Vectors 1C. 6 k) = =0 = =17
|
|
- Susan Stella Williamson
- 5 years ago
- Views:
Transcription
1 Vectors C For each problem, calculate the vector product in the bracet first and then perform the scalar product on the answer. a b c= =4i j 3 a.(b c=(5i+j.(4i j 3 =0 +3= b c a = = 8i+3j+6 b.(c a=(i+j+.( 8i+3j+6 = 8+3+6= c a b= 5 =3i 6j+3 c.(a b=(3i+4.(3i 6j+3 =9+= b c= 3 5 = 8i+8j 8 a.(b c=(i j.( 8i+8j 8 = 8 8+6=0 If a.(b c=0 then a is perpendicular to b c. This means that a is parallel to the plane containing b and c (in fact 3 a= b+ c AB AD= 3 0 = i+ j+ 6 0 Volume of the parallelepiped AE ( AB AD =. = ( i+ j+ 3.( i+ j+ 6 = ++8 =7 Alternatively, volume = e e e 3 b = d d d =(0 (0 +3(6 0=7 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
2 4 Let the vertices A, B, D and E have position vectors a, b, d and e respectively. AB= b a= 4i AD= d a= 3i+ j AE = e a= 3i+ j+ Using the scalar triple product 3 AE. AB AD = 4 0 = 3(0+ 4 ( (8 0 = + 8= 8 3 So volume of parallelpiped = AE. AB AD = 8 5 Let the vertices A, B, C and D have position vectors a, b, c and d respectively. AB= b a= 3i+ j+ AC = c a= j AD= d a= i j+ 3 AB. AC AD = 0 = 3( ( (0 = Volume of tetrahedron = AB. ( AC AD = = a Let the vertices A, B, C and D have position vectors a, b, c and d respectively. BC = c b= i+ j+ BD= d b= j+ BC BD= = i+ 4j Area of face BCD= BC BD= ( 4 = = 3 b The normal to the face BCD is in the direction of BC BD, i.e. in the direction i+ 4j 4 As i+4j 4 = +4 +( 4 =6 The unit vector normal to the face is 6 (i+4j 4= 3 (i+j Multiplying by also gives a vector normal to the face BCD, so (i+j is a solution. 3 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
3 6 c BA= a b= i+ 3j BA. BC BD = i+ j. i+ j = + + = Volume of tetrahedron = AB. ( AC AD = 4 = 7 ( 3 ( a Let the vertices A, B, C and D have position vectors a, b, c and d respectively. (Note a = 0. Now find the length of all edges, as a tetrahedron is regular if all of its edges are the same length. AB= b a= b, so AB = AC = c a= c, so AC = + ( 3 = AD= d a= d = + + = + + = BC = c b= i+ 3 j, so BC = ( + ( 3 = BD, soad d b = + + = = = + +, so BD = ( CD= d c= j+, so CD = + = + = All 6 edges have the same length and so the tetrahedron is regular = 3 0 = = = Volume of tetrahedron = AB. ( AC AD = 4 = b AB. ( AC AD 8 a AB= ( i+ j+ ( i+ j = i j+ 3 AC= ( i j+ ( i+ j = i 3j+ AB AC = 3 = 7i+ 7j+ 7 3 b Area of triangle ABC = AB AC = = = c AO. ( AB AC = ( i j+.(7i+ 7j+ 7 = = 4 Volume of tetrahedron = AO. ( AB AC = 4 = 7 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 3
4 9 a AB= b a= i j+, BC= c b= 3i j AB BC= = 5i j 7 3 BD= d b= i+, DC = c d= i j 4 BD DC= 0 = i 8j+ 4 b i AB BC= BA BC AB BC = BA BC So area of triangle ABC = BA BC = AB BC = 5i+ j+ 7 = = 75= b ii AB BC= BA BC BA BC= AB BC BD. ( BA = ( i+.( 5i+ j+ 7 = 5+ 4= 9 9 So volume of tetrahedron ABCD= BD.( BA = a b c== 3 =i+j 4 5 As a= ( b c,op is perpendicular to OQ and to OR, i.e. OP is perpendicular to the plane OQR. b OP = a = + 4 = 0= 5 Area of OQR = b c = + = 5 Volume of tetrahedron = 3 base height= 3 5 5= 5 3 c Using the scalar triple product: 4 0 a.b ( c = 3 = ( (0 = + 8= a.(b c is 6 volume of tetrahedron (from part b, so result verified. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4
5 a OB OC= (i j 4 ( i+ 4j = 4 = 8i+ j+ 6 4 b OA. ( BA = (3i+ j+.(8i + j+ 6 = = So volume in design units = OA.( OB OC = = Prototype volume = volume in design units = 4 8= cm 3 3 Mass of a prototype model=.3= 7g (to the nearest gram By drawing a rough setch, the three non-parallel edges of the parallelepiped that join at the origin are given by the vectors 0.6i + 0.6j, 0.9i 0.9j, and 0.4i 0.4j The volume in the scientist s scale = = 0.6( (0.9.3 = angstrom (Å = 0. nanometres, so nanometre = 0 Å, and cubic nanometre = 0 3 Å 3 So volume unit cell of crystal in cubic angstroms = = 400Å ( s.f. 3 a The volume of the parallelepiped = AB. ( BD The volume of the tetrahedron = EC. ( EM NC Now from the diagram: