S5_Unit_3_Outcome_1 - Higher Unit 3 Hig h er Vectors and...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Higher Unit 3 Hig h er Vectors and Scalars O utc o m e 1 3D Vectors Properties 3D Section formula Scalar Product Component Form Angle between vectors Perpendicular Properties of Scalar Product Properties of vectors www.m a th s re vis io n .c o m Adding / Sub of vectors Multiplication by a Scalar Unit Vector Position Vector Collinearity Section Formula Exam Type Questions www.mathsrevision.com Ve c to rs & S c a la rs Higher Outcome 1 A ve c to r is a q ua ntity with www.m a th s re vis io n .c o m BOTH m a g nitud e (le ng th ) a nd d ire c tio n. Exa m p le s : G ra vity Ve lo c ity Fo rc e Vectors & Scalars Higher Outcome 1 A scalar is a quantity that has www.m a th s re vis io n .c o m magnitude ONLY. Exa m p le s : T im e S pe e d Ma s s Ve c to rs & S c a la rs Higher Outcome 1 A vector is named using the letters at the end of the directed line segment www.m a th s re vis io n .c o m or using a lowercase bold / underlined letter This vector is named uuu r AB u u uuu r AB or u o r u Ve c to rs & S c a la rs Higher Outcome 1 Also known as column vector A vector may also be www.m a th s re vis io n .c o m represented in component form. uuu r x AB = d = y w uuu r 4 CD = w = 2 uuu r 2 FE = z = - 1 z Magnitude of a Vector Higher Outcome 1 A vector's magnitude (length) is represented by www.m a th s re vis io n .c o m uuu r PQ or u A vector's magnitude is calculated using Pythagoras Theorem. uuu r PQ = u = a 2 + b 2 u Ve c to rs & S c a la rs Higher Outcome 1 Calculate the magnitude of the vector. www.m a th s re vis io n .c o m w = a 2 + b2 w = 42 + 2 2 w = 20 w= 4 5=2 5 uuu r 4 CD = w = 2 w Vectors & Scalars Higher Outcome 1 Calculate the magnitude of the vector. www.m a th s re vis io n .c o m w = a 2 + b2 w = (-4) 2 + 32 uuu 4 r - PQ = 3 w = 16 + 9 w =5 Equal Vectors Higher Outcome 1 Vectors are equal only if they both have www.m a th s re vis io n .c o m the same magnitude ( length ) and direction. Vectors are equal if they have equal components. For vectors a c u = and v = if u = v then a = c and b = d , b d Equal Vectors Higher Outcome 1 By working out the components of each of the vectors determine which are equal. www.m a th s re vis io n .c o m a b d e f g h c Addition of Vectors Higher Outcome 1 Any two vectors can be added in this way www.m a th s re vis io n .c o m Arrows must be nose to tail b 2 a = 4 6 = - 3 8 a + b = 1 Addition of Vectors Higher Outcome 1 Addition of vectors www.m a th s re vis io n .c o m B uuu r uuu 2 r 3 Let AB = and BC = 4 - 5 uuu uuu uuu r r r Then AB + BC = AC A 3 2 5 + = 4 - - 5 1 C uuu 5 r So AC = - 1 Addition of Vectors Higher Outcome 1 In general we have www.m a th s re vis io n .c o m For vectors u a nd v a c If u = and v = then b d a c u + v = + = b d a +c b +d Ze ro Ve c to r Higher Outcome 1 T h e ze ro ve c to r www.m a th s re vis io n .c o m uuu r uuu 1 r 1 - If AB = then BA = 2 - 2 uuu uuu 1 r r 1 - AB + BA = + = 2 - 2 0 0 0 is called the zero vector, written 0 0 Ne g a tive Ve c to r Higher Outcome 1 www.m a th s re vis io n .c o m uuu r uuu r BA is the negative of AB Ne g a tive ve c to r Fo r a ny ve c