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Higher Unit 3 www.mathsrevision.com Higher Outcome 1 Vectors and Scalars Properties of vectors Adding / Sub of vectors Multiplication by a Scalar Unit Vector 3D Vectors Properties 3D Section formula Scalar Product Component Form Position Vector Collinearity Section Formula Exam Type Questions www.mathsrevision.com Angle between vectors Perpendicular Properties of Scalar Product Vectors & Scalars Outcome 1 www.mathsrevision.com Higher A vector is a quantity with BOTH magnitude (length) and direction. Examples : Gravity Velocity Force Vectors & Scalars Outcome 1 www.mathsrevision.com Higher A scalar is a quantity that has magnitude ONLY. Examples : Time Speed Mass Vectors & Scalars Outcome 1 www.mathsrevision.com Higher A vector is named using the letters at the end of the directed line segment or using a lowercase bold / underlined letter u This vector is named AB u AB or u or u Vectors & Scalars Outcome 1 Higher Also known as column vector www.mathsrevision.com A vector may also be represented in component form. w 4 CD w 2 x AB d y 2 FE z 1 z Magnitude of a Vector www.mathsrevision.com Higher Outcome 1 A vector’s magnitude (length) is represented by PQ or u A vector’s magnitude is calculated using Pythagoras Theorem. u PQ u a 2 b2 Vectors & Scalars Outcome 1 Higher www.mathsrevision.com Calculate the magnitude of the vector. w a2 b2 w 4 CD w 2 w 42 22 w 20 w 4 5 2 5 Vectors & Scalars Outcome 1 Higher www.mathsrevision.com Calculate the magnitude of the vector. w a2 b2 w (4) 2 32 4 PQ 3 w 16 9 w 5 Equal Vectors Outcome 1 Higher www.mathsrevision.com Vectors are equal only if they both have 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 Outcome 1 www.mathsrevision.com Higher By working out the components of each of the vectors determine which are equal. a b c d g e f h Addition of Vectors Outcome 1 Higher www.mathsrevision.com Any two vectors can be added in this way b Arrows must be nose to tail 2 a 4 6 3 8 a + b 1 Addition of Vectors Outcome 1 Higher www.mathsrevision.com Addition of vectors 3 2 Let AB and BC 4 5 B Then AB + BC AC 3 2 5 + 4 5 1 A C 5 So AC 1 Addition of Vectors Higher Outcome 1 www.mathsrevision.com In general we have For vectors u and v a c If u and v then b d a c a c u + v + = b d b d Zero Vector Outcome 1 Higher www.mathsrevision.com The zero vector 1 If AB 2 1 then BA 2 1 1 0 AB + BA + = 2 2 0 0 is called the zero vector, written 0 0 Negative Vector Higher Outcome 1 www.mathsrevision.com Negative vector BA is the negative of AB For any vector u u u u u 0 a a If u then u b b Subtraction of Vectors Outcome 1 Higher www.mathsrevision.com Any two vectors can be subtracted in this way Notice arrows nose to nose 6 u 3 5 u - v 0 1 3 v Subtraction of Vectors Outcome 1 Higher www.mathsrevision.com Subtraction of vectors 6 2 Let a and b 5 4 Notice arrows nose to nose a a-b Then a b a (b) b 6 2 4 5 4 1 6 5 2 4 - 4 1 Subtraction of Vectors Higher Outcome 1 www.mathsrevision.com In general we have For vectors u and v a c If u and v then b d a c a c u v = b d b d Multiplication by a Scalar Higher Outcome 1 www.mathsrevision.com Multiplication by a scalar ( a number) 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 Outcome 1 Higher www.mathsrevision.com Multiplication by a scalar Write down a vector b a parallel to a Write down a vector parallel to b Multiplication by a Scalar www.mathsrevision.com Higher Outcome 1 Show that the two vectors are parallel. 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 www.mathsrevision.com Higher Outcome 1 Alternative method. 