EE468G NOTES (1) Reading assignment: Vectors Contents: Unit Vectors, Vectors, Vector operations Vectors in three coordinate systems Vector: A quantity that has magnitude and direction Examples: Force, velocity, field intensity Notation and writting: or A or AA Unit vector: a vector whose magnitude is 1. Examples: ˆ ˆxax is a unit vector pointing in x-direction, ˆyis a unit vector pointing in y-direction. Vector representation in rectangular coordinate system: component format ˆ5Ax ˆ ˆ37B x y ˆ ˆˆ22C x y z   x y x y x y z 2 2 1 -7 3 5 In general, a vector has three components xA, yA, zA: ˆ ˆˆx y zA A x A y A z  , Component xA is the projection of vector A on x-axis.Calculation of magnitude: 222x y zA A A A  . Examples: Direction of a vector: The direction of a vector can be described by three angles that the vector made with the x-axis, y-axis, and z-axis, respectively. In the above examples,  222ˆ ˆ3 7 , 3 7 0 58B x y B       It makes an angle of  1cos 3 58 66.8o with x-axis, It makes an angle of  1cos 7 58 156.8o with y-axis, It makes an angle of  1cos 0 58 90o with z-axis. In general, if ˆ ˆˆx y zA A x A y A z  , the angles it makes with x-, y-, and z-components are: z, , and xy  , where      1 1 1cos , cos , cosx x y y z zA A A A A A       Example: Let ˆ ˆˆ6 2.5 0.8A x y z  , calculate A, z, , and xy  . Solution:     2221116 2.5 0.8 42.89 6.55cos 6 6.55 23.65cos 2.5 6.55 67.56cos 0.8 6.55 97.02oxoyozA         2222 2 2ˆ ˆ3 7 , 3 7 0 58ˆ ˆˆ2 2 , 2 2 1 9 3B x y BC x y z C             Vector operations: Addition and subtraction ˆ ˆˆx y zA A x A y A z  , ˆ ˆˆx y zB B x B y B z   Addition of A and B:    ˆ ˆˆx x y y z zA B A B x A B y A B z       Subtraction of A and B:    ˆ ˆˆx x y y z zA B A B x A B y A B z       The results of addition and subtraction are vectors. The addition and subtraction operation satisfy the commutative law and distributive law. Graphical representation of ABand AB A B A+B B A B A-B Example: Let ˆ ˆˆ6 2.5 0.8A x y z  , ˆ ˆˆ3 2 6B x y z   calculate AB, and AB Solution:      ˆ ˆ ˆ ˆˆ ˆ6 3 2.5 2 6 0.8 3 4.5 5.2A B x y z x y z               ˆ ˆ ˆ ˆˆ ˆ6 3 2.5 2 6 0.8 9 0.5 6.8A B x y z x y z         Vector operation: Scalar product (dot product): ˆ ˆˆx y zA A x A y A z  , ˆ ˆˆx y zB B x B y B z   Scalar product: x x y y z zA B A B A B A B     The result of the dot product of two vectors is a scalar.  The addition and subtraction operation satisfy the commutative law: A B B A   and distributive law:  A B C A B A C       If the two vectors A and B make an angle of AB, then  cosABA B A B Example: Let ˆ ˆˆ6 2.5 0.8A x y z  , ˆ ˆˆ3 2 6B x y z   , ˆ ˆˆ3 10C x y z  , Calculate , , and A B B C A C   Solution:    6 3 2.5 2 0.8 6 17.8AB              3 1 2 3 6 10 51BC             6 1 2.5 3 0.8 10 9.5AC           Two vectors, A and B, are said to be perpendicular to each other if and only if 0AB. B AB AA is perpendicular to B can also be written as: AB. Example: Let ˆ ˆˆ6 1.5A x y a z  , ˆ ˆˆ3 2 6B x y z   , show that when 15/6a , we have AB. Proof: AB requires 0AB Consider  6 3 1.5 2 6 15 6A B a a           Let 0AB, then: 15/6a . In Cartesian coordinate system, ˆ ˆ ˆ ˆ ˆ ˆ,,x y y z z xa a a a a a  Cross product: ˆ ˆˆx y zA A x A y A z  , ˆ ˆˆx y zB B x B y B z   Cross product is defined as:    ˆ ˆˆy z z y z x x z x y y xA B A B A B x A B A B y A B A B z       ->The result of the cross product of two vectors is a vector ->which is perpendicular to both A and B (RH-RULE). ->Cross product satisfies distributive law:  A B C A B A C      ->Cross product does not satisfy commutative law: A B B A   But: (A B B A   ) In Cartesian coordinate system: ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ,,x y z y z x z x ya a a a a a a a a      Hence the unit vectors satisfy the right-hand rule: ˆ ˆ ˆx y za a a ˆza ˆza ˆxa ˆyaCalculation of cross product using determinants: ˆ ˆˆˆ ˆˆy z x yxzx y zy z x yxzx y zx y zA A A AAAA B A A A x y zB B B BBBB B B     Example: Let ˆ ˆˆ6 2.5 0.8A x y z  , ˆ ˆˆ3 2 6B x y z   , ˆ ˆˆ3 10C x y z  , Calculate , A B B C. Solution: ˆ ˆˆ2.5 0.8 6 0.8 6 2.5ˆ ˆˆ6 2.5 0.82 6 3 6 3 23 2 6ˆ ˆˆ16.6 33.6 19.5x y zA B x y zx y z        ˆ ˆˆ2 6 …