to r u u + ( -u ) = u + ( -u ) = 0 a - a If u = then - u = b - b S ub tra c tio n o f Ve c to rs Higher Outcome 1 Any two vectors can be subtracted in this way www.m a th s re vis io n .c o m Notice arrows nose to nose 6 u = 3 1 = 3 v 5 u v = 0 Subtraction of Vectors Higher Outcome 1 Subtraction of vectors www.m a th s re vis io n .c o m Notice arrows nose to nose 6 2 Let a = and b = 5 4 Then a - b = a + (-b) a b 6 - 4 2 = + = 5 - 1 4 a b 6 2 4 = = 5 4 1 Subtraction of Vectors Higher Outcome 1 In general we have www.m a th s re vis io n .c o m For vectors u a nd v a c If u = and v = then b d a c u - v = - = b d a -c b -d Multip lic a tio n b y a S c a la r Higher Outcome 1 Multiplication by a scalar ( a number) www.m a th s re vis io n .c o m x kx If a vector v = then kv = y ky The vector kv is parallel to vector v ( different size ) Hence if u = kv then u is parallel to v Conversely if u is parallel to v then u = kv Multiplication by a Scalar Higher Outcome 1 Multiplication by a scalar www.m a th s re vis io n .c o m Write down a vector b a parallel to a Write down a vector parallel to b Multiplication by a Scalar Higher Outcome 1 Show that the two vectors are parallel. www.m a th s re vis io n .c o m 6 18 w = then z = 9 27 If z = kw then z is parallel to w 18 6 z = = 3 27 9 z = 3w Multiplication by a Scalar Higher Outcome 1 Alternative method. www.m a th s re vis io n .c o m 6 18 w = then z = 9 27 If w = kz then w is parallel to z 6 18 1 w = = 9 27 3 1 w = z 3 Unit Vectors Higher Outcome 1 For ANY vector v th e re e xis ts www.m a th s re vis io n .c o m a p a ra lle l ve c to r u o f m a g nitud e 1 unit. T h is is c a lle d th e u nit ve c to r. Unit Ve c to rs Higher v u Outcome 1 Find th e c o m p o ne nts o f th e unit ve c to r, u , p a ra lle l to ve c to r v , if www.m a th s re vis io n .c o m 3 v = 4 v = 3 +4 2 2 v =5 S o th e unit ve c to r is3 u 3 1 5 v = = 4 4 5 5 A P o s itio n Ve c to rs Higher B Outcome 1 A is th e p o int (3 ,4 ) a nd B is th e p o int (5 ,2 ) . www.m a th s re vis io n .c o m Write d o wn th e c o m p o ne nts o f uuu uuu r 3 r 5 OA = OB = 4 2 uuu 2 r AB = 2 - Ans we rs th e s a m e ! uuu uuu 2 r r 5 3 - = OB - OA = 2 2 4 - A P o s itio n Ve c to rs Higher a 0 b B Outcome 1 www.m a th s re vis io n .c o m uuu r OA is called the position vector of the point A relative to the origin O , written as a uuu r OB is called the position vector of the point B relative to the origin O , written as b r r uuu uuu uuu r AB = AO + OB = -a + b = b - a A P o s itio n Ve c to rs Higher a 0 b B Outcome 1 www.m a th s re vis io n .c o m uuu r AB = b - a where a and b are the position vectors of A and B. P o s itio n Ve c to rs Higher Outcome 1 If P a nd Q h a ve c o o rd ina te s (4 ,8 ) a nd (2 ,3 ) www.m a th s re vis io n .c o m re s p e c tive ly , find th e c o m p o ne nts o f uuu r PQ uuu r 2 4 - 2 - = PQ = q - p = 3 8 - 5 P o s itio n Ve c to rs Higher Outcome 1 G ra p h ic a lly www.m a th s re vis io n .c o m P (4 ,8 ) Q (2 ,3 ) uuu r p Q P - 2 PQ = - 5 q p q O Collinearity Higher Outcome 1 Reminder from chapter 1 www.m a th s re vis io n .c o m Points are said to be collinear if they lie on the same straight line. uuu r uuu r If AB = k