EC = AB EM = BD NC= NB+ BC 6 Using these results and the fact that both the vector product and the scalar product are distributive over vector addition, this gives: Volume of the tetrahedron = AB. ( BD ( NB+ = AB. (( BD NB + ( BD as vector product distributive = AB. ( BD NB + AB. ( BD as scalar product distributive But BD NB is perpendicular to AB so AB. ( BD NB = 0 So the expression for the volume of the tetrahedron simplifies to AB ( BD. Hence the ratio of the two volumes is : b The ratio remains unchanged since the argument in part a does not use any data about N other than it lies on the line AB. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5
6 4 Split the pyramid into two tetrahedrons AEDB and CEDB so the volume of the pyramid is just the combined volume of the two tetrahedrons. AD= i+, AB= i+ j+, AE= 4i+ 0 AD. AB AE = = 6= Volume of AEDB = = 4= = CB= i, CD= i j, CE= i j 0 CB. CD CE = = = 4 Volume of CEDB = 4 7 = 4 = = Hence the combined volume = =4 3 Challenge a Let a=a i+a j+a 3 b=b i+ j+ c=c i+c j+c 3 a a a 3 a.b ( c = b b b = a ( bc bc a ( bc bc + a ( bc bc c c c 3 = a ( bc bc + a ( bc bc + a ( bc bc a b=(a i+a j+a 3 (b i+ j+ =a a j a b +a i+a 3 b j a 3 i=(a a 3 i+(a 3 b a j+(a a b (a b.c=((a a 3 i+(a 3 b a j+(a a b.(c i+c j+c 3 =(a a 3 c +(a 3 b a c +(a a b c 3 =a ( c 3 c +a ( c b c 3 +a 3 (b c c So a.(b c=(a b.c b d.(a (b+c=(d a.(b+c =(d a.b+(d a.c =d.(a b+d.(a c =d.(a b+a c applying part a scalar product is distributive over vector addition applying part a scalar product is distributive over vector addition c Since the equality holds for any choice of vector d, it follows that a ( b+c= a b+a c Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6
MULTIPLE PRODUCTS OBJECTIVES. If a i j,b j k,c i k, = + = + = + then a. ( b c) ) 8 ) 6 3) 4 5). If a = 3i j+ k and b 3i j k = = +, then a. ( a b) = ) 0 ) 3) 3 4) not defined { } 3. The scalar a. ( b c)
More informationb UVW is a right-angled triangle, therefore VW is the diameter of the circle. Centre of circle = Midpoint of VW = (8 2) + ( 2 6) = 100
Circles 6F a U(, 8), V(7, 7) and W(, ) UV = ( x x ) ( y y ) = (7 ) (7 8) = 8 VW = ( 7) ( 7) = 64 UW = ( ) ( 8) = 8 Use Pythagoras' theorem to show UV UW = VW 8 8 = 64 = VW Therefore, UVW is a right-angled
More informationCreated by T. Madas VECTOR PRACTICE Part B Created by T. Madas
VECTOR PRACTICE Part B THE CROSS PRODUCT Question 1 Find in each of the following cases a) a = 2i + 5j + k and b = 3i j b) a = i + 2j + k and b = 3i j k c) a = 3i j 2k and b = i + 3j + k d) a = 7i + j
More information9 Mixed Exercise. vector equation is. 4 a
9 Mixed Exercise a AB r i j k j k c OA AB 7 i j 7 k A7,, and B,,8 8 AB 6 A vector equation is 7 r x 7 y z (i j k) j k a x y z a a 7, Pearson Education Ltd 7. Copying permitted for purchasing institution
More informationMark scheme. 65 A1 1.1b. Pure Mathematics Year 1 (AS) Unit Test 5: Vectors. Pearson Progression Step and Progress descriptor. Q Scheme Marks AOs
Pure Mathematics Year (AS) Unit Test : Vectors Makes an attempt to use Pythagoras theorem to find a. For example, 4 7 seen. 6 A.b 4th Find the unit vector in the direction of a given vector Displays the
More informationCreated by T. Madas 2D VECTORS. Created by T. Madas
2D VECTORS Question 1 (**) Relative to a fixed origin O, the point A has coordinates ( 2, 3). The point B is such so that AB = 3i 7j, where i and j are mutually perpendicular unit vectors lying on the
More informationReview exercise 2. 1 The equation of the line is: = 5 a The gradient of l1 is 3. y y x x. So the gradient of l2 is. The equation of line l2 is: y =
Review exercise The equation of the line is: y y x x y y x x y 8 x+ 6 8 + y 8 x+ 6 y x x + y 0 y ( ) ( x 9) y+ ( x 9) y+ x 9 x y 0 a, b, c Using points A and B: y y x x y y x x y x 0 k 0 y x k ky k x a
More information( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios 9E. b Using the line of symmetry through A. 1 a. cos 48 = 14.6 So y = 29.
