Vectors.

Vectors

5.2 Vectors in Review Connection to the Study Design AOS 4 – Vectors
Addition and subtraction of vectors and their multiplication by a scalar, and position vectors

Key Vocabulary: Vector Scalar Null Associative Magnitude Direction Operation Commutative Identity Inverse Closure Expression Simplify Collinear Midpoint Segment Scalar Quantity Vector Quantity Resultant Tilde Terminal Key Notation: 𝐴𝐵 𝜆∈𝑅\ 0 𝑉

Recap Scalar Quantity: a quantity which can be completely described by its magnitude in a particular unit Vector Quantity: a quantity which can be completely described by its magnitude in a particular unit AND its direction Vector Notation: Geometrically: Use of a directed line segment. Vector going between Point A to Point B: Point A –initial point or tail; Point B – terminal point or head

Addition of two vectors: Follows the triangle rule for addition
Addition of two vectors: Follows the triangle rule for addition. Vectors are placed head to tail. 𝐴𝐵 + 𝐵𝐶 = 𝐴𝐶 𝐴𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡/𝑠𝑢𝑚 Scalar Multiplication of Vectors: 2𝑎=𝑎+𝑎 2a is a vector parallel to a but twice the length

The negative of a vector: Vector of same length but point in opposite direction Head and tail are reversed Subtraction of vectors: Let 𝑎= 𝑂𝐴 and 𝑏= 𝑂𝐵 where O is the origin 𝑎−𝑏=𝑎+ −𝑏 = 𝑂𝐴 − 𝑂𝐵 = 𝑂𝐴 + 𝐵𝑂 = 𝐵𝑂 + 𝑂𝐴 = 𝐵𝐴

A vector which has no magnitude and no direction 𝑎+ −𝑎 =0
The zero or null vector A vector which has no magnitude and no direction 𝑎+ −𝑎 =0 𝑂𝐴 + 𝐴𝐵 + 𝐵𝑂 = 𝑂𝐴 + 𝐴𝐵 + 𝐵𝑂 = 𝑂𝐵 + 𝐵𝑂 = 𝑂𝐵 − 𝑂𝐵 =0 Please note: This is the rule for the addition of three vectors placed head to tail and justified by the associative law. Algebra of Vectors Closure: 𝑎+𝑏∈𝑉 2. Commutative Law: 𝑎+𝑏=𝑏+𝑎 3. Associative Law: 𝑎+𝑏 +𝑐=𝑎+(𝑏+𝑐) 4. Additive Identity Law: 𝑎+0=0+𝑎=𝑎 5. Inverse Law: 𝑎+ −𝑎 = −𝑎 +𝑎=0

Worked Example 1: Simplifying Vector Expressions
Simplify the vector expression 4 𝐴𝐵 − 𝐶𝐵 −4 𝐴𝐶 .

Collinear points Collinear: Three points which all lie on the same straight line. If 𝐴𝐵 and 𝐵𝐶 are parallel vectors then they have Point B in common and must lie on the same straight line. A, B and C are collinear if 𝐴𝐵 =𝜆 𝐵𝐶 where 𝜆∈𝑅\ 0 .

Worked Example 2: Collinear
If 3 𝑂𝐴 −2 𝑂𝐵 − 𝑂𝐶 =0, show that points A, B and C are collinear.

Midpoints: The midpoint, M, of the line segment AB where O is the origin and A and B are points, is given by: Applications to Geometry: Vectors can be used to prove many thermos of geometry. These vector proofs involve the properties of vectors discussed so far.

Worked Example 3: Proof OABC is a parallelogram. P is the midpoint of OA, and the point D divides PC in the ration 1:2. Prove that O, D and B are collinear. Hint: Draw a parallelogram first! 