BC , where k is a scalar, then AB is parallel to BC. For vectors If B is also a point common to both AB and BC then A,B and C are collinear. Collinearity Higher Outcome 1 Prove that the points A(2,4), B(8,6) and C(11,7) are collinear. www.m a th s re vis io n .c o m uuu r 8 2 - = AB = b - a = 6 4 6 2 3 1 uuu r 11 8 - = BC = c - b = 7 6 uuu uuu r 6 3 r AB = 2 2BC = = 2 1 Collinearity Higher Outcome 1 www.m a th s re vis io n .c o m uuu uuu r r Since AB = 2 BC , AB is parallel to BC. B is a point common to both AB and BC so A, B and C are collinear. Section Formula Higher Outcome 1 www.m a th s re vis io n .c o m uuu uuu uuu r r r OS = OA + AS uuu uuu 1 uuu r r r OS = OA + AB 3 1 s = a + (b - a) 3 2 1 s = a+ b 3 3 A B 3 1 S 2 s a O b General Section Formula Higher Outcome 1 www.m a th s re vis io n .c o m uuu uuu uuu r r r OP = OA + AP uuu uuu r r m uuu r OP = OA + AB m+ n m p = a+ (b - a ) m+ n n m p= a+ b m+ n m+n A B m + n m P n p a O b General Section Formula Higher Outcome 1 Summarising we have www.m a th s re vis io n .c o m If p is a position vector of the point P that divides AB in the ratio m : n then A m P n B n m p= a+ b m+ n m+ n General Section Formula Higher Outcome 1 A and B have coordinates (1,5) www.m a th s re vis io n .c o m and (4,10) respectively. Find P if AP : PB is 3:2 P 3 A - 12 2 5 5 + = 2 6 10 2 5 = 8 8 2 B 2 3 p = a+ b 5 5 - 2 1 3 4 p= + = 10 5 5 5 3D Coordinates Higher Outcome 1 In the real world points in space can be located using a 3D coordinate system. www.m a th s re vis io n .c o m For example, air traffic controllers find the location a plane by its height and grid reference. z y O (x, y, z) x 3D Coordinates Higher Outcome 1 Write down the coordinates for the 7 vertices www.m a th s re vis io n .c o m z (0, 0, 2) F (0,0, 0) G E (0, 1, 2) H O y A (6, 1, 2) 2 1 D (6, 1, 0) (6, 0, 2) B 6 C (6, 0, 0) x What is the coordinates of the vertex H so that it makes a cuboid shape. H(0, 1, 0 ) 3D Vectors Higher Outcome 1 3D vectors are defined by 3 components. www.m a th s re vis io n .c o m For example, the velocity of an aircraft taking off can be illustrated by the vector v. z 2 3 O y v 3 7 7 2 (7 , 3 , 2) x 3D Vectors Higher Outcome 1 Any vector can be represented in terms of the i , j a nd k www.m a th s re vis io n .c o m Wh e re i, j a nd k a re un it ve c to rs z k O in th e x, y a nd z d ire c tio n s . 1 0 0 i= j= k = 0 1 0 0 0 1 y j i x 3D Vectors Higher Outcome 1 Any vector can be represented in terms of the i , j a nd k www.m a th s re vis io n .c o m Wh e re i, j a nd k a re un it ve c to rs z in th e x, y a n d z d ire c tio n s . (7 , 3 , 2) y O v 3 7 2 x v = ( 7 i+ 3 j + 2 k ) 7 v = 3 2 3D Vectors Higher Outcome 1 www.m a th s re vis io n .c o m Good News All the rules for 2D vectors apply in the same way for 3D. Magnitude of a Vector Higher Outcome 1 A vector's magnitude (length) is represented by v www.m a th s re vis io n .c o m A 3D vector's magnitude is calculated using Pythagoras Theorem twice. v = x2 + y 2 + z 2 v = 3 + 2 +1 2 2 2 z O y v 2 3 1 v = 14 x Ad d itio n o f Ve c to rs Higher Outcome 1 Addition of vectors www.m a th s re vis io n .c o m 3 2 Let u = and v = 5 4 - 2 1 Then u + v 3 2 5 4 + - = - 5 1 2 3 1 Addition