Trigonometric ratios 9E a b Using the line of symmetry through A y cos.6 So y 9. cos 9. s.f. Using sin x sin 6..7.sin 6 sin x.7.sin 6 x sin.7 7.6 x 7.7 s.f. So y 0 6+ 7.7 6. y 6. s.f. b a Using sin sin
More informationWorksheet A VECTORS 1 G H I D E F A B C
Worksheet A G H I D E F A B C The diagram shows three sets of equally-spaced parallel lines. Given that AC = p that AD = q, express the following vectors in terms of p q. a CA b AG c AB d DF e HE f AF
More informationCCE PR Revised & Un-Revised
D CCE PR Revised & Un-Revised 560 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE 560 00 08 S.S.L.C. EXAMINATION, JUNE, 08 :. 06. 08 ] MODEL ANSWERS : 8-K Date :. 06. 08 ] CODE
More informationCircles, Mixed Exercise 6
Circles, Mixed Exercise 6 a QR is the diameter of the circle so the centre, C, is the midpoint of QR ( 5) 0 Midpoint = +, + = (, 6) C(, 6) b Radius = of diameter = of QR = of ( x x ) + ( y y ) = of ( 5
More informationDraft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 1: Complex numbers 1
1 w z k k States or implies that 4 i TBC Uses the definition of argument to write 4 k π tan 1 k 4 Makes an attempt to solve for k, for example 4 + k = k is seen. M1.a Finds k = 6 (4 marks) Pearson Education
More informationQuantities which have only magnitude are called scalars. Quantities which have magnitude and direction are called vectors.
Vectors summary Quantities which have only magnitude are called scalars. Quantities which have magnitude and direction are called vectors. AB is the position vector of B relative to A and is the vector
More information( y) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios, Mixed Exercise 9. 2 b. Using the sine rule. a Using area of ABC = sin x sin80. So 10 = 24sinθ.
Trigonometric ratios, Mixed Exercise 9 b a Using area of ABC acsin B 0cm 6 8 sinθ cm So 0 4sinθ So sinθ 0 4 θ 4.6 or 3 s.f. (.) As θ is obtuse, ABC 3 s.f b Using the cosine rule b a + c ac cos B AC 8 +
More informationClass IX Chapter 8 Quadrilaterals Maths
Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: Let the common ratio between
More informationSHW 1-01 Total: 30 marks
SHW -0 Total: 30 marks 5. 5 PQR 80 (adj. s on st. line) PQR 55 x 55 40 x 85 6. In XYZ, a 90 40 80 a 50 In PXY, b 50 34 84 M+ 7. AB = AD and BC CD AC BD (prop. of isos. ) y 90 BD = ( + ) = AB BD DA x 60
More informationClass IX Chapter 8 Quadrilaterals Maths
1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Let the common ratio between the angles
More information[BIT Ranchi 1992] (a) 2 (b) 3 (c) 4 (d) 5. (d) None of these. then the direction cosine of AB along y-axis is [MNR 1989]
VECTOR ALGEBRA o. Let a i be a vector which makes an angle of 0 with a unit vector b. Then the unit vector ( a b) is [MP PET 99]. The perimeter of the triangle whose vertices have the position vectors
More informationFACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures
FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 5 VECTORS II. Triple products 2. Differentiation and integration of vectors 3. Equation of a line 4. Equation of a plane.