5.3 Vector Notation: i, j, k Connection to the Study Design
AOS 4 – Vectors Addition and subtraction of vectors and their multiplication by a scalar, and position vectors Linear dependence and independence of a set of vectors and geometric interpretation Magnitude of a vector, unit vector and the orthogonal unit vectors i, j and k

Key Vocabulary: Unit vector Magnitude Circumflex Hat Cartesian coordinates Resolution of vectors Component form Linear dependence Linear combination Linear independence Matrices Key Notation: 𝑎 𝑑 𝑂𝑃 𝐴𝐵 𝑂𝑃

Unit vectors Have a magnitude of 1 Unit Vectors i, j and k
Direction givers Indicated by circumflex/hat above vector 𝑎 = 𝑎 𝑎 Unit Vectors i, j and k Acknowledge that i, j and k are unit vectors therefore do not require circumflex i: unit vector in the positive direction parallel to the x-axis j: unit vector in the positive direction parallel to the y-axis k: unit vector in the positive direction parallel to the z-axis Coefficient in front of each vector represents the magnitude parallel to that particular axis

Position vector Point P has coordinates 𝑥 1 , 𝑦 1 , 𝑧 1 relative to the origin, O 0, 0, 0 𝑟 denotes position vector Vector 𝑟= 𝑂𝑃 can be expressed in terms of three other vectors: One parallel to the: x-axis y-axis z-axis Resolution of vectors: method of splitting a vector up into its components 𝑂𝑃 = 𝑂𝐴 + 𝐴𝐵 + 𝐵𝑃 Where 𝑑 𝑂𝐴 = 𝑥 1 , 𝑑 𝐴𝐵 = 𝑦 1 ,𝑑 𝐵𝑃 = 𝑧 1 ⇒ 𝑂𝐴 = 𝑥 1 𝑖, 𝐴𝐵 = 𝑦 1 𝑗 𝑎𝑛𝑑 𝐵𝑃 = 𝑧 1 𝑘 Therefore; r= 𝑂𝑃 = 𝑥 1 𝑖+ 𝑦 1 𝑗+ 𝑧 1 𝑘 Give answers to questions in terms of i, j and k

Magnitude of a Vector Magnitude of the position vector is give by:
𝑟 = 𝑂𝑃 =𝑑 𝑂𝑃 = 𝑥 𝑦 𝑧 1 2 Distance between O and P is magnitude of vector. Deduced by: Using the triangle in the xy plane, Pythagoras’ Theorem shows: 𝑑 2 𝑂𝑃 = 𝑑 2 𝑂𝐵 + 𝑑 2 𝐴𝐵 = 𝑥 𝑦 1 2 Using triangle OPB, 𝑑 2 𝑂𝑃 = 𝑑 2 𝑂𝐵 + 𝑑 2 𝐴𝐵 + 𝑑 2 𝐵𝑃 = 𝑥 𝑦 𝑧 1 2

Worked Example 4: If the point P has coordinates (3, 2, −4), find:
The vector 𝑂𝑃 A unit vector parallel to 𝑂𝑃

Addition and subtraction of vectors in three dimensions
Only add or subtract vectors in component form Add/subtract vectors i, j, k separately Vector 𝐴𝐵 represents the position vector of B relative to A, that is B as seen from A 𝑑 𝐴𝐵 = 𝐴𝐵 = 𝑥 2 − 𝑥 𝑦 2 − 𝑦 𝑧 2 − 𝑧 1 2

Worked Example 5 Two points, A and B, have the coordinates (1, −2, 1) and (3, 4, −2) respectively. Find a unit vector parallel to 𝐴𝐵 .

Equality of two vectors
Vectors 𝑂𝐴 and 𝑂𝐵 are equal if and only if corresponding coefficients of components are equal Scalar Multiplication of vectors Multiply each components’ coefficient by scalar value

Worked Example 6 If 𝑎=2𝑖−3𝑗+𝑧𝑘 and 𝑏=4𝑖−5−2𝑘, find the value of z if the vector c= 3𝑎−2𝑏 is parallel to the xy plane

𝑎 is parallel to 𝑏 if 𝑎=𝜆𝑏 where 𝜆∈𝑅
Parallel vectors Two vectors are parallel if one is a scalar multiple of the other That is,: 𝑎 is parallel to 𝑏 if 𝑎=𝜆𝑏 where 𝜆∈𝑅

Worked Example 7 Given the vectors r=𝑖−3𝑗+𝑧𝑘 and s=−2𝑖+6𝑗−7𝑘, find the value of z in each case if: The length of the vector r is 8 The vector r is parallel to the vector s