of Vectors Higher Outcome 1 In general we have www.m a th s re vis io n .c o m For vectors u a nd v a d If u = and v = then b e c f a u + v = + b c d e = f a +d b +e + f c Ne g a tive Ve c to r Higher Outcome 1 www.m a th s re vis io n .c o m uuu r uuu r BA is the negative of AB Ne g a tive ve c to r Fo r a ny ve c to r u u + ( -u ) = u + ( -u ) = 0 a - a If u = then - u = b b - c c - S ub tra c tio n o f Ve c to rs Higher Outcome 1 Subtraction of vectors www.m a th s re vis io n .c o m 6 2 Let a = and b = 5 4 3 2 Then a - b 6 2 4 = 5 - 4 = 1 3 2 1 Subtraction of Vectors Higher Outcome 1 For vectors u a nd v www.m a th s re vis io n .c o m a d If u = and v = then b e c f a u - v = - b c d e = f a -d b-e - f c Multip lic a tio n b y a S c a la r Higher Outcome 1 Multiplication by a scalar ( a number) www.m a th s re vis io n .c o m x kx If a vector v = then kv = y ky z kz The vector kv is parallel to vector v ( different size ) Hence if u = kv then u is parallel to v Conversely if u is parallel to v then u = kv Multiplication by a Scalar Higher Outcome 1 Show that the two vectors are parallel. www.m a th s re vis io n .c o m 6 12 w = then z = 9 18 12 24 If z = kw then z is parallel to w 12 6 z = = 2 18 9 24 12 z = 2w Position Vectors Higher Outcome 1 www.m a th s re vis io n .c o m uuu r The position vector of a 3D point A is OA, usually written as a 3 uuu r OA = a = 2 z 1 O y a A (3 ,2 ,1 ) 1 2 3 x P o s itio n Ve c to rs Higher Outcome 1 www.m a th s re vis io n .c o m uur u If R is (2,-5,1) and S is (4,1,-3) then RS = s - r 4 2 2 = 1 5 6 - - = 3 1 4 - - General Section Formula Higher Outcome 1 Summarising we have www.m a th s re vis io n .c o m If p is a p o s itio n ve c to r o f th e p o int P th a t d ivid e s m A n P B n m pth e n = a+ b m+ n m+ n AB in th e ra tio m : n The scalar product Higher Outcome 1 The scalar product is defined as being: tail to tail www.m a th s re vis io n .c o m Must be a = a b cos b 0 180 0 a b The Scalar Product Higher Outcome 1 Find the scalar product for a and b when www.m a th s re vis io n .c o m |a|= 4 , |b|= 5 when (a) = 45o (b) = 90o a = a b cos b 2 2 = 4 5cos 45 2 = 10 2 o 20 10 a= b = 2 The Scalar Product Higher Outcome 1 Find the scalar product for a and b when www.m a th s re vis io n .c o m |a|= 4 , |b|= 5 when (a) = 45o (b) = 90o a = a b cos b = 4 5cos 90 o a = 20 = 0 b 0 Important : If a and b are perpendicular then a . b = 0 Component Form Scalar Product Higher Outcome 1 www.m a th s re vis io n .c o m If a b 1 1 a = 2 and b = 2 then a b a b 3 3 a = a1b1 + a2b2 + a3b3 b Angle between Vectors Higher Outcome 1 www.m a th s re vis io n .c o m To find the angle between two vectors we simply use the scalar product formula rearranged a b cos = a b or a1b1 + a2b2 + a3b3 cos = a b Angle between Vectors Higher Outcome 1 Find the angle between the two vectors below. www.m a th s re vis io n .c o m p = 3i +2j+5k and q = 4i + j+3k 3 4 p = and q = 2 1 5 3 a b cos = a b p = 3 + 2 + 5 = 38 2 2 2 q = 42 + 12 + 32 = 26 Angle between Vectors Higher Outcome 1 Find the angle between the two vectors below. www.m a th s re vis io n .c o m p = 3 4 + 2 + 5 3 = 29 q 1 p = 32 + 22 + 52 = 38 q = 42 + 12 + 32 = 26 a b cos = = a b -1 29 = 0.923 