More informationPAST QUESTIONS ON VECTORS P1
PAST QUESTIONS ON VECTORS P1 1. Diagram shows a solid cylinder standing on a horizontal circular base, centre O and radius 4 units. The line BA is a diameter and the radius OC is at 90 o to OA. Points
More informationVectors Practice [296 marks]
Vectors Practice [96 marks] The diagram shows quadrilateral ABCD with vertices A(, ), B(, 5), C(5, ) and D(4, ) a 4 Show that AC = ( ) Find BD (iii) Show that AC is perpendicular to BD The line (AC) has
More informationVisit: ImperialStudy.com For More Study Materials Class IX Chapter 12 Heron s Formula Maths
Exercise 1.1 1. Find the area of a triangle whose sides are respectively 150 cm, 10 cm and 00 cm. The triangle whose sides are a = 150 cm b = 10 cm c = 00 cm The area of a triangle = s(s a)(s b)(s c) Here
More informationSTRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mathematics SKE, Strand J STRAND J: TRANSFORMATIONS, VECTORS and MATRICES J4 Matrices Text Contents * * * * Section J4. Matrices: Addition and Subtraction J4.2 Matrices: Multiplication J4.3 Inverse Matrices:
More informationMath 9 Chapter 8 Practice Test
Name: Class: Date: ID: A Math 9 Chapter 8 Practice Test Short Answer 1. O is the centre of this circle and point Q is a point of tangency. Determine the value of t. If necessary, give your answer to the
More informationRMT 2013 Geometry Test Solutions February 2, = 51.
RMT 0 Geometry Test Solutions February, 0. Answer: 5 Solution: Let m A = x and m B = y. Note that we have two pairs of isosceles triangles, so m A = m ACD and m B = m BCD. Since m ACD + m BCD = m ACB,
More information( ) = ( ) ( ) = ( ) = + = = = ( ) Therefore: , where t. Note: If we start with the condition BM = tab, we will have BM = ( x + 2, y + 3, z 5)
Chapter Exercise a) AB OB OA ( xb xa, yb ya, zb za),,, 0, b) AB OB OA ( xb xa, yb ya, zb za) ( ), ( ),, 0, c) AB OB OA x x, y y, z z (, ( ), ) (,, ) ( ) B A B A B A ( ) d) AB OB OA ( xb xa, yb ya, zb za)
More informationTriangles. Example: In the given figure, S and T are points on PQ and PR respectively of PQR such that ST QR. Determine the length of PR.
Triangles Two geometric figures having the same shape and size are said to be congruent figures. Two geometric figures having the same shape, but not necessarily the same size, are called similar figures.
More informationTRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions
CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii)
More informationAdvanced Digital Design with the Verilog HDL, Second Edition Michael D. Ciletti Prentice Hall, Pearson Education, 2011
Problem 2-1 Recall that a minterm is a cube in which every variable appears. A Boolean expression in SOP form is canonical if every cube in the expression has a unique representation in which all of the
More informationQ1. The sum of the lengths of any two sides of a triangle is always (greater/lesser) than the length of the third side. askiitians
Class: VII Subject: Math s Topic: Properties of triangle No. of Questions: 20 Q1. The sum of the lengths of any two sides of a triangle is always (greater/lesser) than the length of the third side. Greater
More information9 th CBSE Mega Test - II
9 th CBSE Mega Test - II Time: 3 hours Max. Marks: 90 General Instructions All questions are compulsory. The question paper consists of 34 questions divided into four sections A, B, C and D. Section A
More informationSUMMATIVE ASSESSMENT I, IX / Class IX
I, 0 SUMMATIVE ASSESSMENT I, 0 0 MATHEMATICS / MATHEMATICS MATHEMATICS CLASS CLASS - IX - IX IX / Class IX MA-0 90 Time allowed : hours Maximum Marks : 90 (i) (ii) 8 6 0 0 (iii) 8 (iv) (v) General Instructions:
More informationchapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true?
chapter vector geometry solutions V. Exercise A. For the shape shown, find a single vector which is equal to a)!!! " AB + BC AC b)! AD!!! " + DB AB c)! AC + CD AD d)! BC + CD!!! " + DA BA e) CD!!! " "
More informationNozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch
Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. in the parallelogram, each two opposite
More informationSURA's Guides for 3rd to 12th Std for all Subjects in TM & EM Available. MARCH Public Exam Question Paper with Answers MATHEMATICS
SURA's Guides for rd to 1th Std for all Subjects in TM & EM Available 10 th STD. MARCH - 017 Public Exam Question Paper with Answers MATHEMATICS [Time Allowed : ½ Hrs.] [Maximum Marks : 100] SECTION -
More informationProperties of the Circle
9 Properties of the Circle TERMINOLOGY Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A straight line joining two points on the circumference
More informationClass IX : Math Chapter 11: Geometric Constructions Top Concepts 1. To construct an angle equal to a given angle. Given : Any POQ and a point A.
1 Class IX : Math Chapter 11: Geometric Constructions Top Concepts 1. To construct an angle equal to a given angle. Given : Any POQ and a point A. Required : To construct an angle at A equal to POQ. 1.
More informationYou must have: Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.
