3D Coordinates In the real world points in space can be located using a 3D coordinate system. For example, air traffic controllers find the location a.

Slides:



Advertisements
Similar presentations
Higher Unit 3 Vectors and Scalars 3D Vectors Properties of vectors
Advertisements

Vectors Part Trois.
Co-ordinate geometry Objectives: Students should be able to
Vectors Strategies Higher Maths Click to start Vectors Higher Vectors The following questions are on Non-calculator questions will be indicated Click.
Whiteboardmaths.com © 2004 All rights reserved
Chapter 12 Vectors A. Vectors and scalars B. Geometric operations with vectors C. Vectors in the plane D. The magnitude of a vector E. Operations with.
Section 9.2 Vectors Goals Goals Introduce vectors. Introduce vectors. Begin to discuss operations with vectors and vector components. Begin to discuss.
Distance between 2 points Mid-point of 2 points
Menu Select the class required then click mouse key to view class.
ME 2304: 3D Geometry & Vector Calculus
Higher Unit 2 EF Higher Unit 3 Vectors and Scalars Properties of vectors Adding / Sub of vectors Multiplication.
Higher Maths Revision Notes Vectors Get Started. Vectors in three dimensions use scalar product to find the angle between two directed line segments know.
Scalars A scalar is any physical quantity that can be completely characterized by its magnitude (by a number value) A scalar is any physical quantity that.
H.Melikyan/12001 Vectors Dr.Hayk Melikyan Departmen of Mathematics and CS
Chapter 6 Additional Topics in Trigonometry
Vectors A quantity which has both magnitude and direction is called a vector. Vector notations A B a AB = a AB x and y are the components of vector AB.
Aim: Properties of Square & Rhombus Course: Applied Geo. Do Now: Aim: What are the properties of a rhombus and a square? Find the length of AD in rectangle.
General physics I, lec 1 By: T.A.Eleyan 1 Lecture (2)
Vectors Vectors are represented by a directed line segment its length representing the magnitude and an arrow indicating the direction A B or u u This.
Vectors and Scalars Chapter 8. What is a Vector Quantity? A quantity that has both Magnitude and a Direction in space is called a Vector Quantity.
ch46 Vectors by Chtan FYKulai
Chapter 12 – Vectors and the Geometry of Space 12.2 – Vectors 1.
SAT Prep. A.) Terminology and Notation Lines / Rays / Segments Angles – Classification Straight - 180° Vertical - = Circle – 360°
ME 2304: 3D Geometry & Vector Calculus Dr. Faraz Junejo.
Vectors in Space 11.2 JMerrill, Rules The same rules apply in 3-D space: The component form is found by subtracting the coordinates of the initial.
Mathematics. Cartesian Coordinate Geometry and Straight Lines Session.
Vectors and Scalars and Their Physical Significance.
Proving Properties of Triangles and Quadrilaterals
Physics and Physical Measurement Topic 1.3 Scalars and Vectors.
Vectors Vectors and Scalars Properties of vectors Adding / Sub of vectors Multiplication by a Scalar Position Vector Collinearity Section Formula.
SCALARS & VECTORS. Physical Quantities All those quantities which can be measured are called physical quantities. Physical Quantities can be measured.
Vectors in the Plane 8.3 Part 1. 2  Write vectors as linear combinations of unit vectors.  Find the direction angles of vectors.  Use vectors to model.
CIE Centre A-level Pure Maths
Vectors and Scalars.  A scalar quantity is a quantity that has magnitude only and has no direction in space Examples of Scalar Quantities:  Length 
Coordinate Geometry Midpoint of the Line Joining Two Points Areas of Triangles Parallel and Non-Parallel Lines Perpendicular Lines.
10 Square Roots and Pythagoras’ Theorem
6.4 EQ: What properties do we use to identify special types of parallelograms?
Drawing a sketch is always worth the time and effort involved
Scalars & Vectors – Learning Outcomes
Outline Addition and subtraction of vectors Vector decomposition
Chapter 3 VECTORS.
Physics and Physical Measurement
Vectors and Scalars Chapter 8.
6.3-Vectors in the Plane.
1.3 Vectors and Scalars Scalar: shows magnitude
VECTORS.
What is a Vector? AB and CD have the same distance
Coordinate Geometry – Outcomes
MODULE - 8 ANALYTICAL GEOMETRY.
Recapping: Finding the hypotenuse of a right-angled triangle.
Solving Problems Involving Lines and Points
Angles of Rotation.
Vectors and Scalars.
4-1 Triangles HONORS GEOMETRY.
Definition and Notation
Right Angled Trigonometry
A#40 Vectors Day 6 Paper #1 Review.
Higher Maths Vectors Strategies Click to start.
Recapping: Vector addition.
Vectors and Scalars.
Kinematics & Dynamics in 2 & 3 Dimensions; Vectors
Vector Journeys.
Scalars A scalar quantity is a quantity that has magnitude only and has no direction in space Examples of Scalar Quantities: Length Area Volume Time Mass.
Physics and Physical Measurement
Vectors and Scalars.
VECTORS 3D Vectors Properties 3D Section formula Scalar Product
CHAPTER 3 VECTORS NHAA/IMK/UNIMAP.
Presentation transcript:

3D Coordinates 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 (x, y, z) y O x 1

Write down the coordinates for the vertices x y z A C D E F G H B 1 2 6 (0, 1, 2) (6, 1, 2) (6, 0, 2) (0, 0, 2) (0, 1, 0 ) (6, 1, 0) (0,0, 0) (6, 0, 0) 2

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 O 3 x 7 3

Any vector can be represented in terms of the i , j and k Where i, j and k are unit vectors (one unit long) in the x, y and z directions. z y O k j x i 4

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 2 v = ( 7i+ 3j + 2k ) O 3 x 7 5

Magnitude of a Vector A vector’s magnitude (length) is represented by |v| A 3D vector’s magnitude is calculated using Pythagoras Theorem in 3D (3 , 2 , 1) z v y 1 O 2 x 3

Find the unit vector in the direction of u

Addition of vectors

Addition of Vectors For vectors u and v

Negative vector For any vector u

Subtraction of vectors

Subtraction of Vectors For vectors u and v

Multiplication by a scalar ( a number) Hence if u = kv then u is parallel to v Conversely if u is parallel to v then u = kv

CombiningVectors

If z = kw then z is parallel to w Show that the two vectors are parallel. If z = kw then z is parallel to w

The position vector of the point A(3 , 2 , 1) is OA written as a Position Vectors A (3,2,1) z a y 1 O 2 x 3 The position vector of the point A(3 , 2 , 1) is OA written as a

Position Vectors If R is ( 2 , -5 , 1) and S is (4 , 1 , -3) 4 = 1 -3 = 1 -3 2 – -5 1 2 = 6 -4 Then RS is s – r

y component of RS is -5 to 1 = 6 z component of RS is 1 to -3 = -4 Position Vectors as a journey If R is ( 2 , -5 , 1) and S is (4 , 1 , -3) x component of RS is 2 to 4 = 2 y component of RS is -5 to 1 = 6 z component of RS is 1 to -3 = -4 2 RS = 6 -4

Express VT in terms of f, g and h Vectors as a journey Express VT in terms of f, g and h – h VT = VR + RS + ST VT = – h – f + g – f + g

Collinear Points BC = 2AB A is (0 , -3 , 5), B is (7 , -6 , 9) and C is (21 , -12 , 17). Show that A , B and C are collinear stating the ratio AB:BC. BC = 2AB AB and BC are parallel (multiples of each other) through the common point B, and so must be collinear A 1 B 2 C AB : BC = 1 : 2

Collinear Points Given that the points S(–4, 5, 1), T(–16, –4, 16) and U(–24, –10, 26) are collinear, calculate the ratio in which T divides SU. S 3 T 2 U ST : TU = 3 : 2

The journey from Q to R is the same as the journey from P to S Using Vectors as journeys PQRS is a parallelogram with P(3 , 4 , 0), Q(7 , 6 , -3) and R(8 , 5 , 2). Find the coordinates of S. P(3 , 4 , 0) Q(7 , 6 , -3) Draw a sketch The journey from Q to R is the same as the journey from P to S S R(8 , 5 , 2) From P to S S(3 + 1 , 4 – 1 , 0 + 5) S(4 , 3 , 5)