38 26 o = cos (0.923) = 22.7 Perpendicular Vectors Higher Outcome 1 Show that for a . b =0 www.m a th s re vis io n .c o m 1 3 a = 2 and b = 2 1 - 7 a and b are perpendicular a = a1b1 + a2b2 + a3b3 b a = 3 + 2 2 + 7 ( - 1) b 1 a = 3 + 4 + (-7) = 0 b Perpendicular Vectors Higher Outcome 1 Given a 0 and b 0 and a = 0 b www.m a th s re vis io n .c o m Then a b 0 cos = = =0 a b a b = cos (0) = 90 -1 o If a . b = 0 then a and b are perpendicular Properties of a Scalar Product Higher Outcome 1 Two properties that you need to be aware of www.m a th s re vis io n .c o m a = b b a a + c ) = a + a (b b c Higher Maths Vectors Strategies Click to start Vectors Higher The following questions are on Vectors Non-calculator questions will be indicated You will need a pencil, paper, ruler and rubber. Click to continue Vectors Higher The questions are in groups General vector questions (15) Points dividing lines in ratios Collinear points (8) Angles between vectors (5) Quit Quit Vectors Higher General Vector Questions Continue Quit Quit Back to menu Vectors Higher Vectors u and v are defined by u = 3i + 2 j and v = 2i - 3 j + 4k Determine whether or not u and v are perpendicular to each other. Is Scalar product = 0 2 3 u.v = -3 2 g 4 0 u.v = 3 2 + 2 -3 + 0 4 u.v = 0 u.v = 6 + ( -6 ) + 0 Hence vectors are perpendicular Hint Previous Quit Quit Next Vectors Higher t u = and -2 3 For what value of t are the vectors v 2 = perpendicular ? 10 t Put Scalar product = 0 t 2 u.v = 2 10 - g t 3 u.v = 2t + ( -2 ) 10 + 3t Perpendicular u.v = 0 u.v = 5t - 20 0 = 5t - 20 t=4 Hint Previous Quit Quit Next Vectors VABCD is a pyramid with rectangular base ABCD. uuu uuu r r uuu r The vectors AB, AD and AV are given by uuu r uuu r AB = 8i + 2 j + 2k AD = -2i + 10 j - 2k uuu r AV = i + 7 j + 7k Higher uuu r Express CV in component form. Ttriangle rule ACV Triangle rule ABC uuu uuu uuu r r r uuu uuu uuu r r r Re-arrange CV = AV - AC AC + CV = AV uuu uuu r r uuu uuu uuu r r r AB + BC = AC also BC = AD uuu uuu uuu uuu r r r r CV = AV - AB + AD ( ) Quit uuu r CV = -9i - 5 j + 7 k Previous 1 8 - 2 uuu r CV = 7 - 2 - 10 2 7 2 - - 5 uuu r CV = 5 - 7 Hint Quit Next Vectors The diagram shows two vectors a and b, with | a | = 3 and | b | = 22. These vectors are inclined at an angle of 45 to each other. a) Evaluate i) a.a ii) b.b iii) a.b b) Another vector p is defined by p = 2a + 3b Evaluate p.p and hence write down | p |. Higher i) a a = a a cos 0 3 3 1= 9 3 2 2 ii) b b=2 2 2 2 =8 iii) a b = a b cos 45 b) 1 =6 2 p p = ( 2a + 3b ) a + 3b ) (2 36 + 72 + 72 Previous 4a.a + 12a.b + 9b.b p = 180 6 5 Next Hint 180 Since p.p = p2 Quit Quit Vectors Vectors p, q and r are defined by a) b) c) Express Calculate p.r Find |r| p - q + 2r Higher p = i + j - k, q = i + 4k , and r = 4i - 3 j p - q +in component form 2r a) b) c) ( i + j - k ) - ( i + 4k ) + 2 ( 4i - 3 j ) 8i - 5 j - 5k p.r = 1 p.r = ( i + j - k ) . ( 4i - 3 j ) r = 42 + (-3) 2 p.r = 1 4 + 1 ( -3) + (-1) 0 r =5 r = 16 + 9 Hint Previous Quit Quit Next Vectors Higher The diagram shows a point P with co-ordinates (4, 2, 6) and two points S and T which lie on the x-axis. If P is 7 units from S and 7 units from T, find the co-ordinates of S and T. Use distance formula