Write your name here Surname Other names Pearson Edexcel International GCSE Centre Number Mathematics B Paper 1 Candidate Number Thursday 26 May 2016 Morning Time: 1 hour 30 minutes Paper Reference 4MB0/01
More informationMathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions
Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions Quiz #1. Tuesday, 17 January, 2012. [10 minutes] 1. Given a line segment AB, use (some of) Postulates I V,
More informationMark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
Mark scheme Pure Mathematics Year 1 (AS) Unit Test : Coordinate in the (x, y) plane Q Scheme Marks AOs Pearson 1a Use of the gradient formula to begin attempt to find k. k 1 ( ) or 1 (k 4) ( k 1) (i.e.
More informationProof by induction ME 8
Proof by induction ME 8 n Let f ( n) 9, where n. f () 9 8, which is divisible by 8. f ( n) is divisible by 8 when n =. Assume that for n =, f ( ) 9 is divisible by 8 for. f ( ) 9 9.9 9(9 ) f ( ) f ( )
More informationCHAPTER TWO. 2.1 Vectors as ordered pairs and triples. The most common set of basic vectors in 3-space is i,j,k. where
40 CHAPTER TWO.1 Vectors as ordered pairs and triples. The most common set of basic vectors in 3-space is i,j,k where i represents a vector of magnitude 1 in the x direction j represents a vector of magnitude
More informationCore Mathematics 2 Radian Measures
Core Mathematics 2 Radian Measures Edited by: K V Kumaran Email: kvkumaran@gmail.com Core Mathematics 2 Radian Measures 1 Radian Measures Radian measure, including use for arc length and area of sector.
More informationCollinearity/Concurrence
Collinearity/Concurrence Ray Li (rayyli@stanford.edu) June 29, 2017 1 Introduction/Facts you should know 1. (Cevian Triangle) Let ABC be a triangle and P be a point. Let lines AP, BP, CP meet lines BC,
More informationTriangles. 3.In the following fig. AB = AC and BD = DC, then ADC = (A) 60 (B) 120 (C) 90 (D) none 4.In the Fig. given below, find Z.
Triangles 1.Two sides of a triangle are 7 cm and 10 cm. Which of the following length can be the length of the third side? (A) 19 cm. (B) 17 cm. (C) 23 cm. of these. 2.Can 80, 75 and 20 form a triangle?
More informationPrepared by: M. S. KumarSwamy, TGT(Maths) Page
Prepared by: M S KumarSwamy, TGT(Maths) Page - 119 - CHAPTER 10: VECTOR ALGEBRA QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 06 marks Vector The line l to the line segment AB, then a
More informationCOORDINATE GEOMETRY BASIC CONCEPTS AND FORMULAE. To find the length of a line segment joining two points A(x 1, y 1 ) and B(x 2, y 2 ), use
COORDINATE GEOMETRY BASIC CONCEPTS AND FORMULAE I. Length of a Line Segment: The distance between two points A ( x1, 1 ) B ( x, ) is given b A B = ( x x1) ( 1) To find the length of a line segment joining
More informationSOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)
1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x
More informationReview Exercise 2. , æ. ç ø. ç ø. ç ø. ç ø. = -0.27, 0 x 2p. 1 Crosses y-axis when x = 0 at sin 3p 4 = 1 2. ö ø. æ Crosses x-axis when sin x + 3p è
Review Exercise 1 Crosses y-axis when x 0 at sin p 4 1 Crosses x-axis when sin x + p 4 ö 0 x + p 4 -p, 0, p, p x - 7p 4, - p 4, p 4, 5p 4 So coordinates are 0, 1 ö, - 7p 4,0 ö, - p 4,0 ö, p 4,0 ö, 5p 4,0
More informationTheorem on altitudes and the Jacobi identity
Theorem on altitudes and the Jacobi identity A. Zaslavskiy and M. Skopenkov Solutions. First let us give a table containing the answers to all the problems: Algebraic object Geometric sense A apointa a
More informationSMT 2018 Geometry Test Solutions February 17, 2018
SMT 018 Geometry Test Solutions February 17, 018 1. Consider a semi-circle with diameter AB. Let points C and D be on diameter AB such that CD forms the base of a square inscribed in the semicircle. Given
More informationINMO-2001 Problems and Solutions
INMO-2001 Problems and Solutions 1. Let ABC be a triangle in which no angle is 90. For any point P in the plane of the triangle, let A 1,B 1,C 1 denote the reflections of P in the sides BC,CA,AB respectively.
