v Initial point terminal point A(x 1, y 1 ) B(x 2, y 2 ) V =AB, vector from the point A to the point B, is a directed line segment between A and B.

v A(x 1, y 1 ) B(x 2, y 2 ) MAGNITUDE OF VECTOR ║V║ = (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2

U UNIT VECTOR ║U║ = 1 MAGNITUDE OF VECTOR IS 1

EXAMPLES 1- Vector v has the initial point A(3, 4) and the terminal point B( -2, 5). Magnitude of vector v is (-2-3) 2 + (5-4) 2 = (-5) = 25+1 = 26 ║V║= 26 * V is not a unit vector

2- Vector i has the initial point (0, 0) and the terminal point(1, 0). Magnitude of the vector i is = (1-0) 2 + (0-0) 2 = = 1+ 0 = 1 3- Vector j has the initial point (0, 0) and the terminal point (0,1). Magnitude of the vector j is = (0-0) 2 + (1-0) 2 = = 1+ 0 = 1 * i is a unit vector * j is a unit vector ║ i ║ ║j║║j║

SCALAR MULTIPLICATION OF VECTOR V W −W−W 2W 0.5V −2W k 4k

VECTOR ADDITION V W W+VW+V TRIANGLE METHOD

VECTOR ADDITION V W W+VW+V PARALLELOGRAM METHOD

V W W+VW+V -W V-WV-W VECTOR ADDITION -V W-VW-V

VECTORS IN A COORDINATE PLANE x y A(x 1, y 1 ) B(x 2, y 2 ) C(x 2 − x 1, y 2 − y 1 ) O W=AB V Vectors W and V are equivalent vectors. Since V starts from the origin we use a special notation for V V = x 2 − x 1, y 2 − y 1

VECTORS IN A COORDINATE PLANE x y A(5, 1) B(1, 4) C( − 4, 3) O W = AB V Vectors W and V are equivalent vectors. V = − 4, 3 * − 4 is the x -component,and 3 is the y -component of the vector V

VECTORS IN A COORDINATE PLANE x y A(1, 4) B(-3, 1) C(-4, -3) W = AB V Vectors W and V are equivalent vectors. V = -4,

VECTORS IN A COORDINATE PLANE x y i Vectors i and j are special unit vectors. j = 0, j i = 1, 0

VECTORS IN A COORDINATE PLANE Find a vector that has the initial point (3, -1) and is equivalent to V = -2, 3. x A(3, -1) P(-2, 3) B( 1,2) W = AB Vectors W and V are equivalent vectors. V W If ( x, y) is the terminal point of W, then x − 3 = − 2 → x = 1 and y − ( − 1) = 3 → y = 2 y

BASIC VECTOR OPERATIONS V = a, b and W = c, d are two vectors and k is a real number. 1- ║ V║ = a 2 + b 2 2- v + w = a, b + c, d = a+c, b+d 3- kV =k a, b = ka, kb 4- ║kV║ = k ║V║

BASIC VECTOR OPERATIONS 5V = 5 -2, 3 = -10, 15 ║ v ║ = (-2) = = 13 V +W = -2, 3 + 4, -1 = -2+4, 3-1 = 2, 2 -3W = -3 4, -1 = -12, 3 V = -2, 3, W = 4, -1 5V −3W = -10, , 3 = , 15+3 = -22, 18

ANY VECTOR CAN BE WRITTEN IN TERMS OF THE UNIT VECTORS i AND j If V = a, b is any vector, then by using basic vector operations we get ; V = a, b = a, 0 + 0, b = a 1, 0 + b 0, 1 = a i + b j V = a, b = ai + bj

i j 4i 3j 4 i + 3 j P(4, 3) x y V = 4i+3j = 4, 3 VECTORS WRITTEN IN TERMS OF THE UNIT VECTORS i AND j

VECTORS WRITTEN IN TERMS OF THE UNIT VECTORS i AND j i j 4i -3j 4i-3j P(4, -3) x y V = 4i-3j = 4, -3

DIRECTION ANGLE OF VECTORS x y α direction angle of V β V W V = x, y = V cosα, sinα = P(x, y) V cosα, V sinα Q(a, b) W = a, b = W cos β, sin β = W cosβ, W sinβ V cosα = x sinα = V y

DIRECTION ANGLE OF VECTORS If V = -2i + 3j, then find the direction angle of V. tanα = = = − y x 3 -2 V = -2i+3j = -2, 3 α = tan -1 [ − ] in the second quadrant

DIRECTION ANGLE OF VECTORS If V =, −, then find the direction angle of V. tanα = = − = − y x α = tan -1 [ − ] = − π

DIRECTION ANGLE OF VECTORS If ║V║ = 6 and the direction angle of V i s, then find the x and y-components of V. 6 7π7π V = x, y = V cosα, V sinα V = = 6cos, 6sin 6 7π7π 6 7π7π −6, −6− − 3 3, − 3 V=

UNIT VECTORS ON THE SAME DIRECTION WITH A GIVEN VECTOR If V = x, y, then U =, = V x V y is the unit vector on the same direction with V. V = x, y = V cos θ, sin θ U θ is the direction angle of V