S (a, 0, 0) T (b, 0, 0) 49 = (4 - a ) 2 + 40 9 = (4 - a ) 2 PS 2 = 49 = (4 - a) 2 + 2 2 + 6 2 a=4 3 a = 7 or a = 1 hence there are 2 points on the x axis that are 7 units from P i.e. S and T S (1, 0, 0) and T (7, 0, 0) Hint Previous Quit Quit Next Vectors The position vectors of the points P and Q are p = i +3j+4k and q = 7 i j + 5 k respectively. uuu r a) Express PQ in component form. b) Find the length of PQ. Higher a) uuu r PQ = q - p 7 - uuu 1 r PQ = 1 - - 3 5 4 uuu 8 r PQ = 4 = 8 i - 4 j + k - 1 b) uuu r PQ = 82 + (-4) 2 + 12 = 64 + 16 + 1 = 81 =9 Hint Previous Quit Quit Next Vectors Higher P PQR is an equilateral triangle of side 2 units. uuu r uuu r uuu r PQ = a , PR = b, and QR = c Evaluate a.(b + c) and hence identify two vectors which are perpendicular. a Diagram Q 60 60 b 60 a.(b + c ) = a.b + a.c a.b = a b cos 60 a.c = a c cos120 Hence c R a.b = 2 2 a.c = 2 2 so, 1 2 a.b = 2 NB for a.c vectors must point OUT of the vertex ( so angle is 120 ) 1 - 2 a.c = - 2 Hint a.(b + c ) = 0 Table of Exact Values a is perpendicular to b + c Quit Previous Quit Next Vectors Higher Calculate the length of the vector 2i 3j + 3k Length = 2 + (-3) + 2 2 ( 3) 2 4+9+3 16 4 Hint Previous Quit Quit Next Vectors 1 and 2 1 - 4 - 3 are -1 k Higher Find the value of k for which the vectors perpendicular Put Scalar product = 0 -4 1 0= 3 2 g 1 k -1 - 0 = -4 + 6 - ( k - 1) k =3 0 = 2 - k +1 Hint Previous Quit Quit Next Vectors Higher A is the point (2, 1, 4), B is (7, 1, 3) and C is (6, 4, 2). If ABCD is a parallelogram, find the co-ordinates of D. uuu uuu r r AD = BC = c - b D is the displacement - 7 uuu 6 r BC = 4 - 1 uuu r AD 2 3 - uuu 13 r BC = 3 1 - from A 11 - = 2 3 hence d 13 2 - = 1 3 - + 1 4 - d D ( -11, 2, 3) Hint Previous Quit Quit Next Vectors Higher If u 3 - = 3 and 3 v 1 = write 5 1 - down the components of u + v and u v Hence show that u + v and u v are perpendicular. u+v 2 - = 8 2 u-v 2 4 - - = 8 2 - 2 4 4 - = 2 - 4 look at scalar product ( u + v ) .( u - v ) ( u + v ) .( u - v ) = (-2) (-4) + 8 (-2) + 2 4 = 8 - 16 + 8 =0 Hint Hence vectors are perpendicular Previous Quit Quit Next Vectors Higher The vectors a, b and c are defined as follows: a = 2i k, b = i + 2j + k, c = j + k a) Evaluate a.b + a.c b) From your answer to part (a), make a deduction about the vector b + c a) 2 1 = 0 2 1 - 1 2 0 = 1 0 - 1 - 1 a.b a.b = 2 + 0 - 1 a.b = 1 a.c b) a.c = 0 + 0 - 1 a.c = -1 a.b + a.c = 0 b + c is perpendicular to a Hint Previous Quit Quit Next Vectors Higher A is the point ( 3, 2, 4 ) and B is ( 1, 3, 2 ) Find: uuu r a) the components of AB b) the length of AB uuu r a) AB = b - a - - uuu 1 3 r AB = 2 3 - 2 4 uuu 2 r AB = 1 2 - b) AB = 22 + 12 + (-2) 2 AB = 9 AB = 4 + 1 + 4 AB = 3 Hint Previous Quit Quit Next Vectors Higher In the square based pyramid, all the eight edges are of length 3 units. uuu r uuu r uuu r AV = p, AD = q, AB = r , Evaluate p.(q + r) Triangular faces are all equilateral p.(q + r ) = p.q + p.r p.q = p q cos 60 p.r = p r cos 60 p.(q + r ) = 4 + 4 Previous p.q = 3 3 p.r = 3 3 1 2 1 2 1 2 p.q = 4 p.q = 1 2 1 4 2 Hint 1 2 p.