Linear dependence Linear independence
Vectors are three non-zero vectors Vectors 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 are said to be linearly dependent if there exist non-zero scalars 𝛼,𝛽 𝑎𝑛𝑑 𝛾 such that 𝛼𝑎+𝛽𝑏+𝛾𝑐=0 If 𝛼𝑎+𝛽𝑏+𝛾𝑐=0 , then we can write 𝑐=𝑚𝑎+𝑛𝑏 This implies that one of the vectors is a linear combination of the other two 𝑚=− 𝛼 𝛾 𝑎𝑛𝑑 𝑛=− 𝛽 𝛾 Since 𝛼≠0, 𝛽≠0 and 𝛾≠0, it follows that 𝑚≠0 and 𝑛≠0 Vectors are three non-zero vectors Vectors 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐 are said to be linearly independent if 𝛼𝑎+𝛽𝑏+𝛾𝑐=0 Only if 𝛼=0, 𝛽=0 and 𝛾=0

Worked Example 8 Show that the vectors 𝑎=𝑖−𝑗+4𝑘, b=4𝑖−2𝑗+3𝑘 and c=5𝑖−𝑗+𝑧𝑘 are linearly dependent, and determine the value of z.

Direction cosines 𝛼=𝑎𝑛𝑔𝑙𝑒 𝑐𝑟𝑒𝑎𝑡𝑒𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑂𝑃 𝑎𝑛𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑥−𝑎𝑥𝑖𝑠 𝛽=𝑎𝑛𝑔𝑙𝑒 𝑐𝑟𝑒𝑎𝑡𝑒𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑂𝑃 𝑎𝑛𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑦−𝑎𝑥𝑖𝑠 𝛾=𝑎𝑛𝑔𝑙𝑒 𝑐𝑟𝑒𝑎𝑡𝑒𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑂𝑃 𝑎𝑛𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑧−𝑎𝑥𝑖𝑠 Where 𝑂𝑃 = 𝑥 1 𝑖+ 𝑦 1 𝑗+ 𝑧 1 𝑘 Generalising from the two-dimensional case: cos 𝛼 = 𝑥 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑥 𝑟 cos 𝛽 = 𝑦 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑦 𝑟 cos 𝛾 = 𝑧 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑧 𝑟 cos 𝛼 , cos 𝛽 𝑎𝑛𝑑 cos 𝛾 are called the direction cosines Also: 𝑎= cos 𝛼 𝑖+ cos 𝛽 𝑗+ cos 𝛾 𝑘 cos 2 𝛼 + cos 2 𝛽 + cos 2 𝛾 =1

Worked Example 9 Find the angle to the nearest degree that the vector 2𝑖−3𝑗−4𝑘 makes with the z-axis

Application problems Two-dimensional case: Position vector of moving objects can be found in terms of i, j and k. i: unit vector in the east direction j: unit vector in north direction k: unit vector vertically upward

Worked Example 10 Mary walks 500 metres due south, turns and moves 400 metres due west, and then turns again to move in the direction 𝑁40°𝑊 for a further 200 metres. In all three of those movement she is at the same altitude. At this point, Mary enters a building and travels 20 metres vertically upwards in a lift. Let i, j, and k represent unit vectors of length 1 metre in the directions of east, north and vertically upwards respectively. Find the position vector of Mary when she leaves the lift, relative to her initial position Find her displacement correct to 1 decimal place in metres from her initial point

Column vector notion Vectors have similar properties to matrices
Common to represent vectors as column matrices Vectors in 3D can be represented by unit vectors 𝑖= , 𝑗= 𝑎𝑛𝑑 𝑘= Vector from the origin O to point P with coordinates 𝑥 1 , 𝑦 1 , 𝑧 1 can be express as: 𝑂𝑃 = 𝑥 1 𝑖+ 𝑦 1 𝑗+ 𝑧 1 𝑘 Written in matrix form: 𝑂𝑃 = 𝑥 𝑦 𝑧 = 𝑥 1 𝑦 1 𝑧 1 Please note: Basic operations are performed on matrices in similar ways to those performed on vectors. Vectors and column matrices are called Isometric (Greek word meaning ‘having the same structure’) due to the reason given above.

Worked Example 11 Given the vectors represented as A= −5 , B= 5 3 −2 and C= 3 7 −8 show that the points A, B and C are linearly dependent.

