Maimum height m iii t seconds r q Stretch and challenge sinθ + cos θ sinθ + sin θ sinθ + sin θ sin θ + sinθ + sinθ + sinθ + sinθ or sinθ θ,, 9,,,, 9,, i A r θ r sinθ r r θ r sinθ θ sinθ sinθ θ r

32 P r r q LHS sin + cos sin cos sin cos + cos sin sin + cos sincos sincos RHS r cosθ cosθ θ r iii A r θ r sinθ r θ r sinθ r r sin r r r r %covered % r % 9.% LHS sin cos cos cos + sin cos + cos + sin sin cos cos sin cos cos sin cos sin RHS or or 9 or or or or Eam focus i The equation is sin ; i.e. a, b. c. b. a. The equation is cos. i sin sin k. cos k iii k tan k sin θ LHS + cos θ cos θ + sin θ cos θ cos θ RHS sin or sin or 9 cos or, Area of segment. cm s.f. Perimeter of shaded region + + i θ DE. s.f. iii Shaded area 9. cm s.f.

33 Vectors Eercise. c + a P i c b a a i i j i + j b c iii g j h i j g + h j + i j i j i l. s.f. c l. s.f. iii AB. s.f. b a C iii + a b A B b c i iii OA BC CA iv D is, i iii OB + m n OE + m n BD + m n

34 P iv EC m n. i Unit vector. Unit vector or i j Unit vector is or i k. The unit vector is.. i Eercise. ο i θ 9. d.p. AB i + j k 9 i i + k i + j + k iii i j + k iv v vi CD i + k i j k i + j 9. s.f. ο θ. d.p. θ. d.p. i θ 9.9 d.p. θ.9 d.p. iii a 9 9 θ. d.p. c i PQ i AB AF i + j k QR iii Since QR PQ, iv the are parallel QR is twice the length of PQ QR goes in the opposite direction to PQ. S is,,. BF i + j + k iii θ. d.p. i OM i + k CM i k MB i + j + k CB i + j + k i CD θ. d.p. EC G is,, i Unit vector i j + k Stretch and challenge a b c a b c.. a b c + + a c +

35 a + b + c b + c + There are infinitel man solutions to this sstem of two equations with three unknowns. Choosing the whole number solutions, b, c, a, the vector is or an scalar multiple. p q a + b + + a + a + b a + 9 b b ± Eam focus So the unit vector in the direction of OR is p θ. d.p... P a b. a a ab If b, a + a + 9 a 9 If b, there are no solutions for a. So a 9 and b vb + + m/s v + m/s v v a a a k k vb k k k k + k + k v a 9k k k i j + k

36 P Past eamination questions Algebra i a, b i < or > 9 a, b 9 i S 9 s.f. S i $9 $ iii $ i a 9 b S Co-ordinate geometr i Equation of BC is + or + C is, iii Perimeter +. i Equation of BC is: + Equation of CD is: + 9 C is,9 i Points of intersection are., and, 9 < k < 9 or k < 9 i Equation of CD is: + or + D is, 9 i + u+ u + The coefficient ofthe term is. i a Coefficient of term is i a d b c a $. or $ Functions i b $ or $ f m< or m> i M is, D is, : : i Points are, and, k. Sequences and series i a S i $. S Coefficient of is 9 iii or i a, b 9 i The domain of g is and range is. g +

37 i < and > Range is f. No inverse as f is not one-to-one. gf g + iii + + has b ac so has noreal solutions. The domain of f is 9 i Range of f is f. f P iv g i Range is f. - for iii f for < i a b fg 9 iii Largest value is when A 9. iv g sin i 9 f f does not have an inverse as the function is not one-to-one. iii i The domain of f is < Differentiation i V 9r r r cm iii r cm is a maimum i h r r A r h + r r r r + r r r r + r r r r iii r.cm s.f. iv r. cm is a maimum The point is,. f Angle between the two lines is. d.p.. f + Since + > for all values of, then + is alwas negative. Since f is alwas negative, the curve is alwas decreasing. iii

38 P + f 9 Since + > for all values of, f > for all i values of, hence the function is alwas increasing. d d + + or + iii 9. units/second is a minimum point Integration i Equation is + The intercept is, + V + d + d i Equation of normalatp is: Equation of curveis i + Area 9 i, d d At. units/second iii Area i Equation of curveis + + < or > + i + d + d At, d d + + Equation of tangent at, is: m + c + c c + Co-ordinatesofC iswhen + DC Area of triangle BDC i k Maimum point is, iii The function is decreasing for < < iv A. i When, d d. units/second iii V d d 9 i Shaded area 9 V. s.f. iii Anglebetweenthetangents 9. 9

39 i k Equation is + Using Pthagoras theorem, AB AC + BC P i a, b A. iii c. Trigonometr l l + l + l l AB l l l i b a.,. d.p. iii In triangle ABC tan + tan + + tan tan l l i Area of sector.cm s.f. θ iii AB + cos 9 i Cosine rule: AB AB cm or cos AOB + cos AOB. AOB. s.f. or sin AOB AOB sin AOB. s.f. A cm iii Area of cross-section cm s.f. i tanθ cosθ sinθ cosθ cosθ sinθ cos θ sinθ sin θ sinθ sin θ sin θ + sinθ θ or Since angle AOB, and AO BO, then angle OAB angle OBA So triangle OAB is equilateral so AB cm BC cos + BC 9 BC 9 Perimeter cm i sin θ + cos θ sin θ + sin θ sin θ + sin θ sin θ + sin θ With sin θ wehave + θ,,, i Shaded area.cm s.f. Perimeter.cms.f. i AC cos AC cos l l l DC sin DC sin l l l BC DC l 9

40 P i sin θ or i BD. cm Shaded area.cm or9 cos i cos a, b 9 i Two solutions B f Greatest value is when cos,f Least value is when cos,f iii. or. tan cm O q A cm cm tan. s.f. θ θ So AOB.. s.f. Perimeter.cm s.f. iii Shaded area.cm s.f. i cos i tan sin tan sin sin cos sin cos sin cos cos cos cos cos cos + cos cos + cos. or 9. i Perimeter of shaded region.9cm s.f. Shaded area.cm s.f. i tan θcosθ θ sin cosθ cos θ sin θ cosθ sin θ cosθ cos θ cosθ cos θ cosθ cos θ + cosθ cos θ + cosθ T. or. i LHS sinθ tanθ cosθ sinθ sinθ cosθ sinθ cosθ sin θ cosθ cos θ cosθ cosθ cosθ + cosθ cosθ + cosθ RHS θ. or 9. i Distance from D to AB is sin Distance from E to AB is sin q Equating these distances gives: sinθ sinθ θ sin P. cm s.f. Vectors i MO MB + BO i k MC MO + OC + CC i k + j + k i + j + k θ 9. d.p.

41 i BA a b i + k i j + k i + j k BC c b j + k i j + k i + j + k BA BC i + + SinceBA i BC the vectors are perpendicular. AD d a i + j + k i + k BC i + j + k i + j + k BC : AD : i + j + k i + j + k Since both vectors are multiples of i + j + k the are parallel i The unit vector is p iii q or i θ. iii i Thevectormustbe AC p The unit vectoris Acute angle is.. iii Perimeter. i θ. iii + p + p p p i PQ i + j k RQ i + j + k θ. or i+ j+ k or + p i+ + p j+ p k P

42