More informationClass-IX CBSE Latest Pattern Sample Paper {Mathematics}
Class-IX CBSE Latest Pattern Sample Paper {Mathematics} Term-I Examination (SA I) Time: 3hours Max. Marks: 90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of
More informationTrigonometric Functions 6C
Trigonometric Functions 6C a b c d e sin 3 q æ ö ø 4 tan 6 q 4 æ ö tanq ø cos q æ ö ø 3 cosec 3 q - sin q sin q cos q sin q (using sin q + cos q ) So - sin q sin q æ ö ø 6 4cot 6 q sec q cot q secq cos
More informationA B CDE F B FD D A C AF DC A F
International Journal of Arts & Sciences, CD-ROM. ISSN: 1944-6934 :: 4(20):121 131 (2011) Copyright c 2011 by InternationalJournal.org A B CDE F B FD D A C A BC D EF C CE C A D ABC DEF B B C A E E C A
More informationGeometry 3 SIMILARITY & CONGRUENCY Congruency: When two figures have same shape and size, then they are said to be congruent figure. The phenomena between these two figures is said to be congruency. CONDITIONS
More informationVECTORS IN THE CARTESIAN PLANE WARM-UP
VECTORS IN THE CARTESIAN PLANE LAMC INTERMEDIATE - 4/27/14 WARM-UP (Oldaque P. de Freitas Puzzle) Two ladies are sitting in a street cafe, talking about their children. One lady says that she has three
More informationIntroduction to Vectors Pg. 279 # 1 6, 8, 9, 10 OR WS 1.1 Sept. 7. Vector Addition Pg. 290 # 3, 4, 6, 7, OR WS 1.2 Sept. 8
UNIT 1 INTRODUCTION TO VECTORS Lesson TOPIC Suggested Work Sept. 5 1.0 Review of Pre-requisite Skills Pg. 273 # 1 9 OR WS 1.0 Fill in Info sheet and get permission sheet signed. Bring in $3 for lesson
More information81-E 2. Ans. : 2. Universal set U = { 2, 3, 5, 6, 10 }, subset A = { 5, 6 }. The diagram which represents A / is. Ans. : ( SPACE FOR ROUGH WORK )
81-E 2 General Instructions : i) The question-cum-answer booklet contains two Parts, Part A & Part B. ii) iii) iv) Part A consists of 60 questions and Part B consists of 16 questions. Space has been provided
More informationGeometry Problem Solving Drill 08: Congruent Triangles
Geometry Problem Solving Drill 08: Congruent Triangles Question No. 1 of 10 Question 1. The following triangles are congruent. What is the value of x? Question #01 (A) 13.33 (B) 10 (C) 31 (D) 18 You set
More informationGeometry Problem booklet
Geometry Problem booklet Assoc. Prof. Cornel Pintea E-mail: cpintea math.ubbcluj.ro Contents 1 Week 1: Vector algebra 1 1.1 Brief theoretical background. Free vectors..................... 1 1.1.1 Operations
More information(A) 50 (B) 40 (C) 90 (D) 75. Circles. Circles <1M> 1.It is possible to draw a circle which passes through three collinear points (T/F)
Circles 1.It is possible to draw a circle which passes through three collinear points (T/F) 2.The perpendicular bisector of two chords intersect at centre of circle (T/F) 3.If two arcs of a circle
More informationEGR2013 Tutorial 8. Linear Algebra. Powers of a Matrix and Matrix Polynomial
EGR1 Tutorial 8 Linear Algebra Outline Powers of a Matrix and Matrix Polynomial Vector Algebra Vector Spaces Powers of a Matrix and Matrix Polynomial If A is a square matrix, then we define the nonnegative
More informationSolutionbank Edexcel AS and A Level Modular Mathematics
Page of Exercise A, Question Use the binomial theorem to expand, x
More informationSample Question Paper Mathematics First Term (SA - I) Class IX. Time: 3 to 3 ½ hours
Sample Question Paper Mathematics First Term (SA - I) Class IX Time: 3 to 3 ½ hours M.M.:90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided
More informationReview exercise
Review eercise y cos sin When : 8 y and 8 gradient of normal is 8 y When : 9 y and 8 Equation of normal is y 8 8 y8 8 8 8y 8 8 8 8y 8 8 8 8y 8 8 8 y e ln( ) e ln e When : y e ln and e Equation of tangent
More information10. Circles. Q 5 O is the centre of a circle of radius 5 cm. OP AB and OQ CD, AB CD, AB = 6 cm and CD = 8 cm. Determine PQ. Marks (2) Marks (2)
10. Circles Q 1 True or False: It is possible to draw two circles passing through three given non-collinear points. Mark (1) Q 2 State the following statement as true or false. Give reasons also.the perpendicular
More informationPart (1) Second : Trigonometry. Tan
Part (1) Second : Trigonometry (1) Complete the following table : The angle Ratio 42 12 \ Sin 0.3214 Cas 0.5321 Tan 2.0625 (2) Complete the following : 1) 46 36 \ 24 \\ =. In degrees. 2) 44.125 = in degrees,
More informationP1 Vector Past Paper Questions
P1 Vector Past Paper Questions Doublestruck & CIE - Licensed to Brillantmont International School 1 1. Relative to an origin O, the position vectors of the points A and B are given by OA = 2i + 3j k and
More information3D GEOMETRY. 3D-Geometry. If α, β, γ are angle made by a line with positive directions of x, y and z. axes respectively show that = 2.