UNIT VECTORS ON THE SAME DIRECTION WITH A GIVEN VECTOR If V = -3, 4, then find the unit vector on the same direction. U =, = V x V y v = (-3) = = 25 = , Now find the vector W on the same direction with magnitude 6. W = 6U = 6 = See the illustrations on the next slide , ,

x y P(-3, 4) O U V 4 -3 W V = 5U W = 6U X = -3U X ║W║= 6 ║V║ = 5 ║X║=3

If V = −2 i + 4 j, then find the vector on the opposite direction with magnitude 6. First, find the unit vector on the direction of V. U = =, Now, multiply the unit vector with −6. That will give you the answer. W = −6U The vector on the opposite direction with magnitude 6. V x V y −2− EXAMPLE

DOT PRODUCT OF VECTORS V = a, b and W = c, d V∙W = ac + bd V∙V = a 2 + b 2 = ║V║ 2

DOT PRODUCT OF VECTORS 1− V ∙ W = W ∙ V dot product is commutative 2− U ∙ (V + W) = U ∙ V + U ∙ W distributive 3− a (V ∙ W) = (aV ) ∙ W=V ∙ (aW),a is a scalar 4− V ∙ V = ║V║ 2 5− 0 ∙ W = 0 zero vector 6− i ∙ i = j ∙ j = 1 7− i ∙ j = j ∙ i = 0

DOT PRODUCT OF VECTORS ║U+V║ 2 = (U+V).(U+V) = U.U+U.V+V.U+V.V ║U+V║ 2 = ║U║ 2 + 2U.V + ║V║ 2 SIMILARLY ║U−V║ 2 = (U−V).(U−V) = U.U−U.V−V.U+V.V ║U−V║ 2 = ║U║ 2 − 2U.V + ║V║ 2

ANGLE BETWEEN TWO VECTORS V W θ angle between V and W θ V∙W = ║V║║W║cos θ cos θ = __________ ║V║║W║ V∙W

EXAMPLE V = − 2, 3, W = 4, − 1 V∙W = ac + bd = − ( − 1) = − 8 − 3 = − 11 V∙V = a 2 + b 2 = ( − 2) = 13 = ║V║ 2 cos θ = __________ = _______ ║V║║W║ V∙WV∙W −

EXAMPLE V = 3, −6, W = −1, 2 V∙W = 3. ( − 1) − 6. 2 = − 3 − 12 = − 15 ║V║= 45, ║W║= 5 cos θ = __________ = _______ = ______ = _____ = − 1 ║V║║W║ V∙WV∙W − − − cos θ = − 1, then θ = cos -1 ( − 1) = π

EXAMPLE V W If V∙W = 0,then V and W are perpendicular θ is 90 ◦, cos θ = 0 θ

EXAMPLE If V = 4i-3j, then find a vector that is perpendicular to V V∙W = 4x – 3y = 0 If W = xi + yj, then Any choice of x and y that satisfies the equation above is an answer Since 3 and 4 satisfy the equation W = 3, 4 is one of the vectors

V W V and W are parallel V W θ is 0 ◦, cos θ = 1 θ is 180 ◦, cos θ = -1 Same direction Opposite direction

EXAMPLE V = 3, −6, W = −1, 2 cos θ = __________ = _______ = ______ = _____ = − 1 ║V║║W║ V∙WV∙W − − − cos θ = − 1, then θ = cos -1 ( − 1) = π V and W are parallel with opposite direction

EXAMPLE V = 3, −6, W = 1, −2−2 cos θ = __________ = _______ = ______ = ___ = 1 ║V║║W║ V∙WV∙W cos θ = 1, then θ = cos -1 (1) = 0 V and W are parallel with same direction

EXAMPLE V = 3, −6, W = 4, 2 cos θ = __________ = _______ = 0 ║V║║W║ V∙WV∙W cos θ = 0, then θ = cos -1 (0) = 90 ◦ V and W are perpendicular ( orthogonal ) vectors

EXAMPLE If ║U + V║ = 7, U is a unit vector and cos θ = ____ where θ is the angle between U and V, then find the magnitude of V. 1 2 ║U+V║ 2 = ║U║ 2 + 2U.V + ║V║ 2 = U.V + ║V║ 2 = 7, ║V║ 2 + 2U.V − 6 = 0 cos θ = __________ = ___, 2U.V = ║V║ ║U║║V║ U∙V 1 2 ║V║ 2 + 2║V║ − 6 = 0 (║V║−2)(║V║+3) = 0 ║V║= 2 or ║V║= −3, ║V║≥ 0 so ║V║= 2

SCALAR PROJECTION V W θ proj W V = ║V║cos θ = _______ ║W║║W║ V∙WV∙W

EXAMPLE If V = 2i 2i + 2j 2j, and W = −4i−2j −4i−2j, then find proj W V proj W V = _______ = _____ = ______ = ____ ║W║║W║ V∙WV∙W

EXAMPLE If V = kW for any nonzero number k, then proj W V = _______ = _________ = _________ = ________ ║W║║W║ V∙WV∙W(kW)∙W ║W║║W║ k(W∙W) ║W║║W║ k ║W║ 2 ║W║ proj W V = k║W║ How about proj V W ?.