Find the co-ordinates of Q. 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. P(-1 , -1 , 0) R(5 , 2 , -3) PQ = 2/3 PR Q 1 2 6 PQ = 2/3 3 -3 4 = 2 -2 The journey from P to Q Q(-1 + 4, -1 + 2, 0 – 2) Q(3 , 1 , – 2)

P divides the line joining S(1,0,2) and T(5,4,10) in the ratio 1:3 P divides the line joining S(1,0,2) and T(5,4,10) in the ratio 1:3. Find the coordinates of P. SP = 1/4 ST S(1 , 0 , 2) T(5 , 4 , 10) 4 SP = 1/4 4 8 1 = 1 2 The journey from S to P P 3 1 P(1 + 1, 0 + 1, 2 + 2) P(2 , 1 , 4)

a and b must be divergent, ie joined tail to tail The scalar product The scalar product is defined as being: a . b = |a| |b| cos θ 0 ≤ θ ≤ 180 a θ a and b must be divergent, ie joined tail to tail b

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 θ  a . b = 4 × 5 cos 45o  a . b = 20 × 1/√2 × √2/√2  a . b = 10√2 a . b = |a| |b| cos θ  a . b = 4 × 5 cos 90o  a . b = 20 × 0  a . b = 0 When θ = 90o

This equilateral triangle has sides of 3 units. p . q The Scalar Product This equilateral triangle has sides of 3 units. p . q p . q = |p| |q| cos θ  p . q = 3 × 3 cos 60o  p . q = 9 × 1/2  p . q = 41/2 27

This equilateral triangle has sides of 3 units. p . (q + r) The Scalar Product This equilateral triangle has sides of 3 units. p . (q + r) r r p . (q + r)= p . q + p . r 60o  p . q = 3 × 3 cos 60o p and r are not divergent so move r  p . q = 3 × 3 × ½  p . q = 41/2  p . r = 3 × 3 cos 60o  p . r = 9  ½ p . (q + r) = p . q + p . r = 9  p . r = 4½ 28

If a and b are perpendicular then a . b = 0

a . b = a1b1 + a2b2 + a3b3 Component Form Scalar Product a1 b1 If a = a2 and b = b2 a3 b3 a . b = a1b1 + a2b2 + a3b3

Angle between Vectors To find the angle between two vectors we simply use the scalar product formulae rearranged a . b = |a| |b| cos θ a . b = a1b1 + a2b2 + a3b3 a . b |a| |b| a1b1 + a2b2 + a3b3 |a| |b| cos θ = cos θ =

Find the angle between the two vectors below. p = 3i + 2j + 5k and q = 4i + j + 3k 3 p = 2 5 4 q = 1 3 |p| = √(32 + 22 + 52) |q| = √(42 + 12 + 32) |q| = √26 |p| = √38 a1b1 + a2b2 + a3b3 |a| |b| a . b = a1b1 + a2b2 + a3b3 cos θ = = 3×4 + 2×1 + 5×3 29 √38 × √26 = = 0∙923 = 29 θ = cos-1 0∙923 = 22∙7o

Perpendicular Vectors a . b = | a | | b | cosƟ If a and b are perpendicular then a . b = 0 cos 90o = 0 If a and b are perpendicular then a1b1 + a2b2 + a3b3 = 0

If a and b are perpendicular then a . b = 0 Show that a and b are perpendicular 3 1 if a = 2 and b = 2 -1 7 a . b = a1b1 + a2b2 + a3b3  a . b = 3×1 + 2×2 + (-1)×7  a . b = 3 + 4 – 7  a . b = 0  a and b are perpendicular 34

Two properties that you need to be aware of Properties of a Scalar Product Two properties that you need to be aware of a . b = b . a a .( b + c)= a . b + a . c

If | p | = 5 and | q | = 4, find p . (p + q) 60o q p . (p + q) = p . p + p . q = | p | × | p |cos 0o + | p | × | q |cos60o = 5 × 5 × 1 + 5 × 4 × ½ = 25 + 10 = 35 36