D GEOMETRY ) If α β γ are angle made by a line with positive directions of x y and z axes respectively show that i) sin α + sin β + sin γ ii) cos α + cos β + cos γ + 0 Solution:- i) are angle made by a
More informationCO-ORDINATE GEOMETRY. 1. Find the points on the y axis whose distances from the points (6, 7) and (4,-3) are in the. ratio 1:2.
UNIT- CO-ORDINATE GEOMETRY Mathematics is the tool specially suited for dealing with abstract concepts of any ind and there is no limit to its power in this field.. Find the points on the y axis whose
More informationFor more information visit here:
The length or the magnitude of the vector = (a, b, c) is defined by w = a 2 +b 2 +c 2 A vector may be divided by its own length to convert it into a unit vector, i.e.? = u / u. (The vectors have been denoted
More informationMark Scheme (Results) Summer Pearson Edexcel International GCSE in Mathematics B Paper 2R (4MB0/02R)
Mark Scheme (Results) Summer 04 Pearson Edexcel International GCSE in Mathematics B Paper R (4MB0/0R) Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading
More informationIIT-JEE (2012) (Vector+3D+Probability) Solutions
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 PAPER -A IIT-JEE (0) (Vector+D+Probability) Solutions TOWARDS IIT- JEE IS NOT A JOURNEY, IT S A BATTLE, ONLY THE TOUGHEST
More informationHigher Order Thinking Skill questions
Higher Order Thinking Skill questions TOPIC- Constructions (Class- X) 1. Draw a triangle ABC with sides BC = 6.3cm, AB = 5.2cm and ÐABC = 60. Then construct a triangle whose sides are times the corresponding
More informationPage 1 of 15. Website: Mobile:
Exercise 10.2 Question 1: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5
More information( ) ( ) or ( ) ( ) Review Exercise 1. 3 a 80 Use. 1 a. bc = b c 8 = 2 = 4. b 8. Use = 16 = First find 8 = 1+ = 21 8 = =
Review Eercise a Use m m a a, so a a a Use c c 6 5 ( a ) 5 a First find Use a 5 m n m n m a m ( a ) or ( a) 5 5 65 m n m a n m a m a a n m or m n (Use a a a ) cancelling y 6 ecause n n ( 5) ( 5)( 5) (
More informationQuestion Bank Tangent Properties of a Circle
Question Bank Tangent Properties of a Circle 1. In quadrilateral ABCD, D = 90, BC = 38 cm and DC = 5 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 7 cm. Find
More informationQ Scheme Marks AOs Pearson Progression Step and Progress descriptor. and sin or x 6 16x 6 or x o.e
1a A 45 seen or implied in later working. B1 1.1b 5th Makes an attempt to use the sine rule, for example, writing sin10 sin 45 8x3 4x1 States or implies that sin10 3 and sin 45 A1 1. Solve problems involving
More informationAnswer Key. 9.1 Parts of Circles. Chapter 9 Circles. CK-12 Geometry Concepts 1. Answers. 1. diameter. 2. secant. 3. chord. 4.
9.1 Parts of Circles 1. diameter 2. secant 3. chord 4. point of tangency 5. common external tangent 6. common internal tangent 7. the center 8. radius 9. chord 10. The diameter is the longest chord in
More informationMEP Pupil Text 13-19, Additional Material. Gradients of Perpendicular Lines
Graphs MEP Pupil Text -9, Additional Material.B Gradients of Perpendicular Lines In this section we explore the relationship between the gradients of perpendicular lines and line segments. Worked Example
More informationIt is known that the length of the tangents drawn from an external point to a circle is equal.
CBSE -MATHS-SET 1-2014 Q1. The first three terms of an AP are 3y-1, 3y+5 and 5y+1, respectively. We need to find the value of y. We know that if a, b and c are in AP, then: b a = c b 2b = a + c 2 (3y+5)
More informationGS trapezoids in GS quasigroups
Mathematical Communications 7(2002), 143-158 143 GS trapezoids in GS quasigroups Vladimir Volenec and Zdenka Kolar Abstract. In this paper the concept of a GS trapezoid in a GS quasigroup is defined and
More informationSection 5.5: Matrices and Matrix Operations
Section 5.5 Matrices and Matrix Operations 359 Section 5.5: Matrices and Matrix Operations Two club soccer teams, the Wildcats and the Mud Cats, are hoping to obtain new equipment for an upcoming season.