(q + r ) = 9 Quit Quit Table of Next Vectors Higher You have completed all 15 questions in this section Previous Quit Quit Back to start Vectors Higher Points dividing lines in ratios Collinear Points Continue Quit Quit Back to menu Vectors Higher A and B are the points (-1, -3, 2) and (2, -1, 1) respectively. B and C are the points of trisection of AD. That is, AB = BC = CD. Find the coordinates of D uuu r AB 1 uuu = r AD 3 uuu uuu r r 3AB = AD 3( b - a ) = d - a 3b - 3a = d - a d 1 2 - = 3 1 2 3 - - - 1 2 Previous d = 3b - 2a d 8 = 3 1 - Quit D(8, 3, 1) Hint Quit Next Vectors Higher The point Q divides the line joining P(1, 1, 0) to R(5, 2 3) in the ratio 2:1. Find the co-ordinates of Q. Diagram P 2 Q 1 R uuu r PQ 2 uuu = r QR 1 uuu r uuu r PQ = 2QR q - p = 2r - 2q 3q = 2r + p 1 9 q = 3 3 6 - Hint 3q = 1 5 - 2 2 1 + - 3 - 0 Q(3, 1, - 2) Previous Quit Quit Next Vectors a) Roadmakers look along the tops of a set of T-rods to ensure that straight sections of road are being created. Relative to suitable axes the top left corners of the T-rods are the points A(8, 10, 2), B(2, 1, of road ABC has Determine whether or not the section1) and C(6, 11, 5).been built in a straight line. b) A further T-rod is placed such that D has co-ordinates (1, 4, 4). Higher 6 2 14 2 uuu uuu r r uuu r Show that DB is perpendicular to AB. a) AB = b - a AB = 3 AC = 21 = 7 3 9 = 3 b) uuu r uuu r AB and AC are scalar multiples, so are parallel. A is common. A, B, C are collinear 6 uuu uuu 3 uuu uuu r r r r Use scalar product AB.BD = 3 AB.BD = 18 - 27 + 9 = 0 9 . - 3 3 3 1 7 1 Hence, DB is perpendicular to AB Next Hint Previous Quit Quit Vectors VABCD is a pyramid with rectangular base ABCD. Relative to some appropriate axis, uu r VA uuu represents 7i 13j 11k r AB represents 6i + 6j 6k uuu r 8i 4j 4k AD represents K divides BC in the ratio 1:3 Higher uuu r Find VK in component form. uu uuu uur r r VA + AB = VB uuu uur uuu r r VK = VB - KB uuu uuu uur r r VK + KB = VB 4 uuu 1 uuu 1 uuu r r r r 1 uuu KB = CB = DA = - AD 4 4 4 uuu uu uuu 1 uuu r r r r VK = VA + AB + AD uuu 1 r VK = -8 18 - Hint - uuu 7 6 1 8 r VK = 13 6 4 - + + - 11 6 4 4 - - - Quit Previous Quit Next Vectors The line AB is divided into 3 equal parts by the points C and D, as shown. A and B have co-ordinates (3, 1, 2) and (9, 2, 4). uuu r uuu r AB and AC a) Find the components of b) Find the co-ordinates of C and D. Higher a) uuu r AB = b - a uuu 6 r AB = 3 6 - uuu 1 uuu 2 r r AC = AB = 1 3 2 - b) uuu r C is a displacement of AC from A similarly c d 5 2 = 1 0 + 2 0 - 3 2 = 1 1 - + 2 2 - C (5, 0, 0) D(7, 1, - 2) Hint Quit Previous Quit Next Vectors Higher Relative to a suitable set of axes, the tops of three chimneys have co-ordinates given by A(1, 3, 2), B(2, 1, 4) and C(4, 9, 8). Show that A, B and C are collinear uuu r AB = b - a uuu 1 r AB = 4 - 2 uuu 3 1 r AC = 12 3 4 - = - 6 2 uuu r uuu r AB and AC are scalar multiples, so are parallel. A is common. A, B, C are collinear Hint Previous Quit Quit Next Vectors Higher A is the point (2, 5, 6), B is (6, 3, 4) and C is (12, 0, 1). Show that A, B and C are collinear and determine the ratio in which B divides AC uuu r AB = b - a uuu r uuu r AB and BC uuu r AB 2 uuu = r BC 3 uuu 4 2 r AB = 2 2 = 1 2 - 1 - uuu 6 2 r BC = 3 3 = 1 3 - 1 - are scalar multiples, so are parallel. B is common. A, B, C are collinear 3 C A 2 B B