6 18 w then z 9 27 If w = kz then w is parallel to z 6 1 18 w = 3 27 9 1 w z 3 Unit Vectors www.mathsrevision.com Higher Outcome 1 For ANY vector v there exists a parallel vector u of magnitude 1 unit. This is called the unit vector. Unit Vectors u Outcome 1 Higher www.mathsrevision.com v Find the components of the unit vector, u , parallel to vector v , if 3 v 4 v 3 4 2 v 5 2 So the unit vector is u 1 3 u 5 4 3 5 4 5 A Position Vectors Outcome 1 Higher www.mathsrevision.com B A is the point (3,4) and B is the point (5,2). Write down the components of 3 OA 4 OB OB OA 5 2 AB 5 3 2 2 4 2 2 2 Answers the same ! A Position Vectors Higher a Outcome 1 www.mathsrevision.com 0 B b OA is called the position vector of the point A relative to the origin O, written as a OB is called the position vector of the point B relative to the origin O, written as b AB AO OB a b b a A Position Vectors Higher a Outcome 1 www.mathsrevision.com 0 B b AB b a where a and b are the position vectors of A and B. Position Vectors www.mathsrevision.com Higher Outcome 1 If P and Q have coordinates (4,8) and (2,3) respectively, find the components of PQ 2 4 2 PQ q p 3 8 5 Position Vectors Outcome 1 www.mathsrevision.com Higher P Graphically P (4,8) Q (2,3) p 2 PQ = 5 q-p Q q O Collinearity Outcome 1 www.mathsrevision.com Higher Reminder from chapter 1 Points are said to be collinear if they lie on the same straight line. For vectors If AB kBC , where k is a scalar, then AB is parallel to BC. If B is also a point common to both AB and BC then A,B and C are collinear. Collinearity www.mathsrevision.com Higher Outcome 1 Prove that the points A(2,4), B(8,6) and C(11,7) are collinear. 8 2 6 AB b a 6 4 2 11 8 3 BC c b 7 6 1 6 3 AB 2 2BC 2 1 Collinearity www.mathsrevision.com Higher Outcome 1 Since AB 2BC , AB is parallel to BC. B is a point common to both AB and BC so A, B and C are collinear. Section Formula Outcome 1 www.mathsrevision.com Higher B OS OA AS 3 1 OS OA AB 3 1 s a (b a) 3 2 1 s a b 3 3 1 2 S s A a b O General Section Formula Outcome 1 www.mathsrevision.com Higher B OP OA AP m+n m OP OA AB mn m p a (b a ) mn n m p a b mn mn m n P p A a b O General Section Formula Outcome 1 www.mathsrevision.com Higher Summarising we have B n If p is a position vector of the point P that divides AB in the ratio m : n then n m p a b mn mn A m P General Section Formula Higher Outcome 1 www.mathsrevision.com A and B have coordinates (-1,5) and (4,10) respectively. P Find P if AP : PB is 3:2 2 3 p a b 5 5 2 1 3 4 p 5 5 5 10 2 3 A 2 12 5 5 2 6 10 5 2 8 8 B 3D Coordinates Outcome 1 www.mathsrevision.com Higher In the real world points in space can be located using a 3D coordinate system. 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 Outcome 1 Higher www.mathsrevision.com Write down the coordinates for the 7 vertices z (0, 0, 2) F (0,0, 0) G E(0, 1, 2) H O y (6, 0, 2)B 6 A(6, 1, 2) 2 1 D(6, 1, 0) C (6, 0, 0) What is the coordinates of the vertex H so that it makes a cuboid shape. x H(0, 1, 0 ) 3D Vectors Outcome 1 www.mathsrevision.com Higher 3D vectors are defined by 3 components. For example, the velocity of an aircraft taking off can be illustrated by the vector v. z (7, 3, 2) 2 v y 2 3 O 3 7 7 x 3D Vectors Outcome 1 Higher www.mathsrevision.com Any vector can be represented in terms of the i , j and k Where i, j and k are unit vectors z in the x, y and z directions. 