More informationSt Andrew s Academy Mathematics Department Higher Mathematics VECTORS
St Andrew s Academy Mathematics Department Higher Mathematics VECTORS hsn.uk.net Higher Mathematics Vectors Contents Vectors 1 1 Vectors and Scalars EF 1 Components EF 1 Magnitude EF 4 Equal Vectors EF
More informationTHE THEOREM OF DESARGUES
THE THEOREM OF DESARGUES TIMOTHY VIS 1. Introduction In this worksheet we will be exploring some proofs surrounding the Theorem of Desargues. This theorem plays an extremely important role in projective
More informationMATH1050 Greatest/least element, upper/lower bound
MATH1050 Greatest/ element, upper/lower bound 1 Definition Let S be a subset of R x λ (a) Let λ S λ is said to be a element of S if, for any x S, x λ (b) S is said to have a element if there exists some
More informationOne Theorem, Six Proofs
50/ ONE THEOREM, SIX PROOFS One Theorem, Six Proofs V. Dubrovsky It is often more useful to acquaint yourself with many proofs of the same theorem rather than with similar proofs of numerous results. The
More informationADDITIONAL MATHEMATICS
005-CE A MATH HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 005 ADDITIONAL MATHEMATICS :00 pm 5:0 pm (½ hours) This paper must be answered in English 1. Answer ALL questions in Section A and any FOUR
More informationPROBLEMS 07 - VECTORS Page 1. Solve all problems vectorially: ( 1 ) Obtain the unit vectors perpendicular to each of , 29 , 29
PROBLEMS 07 VECTORS Page Solve all problems vectorially: ( ) Obtain the unit vectors perpendicular to each x = ( ) d y = ( 0 ). ± 9 9 9 ( ) If α is the gle between two unit vectors a d b then prove th
More informationVECTORS IN THREE DIMENSIONS
Mathematics Revision Guides Vectors in Three Dimensions Page of MK HOME TUITION Mathematics Revision Guides Level: ALevel Year VECTORS IN THREE DIMENSIONS Version: Date: Mathematics Revision Guides Vectors
More information1.1 Exercises, Sample Solutions
DM, Chapter, Sample Solutions. Exercises, Sample Solutions 5. Equal vectors have the same magnitude and direction. a) Opposite sides of a parallelogram are parallel and equal in length. AD BC, DC AB b)
More informationCBSE X Mathematics 2012 Solution (SET 1) Section D
Section D Q 9. A shopkeeper buys some books for Rs 80. If he had bought 4 more books for the same amount, each book would have cost Rs 1 less. Find the number of books he bought. Let the number of books
More information(b) the equation of the perpendicular bisector of AB. [3]
HORIZON EDUCATION SINGAPORE Additional Mathematics Practice Questions: Coordinate Geometr 1 Set 1 1 In the figure, ABCD is a rhombus with coordinates A(2, 9) and C(8, 1). The diagonals AC and BD cut at
More informationVectors. Teaching Learning Point. Ç, where OP. l m n
Vectors 9 Teaching Learning Point l A quantity that has magnitude as well as direction is called is called a vector. l A directed line segment represents a vector and is denoted y AB Å or a Æ. l Position
More informationTriangle Congruence and Similarity Review. Show all work for full credit. 5. In the drawing, what is the measure of angle y?
Triangle Congruence and Similarity Review Score Name: Date: Show all work for full credit. 1. In a plane, lines that never meet are called. 5. In the drawing, what is the measure of angle y? A. parallel
More informationClass IX Chapter 7 Triangles Maths. Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure).
Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD, AC = AD (Given) CAB = DAB (AB bisects
More informationMatrix & Linear Algebra
Matrix & Linear Algebra Jamie Monogan University of Georgia For more information: http://monogan.myweb.uga.edu/teaching/mm/ Jamie Monogan (UGA) Matrix & Linear Algebra 1 / 84 Vectors Vectors Vector: A
More informationKARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE G È.G È.G È.. Æ fioê, d È 2018 S. S. L. C. EXAMINATION, JUNE, 2018
CCE RR REVISED & UN-REVISED O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl 560 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE 560 00 G È.G È.G È.. Æ fioê, d È 08 S. S. L.
More informationFIXED BEAMS CONTINUOUS BEAMS
FIXED BEAMS CONTINUOUS BEAMS INTRODUCTION A beam carried over more than two supports is known as a continuous beam. Railway bridges are common examples of continuous beams. But the beams in railway bridges
More information