divides AB in ratio 2 : 3 Quit Quit Hint Previous Next Vectors Higher Relative to the top of a hill, three gliders have positions given by R(1, 8, 2), S(2, 5, 4) and T(3, 4, 6). Prove that R, S and T are collinear uuu r RS = s - r uuu r uuu r RS and RT 3 1 uuu r RS = 3 3 = 1 6 2 4 1 uuu r RT = 4 4 = 1 8 2 are scalar multiples, so are parallel. R is common. R, S, T are collinear Hint Previous Quit Quit Next Vectors Higher You have completed all 8 questions in this section Previous Quit Quit Back to start Vectors Higher Angle between two vectors Continue Quit Quit Back to menu Vectors Higher The diagram shows vectors a and b. If |a| = 5, |b| = 4 and a.(a + b) = 36 Find the size of the acute angle between a and b. cos = a.b a b a.(a + b) = 36 25 + a.b = 36 11 = cos 20 -1 a.a + a.b = 36 a.b = 11 a.a = a a = 25 11 cos = 5 4 Previous = 56.6 Hint Quit Quit Next Vectors The diagram shows a square based pyramid of height 8 units. Square OABC has a side length of 6 units. The co-ordinates of A and D are (6, 0, 0) and (3, 3, 8). C lies on the y-axis. a) Write down the co-ordinates of B uuu r uuu r DA and DB b) Determine the components of c) Calculate the size of angle ADB. uuu r 3 a) b) DA = 3 B(6, 6, 0) - 8 - Higher uuu 3 r DB = 3 8 - c) uuu uuu r r DA.DB cos = uuu uuu r r DA DB 64 82 82 uuu uuu 3 3 r r DA.DB = 3 3 64 - . = 8 8 - - cos = Previous = 38.7 Quit Quit Hint Next Vectors A box in the shape of a cuboid designed with circles of different sizes on each face. The diagram shows three of the circles, where the origin represents one of the corners of the cuboid. The centres of the circles are A(6, 0, 7), B(0, 5, 6) and C(4, 5, 0) 6 Find the size of angle ABC uuu r Vectors to point BA = 5 - 1 away from vertex 4 uuu r BC = 0 6 - Higher uuu uuu r r BA.BC = 24 + 0 - 6 = 18 uuu r BA = 36 + 25 + 1 = 62 uuu r BC = 16 + 36 = 52 cos = 18 62 52 = 71.5 Quit Quit Hint Previous Next Vectors A cuboid measuring 11cm by 5 cm by 7 cm is placed centrally on top of another cuboid measuring 17 cm by 9 cm by 8 cm. Co-ordinate axes are taken as shown. a) The point A has co-ordinates (0, 9, 8) and C has co-ordinates (17, 0, 8). Write down the co-ordinates of B 15 b) Calculate the size of angle ABC. uuu r a) B (3, 2, 15) b) BC = 2 - 7 - - 3 uuu r BA = 7 7 - Higher uur uuu u r BA.BC = -45 - 14 + 49 = - 10 uuu r BC = 225 + 4 + 49 = 278 = 93.3 Hint uuu r BA = 9 + 49 + 49 = 107 Previous -10 cos = 278 107 Quit Quit Next Vectors A triangle ABC has vertices A(2, 1, 3), B(3, 6, 5) and C(6, 6, 2). a) Higher uuu r Find AB and uuu r AC b) Calculate the size of angle BAC. c) Hence find the area of the triangle. 1 4 uuu r uuu r 7 7 a) AB = b - a = AC = c - a = 5 2 - b) uuu r AB = 12 + 7 2 + 22 = 54 uuu r AC = 90 uuu uuu r r AB. AC = 4 + 49 - 10 = 43 cos = 43 = 0.6168 54 90 = cos -1 0.6168 = 51.9 = 1 90 54 sin 51.9 2 BAC = 51.9 = c) Area of ABC = 1 ab sin C 2 27.43 unit 2 Hint Previous Quit Quit Next Vectors Higher You have completed all 5 questions in this section Previous Quit Quit Back to start ...
View Full Document

This note was uploaded on 02/13/2012 for the course MAT 205 math 205 taught by Professor Google during the Spring '10 term at University of Phoenix.

Ask a homework question - tutors are online