1 i= 0 0 y k O j i x 0 j = 1 0 0 k=0 1 3D Vectors Outcome 1 Higher www.mathsrevision.com Any vector can be represented in terms of the i , j and k Where i, j and k are unit vectors in the x, y and z directions. z (7, 3, 2) v y 3 O 7 2 v = ( 7i+ 3j + 2k ) x 7 v = 3 2 3D Vectors www.mathsrevision.com Higher Outcome 1 Good News All the rules for 2D vectors apply in the same way for 3D. Magnitude of a Vector Outcome 1 www.mathsrevision.com Higher A vector’s magnitude (length) is represented by v A 3D vector’s magnitude is calculated using Pythagoras Theorem twice. v x2 y 2 z 2 v 3 2 1 2 2 v 14 2 z v y 1 2 O 3 x Addition of Vectors Higher Outcome 1 www.mathsrevision.com Addition of vectors 3 2 Let u 4 and v 5 1 2 Then u + v 3 2 5 4 + 5 1 1 2 3 Addition of Vectors Higher Outcome 1 www.mathsrevision.com In general we have For vectors u and v a d If u b and v e then f c a d a d u + v b + e = b e c f c f Negative Vector Higher Outcome 1 www.mathsrevision.com Negative vector BA is the negative of AB For any vector u u u u u 0 a a If u b then u b c c Subtraction of Vectors Higher Outcome 1 www.mathsrevision.com Subtraction of vectors 6 2 Let a 5 and b 4 2 3 Then a b 6 5 3 2 4 - 4 1 2 1 Subtraction of Vectors Higher Outcome 1 www.mathsrevision.com For vectors u and v a d If u b and v e then c f a d a d u v b e = b e c f c f Multiplication by a Scalar Higher Outcome 1 www.mathsrevision.com Multiplication by a scalar ( a number) x kx If a vector v y then kv ky kz z 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 www.mathsrevision.com Higher Outcome 1 Show that the two vectors are parallel. 6 12 w 9 then z 18 24 12 If z = kw then z is parallel to w 12 6 z 18 = 2 9 24 12 z 2w Position Vectors www.mathsrevision.com Higher Outcome 1 The position vector of a 3D point A is OA, usually written as a 3 OA = a = 2 1 A (3,2,1) z a y 1 2 O 3 x Position Vectors www.mathsrevision.com Higher Outcome 1 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 Outcome 1 www.mathsrevision.com Higher Summarising we have B n If p is a position vector of the point P that divides AB in the ratio m : n then n m p a b mn mn A m P The scalar product Higher Outcome 1 Must be www.mathsrevision.com The scalar product is defined being: tail as to tail a a b a b cos 0 180 0 θ b The Scalar Product www.mathsrevision.com Higher Outcome 1 Find the scalar product for a and b when |a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o a b a b cos = 4 5cos 45 20 10 2 2 a b 10 2 2 2 o The Scalar Product www.mathsrevision.com Higher Outcome 1 Find the scalar product for a and b when |a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o a b a b cos = 4 5cos90 o a b 20 0 0 Important : If a and b are perpendicular then a.b=0 Component Form Scalar Product Outcome 1 www.mathsrevision.com Higher If a1 b1 a a2 and b b2 then a b 3 3 a b a1b1 a2b2 a3b3 Angle between Vectors Outcome 1 www.mathsrevision.com Higher 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 = ab Angle between Vectors www.mathsrevision.com Higher Outcome 1 Find the angle between the two vectors below. p 3i +2j+5k and q 4i +j+3k 3 4 p 2 and q 1 5 3 a b cos = a b p 32 22 52 38 q 42 12 32 26 Angle between Vectors Outcome 1 www.mathsrevision.com Higher Find the angle between the two vectors below. p q 3 4 2 1 5 3 29 a b cos = = a b q 42 12 32 26 29 = 0.923 38 26 = cos (0.923) = 22.7 -1 p 32 22 52 38 o Perpendicular Vectors Outcome 1 www.mathsrevision.com Higher Show that for a.b =0 3 1 a 2 and b 2 1 7 a and b are perpendicular a b a1b1 a2b2 a3b3 a b 31 2 2 7 (1) a b 3 4 (-7) 0 Perpendicular Vectors Outcome 1 www.mathsrevision.com Higher Given a 0 and b 0 and a b 0 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 www.mathsrevision.com Two properties that you need to be aware of a b ba a (b c) a b a 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 3 u.v 2 0 u.v 3 2 2 3 0 4 u.v 0 2 3 4 u.v 6 6 0 Hence vectors are perpendicular Hint Previous Quit Quit Next Vectors Higher For what value of t are the vectors Put Scalar product = 0 t u 2 3 t u.v 2 3 u.v 2t 2 10 3t Perpendicular u.v = 0 and v 2 10 t perpendicular ? 