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 be  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
Noncalculator 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 Rearrange 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 coordinates (4, 2, 6) and two points S and T which lie on the xaxis. If P is 7 units from S and 7 units from T, find the coordinates 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 coordinates 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 uv 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 coordinates 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 Trods to ensure that straight sections of road are being created. Relative to suitable axes the top left corners of the Trods 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 Trod is placed such that D has coordinates (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 coordinates (3, 1, 2) and (9, 2, 4). uuu r uuu r AB and AC a) Find the components of b) Find the coordinates 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 coordinates 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 coordinates of A and D are (6, 0, 0) and (3, 3, 8). C lies on the yaxis. a) Write down the coordinates 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. Coordinate axes are taken as shown. a) The point A has coordinates (0, 9, 8) and C has coordinates (17, 0, 8). Write down the coordinates 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 ...
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 Spring '10
 Linear Algebra, Vectors, Vector Space, Dot Product, scalar product

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