Find the acute angle between the diagonals of PQRS T divides PR in the ratio 5:4 P(-2,-1,-4) Show that Q, T and S are collinear, and find the ratio in which T divides QS S(7,2,17) Q(1,5,-7) Find the acute angle between the diagonals of PQRS R(7,8,5) TS = 2QT, vectors are parallel through the common point T, so Q , T , S are collinear T(-2+5,-1+5,-4+5) T(3 , 4 , 1) QT : TS = 1 : 2 37

Find the acute angle between the diagonals of PQRS 38

Hence vectors are perpendicular Vectors u and v are defined by u = 3i + 2j and v = 2i – 3j + 4k Determine whether or not u and v are perpendicular. 3 2 u . v = 2 . -3 0 4  u . v = 3×2 + 2×(-3) + 0×4  u . v = 6 – 6 + 0  u . v = 0 Hence vectors are perpendicular

u . v = 0 if vectors perpendicular For what value of t are the vectors u and v perpendicular ? u . v = 0 if vectors perpendicular t 2 u . v = -2 . 10 3 t  u . v = t ×2 + (-2)×10 + 3×t  u . v = 5t – 20 u . v = 0  5t – 20 = 0 t = 4

VABCD is a pyramid with rectangular base ABCD. The vectors AB, AD and AV are given by AB = 8i + 2j + 2k, AD = -2i + 10j – 2k and AV = i + 7j + 7k Express CV in component form. Look for an alternative route from C to V along vectors you know. Note BC = AD and AB = DC CV = CB + BA + AV 8 AB = 2 2 -2 AD = 10 1 AV = 7 7 CV = -AD – AB + AV 2 -10 = 8 - + 1 7 CV = –5i – 5j + 7k -5 7 =

The diagram shows two vectors a and b, with | a | = 3 and | b | = 22. 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) ii) iii) b) p2

Vectors p, q and r are defined by p = i + j – k, q = i + 4k, r = 4i – 3j a) Express p – q + 2r in component form b) Calculate p.r c) Find |r| a) b) c)

hence there are 2 points on the x axis that are 7 units from P 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. hence there are 2 points on the x axis that are 7 units from P i.e. S and T and

The position vectors of the points P and Q are p = –i + 3j + 4k and q = 7i – j + 5k respectively. a) Express PQ in component form. b) Find the length of PQ.

120o so, a is perpendicular to b + c Q 60° a b c PQR is an equilateral triangle of side 2 units. Evaluate a.(b + c) and hence identify two vectors which are perpendicular. 120o NB for a.c vectors must diverge ( so angle is 120° ) so, a is perpendicular to b + c Hence

Calculate the length of the vector 2i – 3j + 3k

Find the value of k for which the vectors and are perpendicular Put Scalar product = 0

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. The journey B to A is the same as the journey from C to D That journey from C gives: D(–6 + [-5], 4 + [-2], 2 + 1) D(–11, 2 , 3)

b + c is perpendicular to a 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 (a), make a deduction about the vector b + c a) b) b + c is perpendicular to a

Triangular faces are all equilateral In the square based pyramid, all the eight edges are of length 3 units. Evaluate p.(q + r) Triangular faces are all equilateral

(-1,-3,2) (2,-1,1)  D is (-1 + 9, -3 + 6, 2 – 3)  D is (8, 3, –1) 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 (-1,-3,2) (2,-1,1)  D is (-1 + 9, -3 + 6, 2 – 3)  D is (8, 3, –1)

Find the co-ordinates of Q. 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. P Q R 2 1 (5,2,-3) (-1,-1,0)  Q is (-1 + 4, -1 + 2, 0 – 2)  Q is (3, 1, –2)

VABCD is a pyramid with rectangular base ABCD. VA = – 7i – 13j – 11k, AB = 6i + 6j – 6k, AD = 8i – 4j – 4k ; K divides BC in the ratio 1:3. Find VK in component form. + ¼BC = VA + AB VK + ¼AD = VA + AB = – 7i – 13j – 11k 6i + 6j – 6k 2i – j – k = i – 8j – 18k

AB and BC are scalar multiples, so are parallel. 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 and BC are scalar multiples, so are parallel. B is common. A, B, C are collinear A B C 2 3 B divides AB in ratio 2 : 3