2 10 t u.v 5t 20 0 5t 20 t4 Hint Previous Quit Quit Next Vectors Higher VABCD is a pyramid with rectangular base ABCD. The vectors AB, AD and AV are given by AB 8i 2 j 2k AD 2i 10 j 2k AV i 7 j 7k Express CV in component form. Ttriangle rule ACV AC CV AV Triangle rule ABC AB BC AC also CV AV AB AD Re-arrange CV AV AC BC AD 1 8 2 CV 7 2 10 7 2 2 5 CV 5 7 CV 9i 5 j 7k Previous Quit Hint Quit Next Vectors Higher 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 |. i) a a a a cos0 3 3 1 9 iii) a b a b cos 45 3 2 2 b) p p 2a 3b 2a 3b 36 72 72 180 Previous 8 1 6 2 4a.a 12a.b 9b.b Since p.p = p2 Quit bb 2 2 2 2 ii) Quit p 180 6 5 Next Hint Vectors Higher Vectors p, q and r are defined by p i j - k, a) Express p q 2r in component form b) Calculate p.r c) Find |r| q i 4k , i j - k i 4k 2 4i 3 j and r 4i 3 j 8i 5 j - 5k a) p q 2r b) p.r i j - k . 4i 3 j p.r 1 4 1 (3) (1) 0 p.r 1 c) r 42 (3) 2 r 16 9 r 5 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) PS 2 49 (4 a)2 22 62 a 43 T (b, 0, 0) 49 (4 a)2 40 9 (4 a)2 a 7 or a 1 hence there are 2 points on the x axis that are 7 units from P S (1, 0, 0) and i.e. S and T T (7, 0, 0) Hint Previous Quit Quit Next Vectors Higher The position vectors of the points P and Q are p = –i +3j+4k and q = 7 i – j + 5 k respectively. a) Express PQ in component form. b) Find the length of PQ. a) b) PQ q - p PQ 7 1 1 - 3 5 4 PQ 82 (4)2 12 PQ 64 16 1 8 4 1 81 8i 4 j k 9 Hint Previous Quit Quit Next Vectors Higher PQR is an equilateral triangle of side 2 units. P PQ a, PR b, and QR c Evaluate a.(b + c) and hence identify a b Diagram two vectors which are perpendicular. Q a.(b c) a.b a.c a.b a b cos60 60° a.b 2 2 1 2 60° 60° c R a.b 2 NB for a.c vectors must point OUT of the vertex ( so angle is 120° ) 1 a.c 2 2 a.c a c cos120 2 Hence a.(b c) 0 a.c 2 so, a is perpendicular to b + c Hint Previous Table of Exact Values Quit Quit Next Vectors Higher Calculate the length of the vector 2i – 3j + 3k Length 2 (3) 2 2 3 2 493 16 4 Hint Previous Quit Quit Next Vectors Higher Find the value of k for which the vectors Put Scalar product = 0 1 2 1 1 0 2 1 and 4 3 k 1 are perpendicular 4 3 k 1 0 4 6 (k 1) 0 2 k 1 k 3 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. AD BC c b D is the displacement hence d BC AD 2 13 1 3 4 1 6 7 4 1 2 3 BC 13 3 1 from A d 11 2 3 D 11, 2, 3 Hint Previous Quit Quit Next Vectors If u 3 3 3 Higher and v 1 5 1 write down the components of u + v and u – v Hence show that u + v and u – v are perpendicular. uv 2 8 2 u v . u v uv 2 4 8 2 2 4 4 2 4 look at scalar product u v . u v (2) (4) 8 (2) 2 4 8 16 8 0 Hence vectors are perpendicular Previous Quit Quit Hint 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 a.b 0 2 1 1 a.b 2 0 1 a.b 1 2 0 0 1 1 1 a.c 0 0 1 a.c 1 a.c b) 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: a) the components of AB b) the length of AB a) b) AB b a AB 1 3 3 2 2 4 AB 22 12 (2) 2 AB 2 1 2 AB 4 1 4 AB 3 AB 9 Hint Previous Quit Quit Next Vectors Higher In the square based pyramid, all the eight edges are of length 3 units. 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 cos60 p.r p r cos60 1 2 p.(q r ) 4 4 Previous 1 2 Table of p.q 1 3 3 2 p.r 1 3 3 2 p.q 4 p.q 1 2 1 4 2 p.