5.4 Scalar product and applications
Connection to the Study Design AOS 4 – Vectors Scalar (dot) product of two vectors, deduction of dot product for i, j and k systems; its use to find scalar vector resolute Magnitude of a vector, unit vector and the orthogonal unit vectors i, j, k Parallel and perpendicular vectors

Key Vocabulary: Key Notation:

Multiplying Vectors Think, Pair, Share
When a vector is multiplied by a scalar, the resultant is a vector. Think: When multiplying two vectors together, what is the resultant? Use drawing of vectors to aid you in your discovery. Pair: In partners, one person is A and the other is B. Share: Person A goes first to share their ideas. Person B does not talk. Only listens. Person B goes next to share their ideas. Person A does not talk. Only listens Share with the class.

Definition of the Scalar Product
Scalar Product/Dot Product 𝑎∙𝑏= 𝑎 𝑏 cos 𝜃 Read as a dot b

Worked Example 12 Given the diagram below, find 𝑎∙𝑏

Properties of the Scalar Product
Resultant is a number Number can be positive, negative or zero 𝑎∙𝑏= 𝑎 𝑏 cos 𝜃 is for non zero vectors both 𝑎 >0 and 𝑏 >0. This implies that the sign of 𝑎∙𝑏 depends on the sign of cos 𝜃 𝑎∙𝑏>0 if cos 𝜃 > 0; therefore 𝜃 is 𝑎∙𝑏<0 if cos 𝜃 < 0; therefore 𝜃 is 𝑎∙𝑏=0 if cos 𝜃 = 0; therefore 𝜃 is It’s commutative which means Which follows as the angle between b and a is and = The scalar product of a vector with itself is the squared magnitude of the vector. That means Scalar or common factors in a vector are merely multiples. That is, if 𝜆∈𝑅, then 𝑎∙ 𝜆𝑏 = 𝜆𝑎 ∙𝑏=𝜆(𝑎∙𝑏)

Component forms i, j and k are unit vectors
𝑖 =1; 𝑗 =1; 𝑘 =1 The angle between vectors i and i is zero implies cos 𝜃 =1. Same applies for j and k Unit vectors are mutually perpendicular If 𝑎= 𝑥 1 𝑖+ 𝑦 1 𝑗+ 𝑧 1 𝑘 and b= 𝑥 2 𝑖+ 𝑦 2 𝑗+ 𝑧 2 𝑘 then 𝑎∙𝑏=

Worked Example 13 If 𝑎=2𝑖+3𝑗−5𝑘 and 𝑏=4𝑖−5𝑗− 2𝑘, find 𝑎∙𝑏:

Orthogonal Vectors Orthogonal means perpendicular
If two vectors are orthogonal then they at 90° of each other Their dot product of two orthogonal vectors is zero

Worked Example 14 If the two vectors 𝑎=3𝑖−2𝑗−4𝑘 and 𝑏=2𝑖−5𝑗+𝑧𝑘 are orthogonal, find the value of z

Angle between two vectors
Previously the angle made by a single vector with the x- y- or z-axis was found by using direction cosines Rearranging the Dot Product to solve for 𝜃

Worked Example 15 Given the vectors 𝑎=𝑖−2𝑗+3𝑘 and 𝑏=2𝑖−3𝑗−4𝑘, find the angle in decimal degrees between the vectors 𝑎 and 𝑏

Finding magnitudes of vectors
The magnitude and the sum or differences of vectors can be found using the properties of the scalar product 𝑎∙𝑏= 𝑎 2 𝑎∙𝑏=𝑏∙𝑎

Worked Example 16 If 𝑎 =3, 𝑏 =5 and 𝑎∙𝑏 =−4, find 𝑎+𝑏

Projections Scalar resolute
Vectors can be resolved either parallel or perpendicular to the x- or y- axis Projections discuss a generalisation of the above process in which one vector is resolved parallel and perpendicular to another vector The projection of vector 𝑎= 𝑂𝐴 onto vector 𝑏= 𝑂𝐵 is defined by dropping the perpendicular from the end of a onto b (at Point C). The projection is defined as this distance along b in the direction of b. The distance OC is called the Scalar Resolute of 𝑎 onto 𝑏 or the Scalar Resolute of 𝑎 parallel to 𝑏