(q r ) 9 Quit Quit Hint 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 AB 1 AD 3 3AB AD 3b 3a d a d 2 1 3 1 2 3 1 2 Previous 3 b a d a 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 PQ 2 QR 1 3q P 1 R q p 2r 2q PQ 2QR 5 1 2 2 1 3 0 2 Q 3q 2r p 1 9 q 3 3 6 Q(3, 1, 2) Hint Previous Quit Quit Next Vectors a) Roadmakers look along the tops of a set of T-rods to ensure Higher 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, 1) and C(6, 11, 5). Determine whether or not the section of road ABC has been built in a straight line. b) A further T-rod is placed such that D has co-ordinates (1, –4, 4). Show that DB is perpendicular to AB. a) AB b a AB 6 2 9 3 3 3 1 AC 14 2 21 7 3 7 1 AB and AC are scalar multiples, so are parallel. A is common. A, B, C are collinear b) Use scalar product Previous AB.BD 6 3 9 . 3 3 3 Quit AB.BD 18 27 9 0 Hence, DB is perpendicular to AB Quit Next Hint Vectors VABCD is a pyramid with rectangular base ABCD. Higher Relative to some appropriate axis, VA represents – 7i – 13j – 11k 6i + 6j – 6k AB represents 8i – 4j – 4k AD represents K divides BC in the ratio 1:3 Find VK in component form. VA AB VB VK VB KB VK 1 4 1 4 1 4 VK VA AB AD 7 6 8 1 13 6 4 11 6 4 4 Previous 1 4 KB CB DA AD VK KB VB VK Quit 1 8 18 Quit Hint Next Vectors Higher 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). a) Find the components of AB and AC b) Find the co-ordinates of C and D. a) b) AB b a AB 6 3 6 C is a displacement of AC from A similarly Previous d 5 2 0 1 0 2 2 1 AC AB 1 3 2 c 3 2 1 1 2 2 C (5, 0, 0) D(7, 1, 2) Hint Quit 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 AB b a AB 1 4 2 AC 3 1 12 3 4 6 2 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 AB b a AB and BC AB 2 BC 3 Previous AB 4 2 2 2 1 2 1 BC 6 2 3 3 1 3 1 are scalar multiples, so are parallel. B is common. A, B, C are collinear A 2 B 3 C B divides AB in ratio 2 : 3 Quit Quit Hint 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 RS s r RS and RT RS 3 1 3 3 1 6 2 RT 4 1 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 a a 25 cos 11 5 4 Previous a.(a b) 36 a.a a.b 36 25 a.b 36 11 cos 20 56.6 1 Quit a.b 11 Hint Quit Next Vectors The diagram shows a square based pyramid of height 8 units. Higher 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 b) Determine the components of DA and DB c) Calculate the size of angle ADB. a) c) B(6, 6, 0) cos DA.DB DA DB cos Previous b) 3 DA 3 8 64 82 82 DA.DB DB 3 3 8 3 3 3 . 3 64 8 8 38.7 Quit Hint Quit Next Vectors A box in the shape of a cuboid designed with circles of different Higher 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) Find the size of angle ABC 6 Vectors to point BA 5 1 away from vertex 4 BC 0 6 BA 36 25 1 62 cos 18 62 52 Previous BA.BC 24 0 6 18 BC 16 36 52 71.5 Hint Quit Quit 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. Higher 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 b) Calculate the size of angle ABC. a) B(3, 2, 15) b) 15 BC 2 7 BA.BC 45 14 49 10 BA 9 49 49 107 Previous 3 BA 7 7 BC 225 4 49 278 10 cos 278 107 Quit Quit 93.3 Hint Next Vectors Higher A triangle ABC has vertices A(2, –1, 3), B(3, 6, 5) and C(6, 6, –2). a) Find AB and AC b) Calculate the size of angle BAC. c) Hence find the area of the triangle. 1 4 AC c a 7 a) AB b a 7 2 5 b) AB 12 72 22 54 cos c) 43 0.6168 54 90 Area of ABC = Previous AC 90 AB.AC 4 49 10 43 cos1 0.6168 51.9 1 ab sin C 2 Quit 1 90 54 sin 51.9 2 Quit BAC = 27.43 51.9 unit 2 Hint Next Vectors Higher You have completed all 5 questions in this section Previous Quit Quit Back to start Are you on Target ! Outcome 1 www.mathsrevision.com Higher • Update you log book • Make sure you complete and correct ALL of the Vector questions in the past paper booklet.