Parallel vector resolute
Perpendicular vector resolute Vector resolute of 𝑎= 𝑂𝐴 onto the vector 𝑏= 𝑂𝐵 is defined as the vector along 𝑏 Vector 𝑂𝐶 has a length of 𝑎∙ 𝑏 and its direction is in the direction of the unit vector 𝑏 Vector resolute is given by 𝑂𝐶 = 𝑎∙ 𝑏 𝑏 Called the component of the vector onto the vector or parallel to vector Vector resolute or component of 𝑎= 𝑂𝐴 perpendicular to the vector is 𝑏= 𝑂𝐵 is vector 𝑂𝐶 Is obtained by subtracting the vectors: 𝑂𝐶 + 𝐶𝐴 = 𝑂𝐴 ; 𝐶𝐴 = 𝑂𝐴 − 𝑂𝐶 𝐶𝐴 is obtained by:

Worked Example 17 Given the vectors 𝑢=3𝑖−𝑗+2𝑘 and 𝑣=2𝑖−3𝑗−𝑘, find:
The scalar resolute of 𝑢 in the direction of 𝑣 The vector resolute of 𝑢 in the direction of 𝑣 a) The scalar resolute of 𝑢 perpendicular of 𝑣

5.5 Vector proofs using the scalar product
Connection to the Study Design AOS 4 – Vectors Vector proofs of simple geometric results, for example the diagonals of a rhombus are perpendicular, the medians of triangle are concurrent, the angle subtended by a diameter in circle is a right angle

Geometrical shapes Quadrilaterals Trapeziums Four sided figure
No two sides are necessarily parallel nor equal in length Trapeziums Four sided figure with on pair of sides parallel but not equal Trapezium ABCD since AB is parallel to DC 𝐴𝐵 =𝜆 𝐷𝐶

Parallelograms Four sided figure with two sets of parallel sides of equal length Parallelogram ABCD, 𝐴𝐵 = 𝐷𝐶 and 𝐴𝐷 = 𝐵𝐶 Rectangles A parallelogram with all angles at 90° Rectangle ABCD, 𝐴𝐵 = 𝐷𝐶 , 𝐴𝐷 = 𝐵𝐶 Thus 𝐴𝐵 ∙ 𝐵𝐶 =0, 𝐵𝐶 ∙ 𝐶𝐷 =0, 𝐶𝐷 ∙ 𝐷𝐴 = 0 and 𝐷𝐴 ∙ 𝐴𝐵 =0 hence all sides are perpendicular

Rhombuses A parallelogram with all sides equal in length Squares A Rhombus with all angles at 90°

Triangles Median of a triangle is the line segment from a vertex to the midpoint of the opposite side Centroid of a triangle is the point of intersection of the three medians G is the centroid of the triangle ABC and O is the origin, it can be shown that 𝑂𝐺 = 𝑂𝐴 + 𝑂𝐵 + 𝑂𝐶

Using vectors to prove geometrical theorems
Vector properties can be used to prove geometrical theorems Following statements are useful for proving geometrical theorems O is the origin and A and B are points, midpoint M of line segment AB given by: 𝑂𝑀 = 𝑂𝐴 + 𝐴𝑀 = 𝑂𝐴 𝐴𝐵 = 𝑂𝐴 𝑂𝐵 − 𝑂𝐴 = 𝑂𝐴 + 𝑂𝐵 If two vectors 𝐴𝐵 and 𝐶𝐷 are parallel, then 𝐴𝐵 =λ 𝐶𝐷 where 𝜆∈𝑅 is a scalar If two vectors 𝐴𝐵 and 𝐶𝐷 are perpendicular, then 𝐴𝐵 ∙ 𝐶𝐷 =0 If two vectors 𝐴𝐵 and 𝐶𝐷 are equal, then 𝐴𝐵 is parallel to 𝐶𝐷 ; furthermore, these two vectors are equal in length, so that 𝐴𝐵 = 𝐶𝐷 ⟹ 𝐴𝐵 = 𝐶𝐷 5. If 𝐴𝐵 =𝜆 𝐵𝐶 , then the points A, B and C are collinear; that is, A, B and C all lie on a straight line

Worked Example 18 Prove that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Parametric equations Connection to the Study Design AOS 4 – Vectors

Parametric equations

Worked Example 19 Given the vector equation , find and sketch the Cartesian equation of the path, and state the domain and range

Eliminating the parameter

Parametric representation

Worked Example 22 Show that the parametric equations and where represent the hyperbola

Sketching parametric curves

Vectors.

Similar presentations

Presentation on theme: "Vectors."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Vectors.

Similar presentations

Presentation on theme: "Vectors."— Presentation transcript:

Similar presentations

About project

Feedback