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Initial point

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

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Initial point

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  1. terminal point B(x2 , y2) v Initial point A(x1 , y1) V=AB, vector from the point A to the point B, is a directed line segment between A and B.

  2. MAGNITUDE OF VECTOR B(x2, y2) v A(x1 , y1) ║V║ = (x2−x1)2+ (y2−y1)2

  3. EQUIVALENT VECTORS SAME MAGNITUDE SAME DIRECTION

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

  5. 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)2 + 12 = 25+1 = 26 ║V║= 26 *V is not a unit vector

  6. 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 = 12 + 02 = 1+ 0 = 1 ║ i║ *iis a unit vector 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 = 12 + 02 = 1+ 0 = 1 ║j║ *jis a unit vector

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

  8. VECTOR ADDITION W+V V W TRIANGLE METHOD

  9. VECTOR ADDITION V W+V W PARALLELOGRAM METHOD

  10. VECTOR ADDITION V W+V V-W -W W W-V -V

  11. VECTORS IN A COORDINATE PLANE y C(x2−x1, y2−y1) B(x2, y2) V W=AB x O V = x2−x1 , y2−y1 A(x1 , y1) Vectors W and V are equivalent vectors. Since V starts from the origin we use a special notation for V

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

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

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

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

  16. BASIC VECTOR OPERATIONS V = a , b and W = c , d are two vectors and k is a real number. 1-║V║ = a2 + b2 2- v+w= a , b +c , d = a+c , b+d 3- kV =k a , b = ka , kb 4- ║kV║ = k║V║

  17. BASIC VECTOR OPERATIONS V = -2 , 3 , W = 4 , -1 ║v║= (-2)2 + 32 = 4 + 9 = 13 V +W = -2 , 3 + 4 , -1 = -2+4, 3-1 = 2 , 2 5V = 5 -2 , 3 = -10 , 15 -3W = -3 4 , -1 = -12 , 3 5V −3W = -10 , 15 + -12 , 3 = -10-12 , 15+3 = -22 , 18

  18. ANY VECTOR CAN BE WRITTEN IN TERMS OFTHE UNIT VECTORSi ANDj If V = a , b is any vector, then by using basic vector operations we get ; V = a , b = a , 0 + 0 , b = a1 , 0 + b 0 , 1 = ai+ bj V = a , b = ai + bj

  19. VECTORS WRITTEN IN TERMS OFTHE UNIT VECTORSi ANDj y P(4 , 3) 3j 4i+3j j i 4i x V = 4i+3j = 4 , 3

  20. VECTORS WRITTEN IN TERMS OFTHE UNIT VECTORSi ANDj y j i 4i x 4i-3j -3j P(4 , -3) V = 4i-3j = 4 , -3

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

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

  23. DIRECTION ANGLE OF VECTORS 1 3 If V = , −, then find the direction angle of V. 2 2 3 y 2 tanα = = − = − 3 x 1 2 π α = tan-1[−] = − 3 3

  24. DIRECTION ANGLE OF VECTORS 7π If ║V║ = 6 and the direction angle of V i s , then find the x and y-components of V. 6 cosα , V = x , y = sinα V V 3 1 7π 7π V = = 6cos , 6sin −6 , −6 6 6 2 2 V= −3 3 , −3

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

  26. UNIT VECTORS ON THE SAME DIRECTION WITH A GIVEN VECTOR If V = -3,4,then find the unit vector on the same direction. v = (-3)2 + 42 = 9 + 16 = 25 = 5 -3 4 x y U = , = , 5 5 V V Now find the vector W on the same direction with magnitude 6. -3 4 -18 24 W = 6U = 6 = See the illustrations on the next slide , , 5 5 5 5

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

  28. EXAMPLE If V = −2i +4j, 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. −2 4 x y 20 20 V V

  29. DOT PRODUCT OF VECTORS V = a , b and W = c , d V∙W = ac + bd V∙V = a2 + b2= ║V║2

  30. 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

  31. 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

  32. ANGLE BETWEEN TWO VECTORS V W θ θ angle between Vand W V∙W V∙W = ║V║║W║cos θ cos θ = __________ ║V║║W║

  33. EXAMPLE V = −2 , 3 , W = 4 , − 1 V∙W = ac + bd = −2.4 + 3.(−1) = −8 − 3 = −11 V∙V = a2 + b2= (−2)2 + 32= 13 = ║V║2 V∙W −11 cos θ = __________= _______ ║V║║W║ 13 17

  34. EXAMPLE V = 3 , −6 , W = −1 , 2 V∙W = 3. (−1)− 6.2 = − 3−12 = −15 ║V║= 45 , ║W║= 5 V∙W −15 −15 −15 cos θ = __________= _______ = ______ = _____ = −1 ║V║║W║ 45 5 225 15 cos θ= −1, then θ = cos-1(−1) = π

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

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

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

  38. EXAMPLE V = 3 , −6 , W = −1 , 2 V∙W −15 −15 −15 cos θ = __________= _______ = ______ = _____ = −1 ║V║║W║ 45 5 225 15 cos θ= −1, then θ = cos-1(−1) = π V and W are parallel with opposite direction

  39. EXAMPLE V = 3 , −6 , W = 1 , −2 V∙W 15 15 15 cos θ = __________= _______ = ______ = ___ = 1 ║V║║W║ 45 5 225 15 cos θ= 1, then θ = cos-1(1) = 0 V and W are parallel with same direction

  40. EXAMPLE V = 3 , −6 , W = 4 , 2 V∙W 0 cos θ = __________= _______ = 0 ║V║║W║ 45 20 cos θ= 0 , then θ = cos-1(0) = 90◦ V and W are perpendicular ( orthogonal ) vectors

  41. EXAMPLE 1 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. 2 ║U+V║2= ║U║2 + 2U.V + ║V║2 = 7 1 + 2U.V + ║V║2 = 7 , ║V║2 + 2U.V − 6 = 0 cos θ = __________ = ___,2U.V = ║V║ U∙V 1 2 ║U║║V║ ║V║2 + 2║V║ − 6 = 0 (║V║−2)(║V║+3) = 0 ║V║= 2 or ║V║= −3, ║V║≥ 0 so ║V║= 2

  42. SCALAR PROJECTION W V θ V∙W projWV = ║V║cos θ = _______ ║W║

  43. EXAMPLE If V = 2i + 2j , and W = −4i−2j , then find projWV projWV = _______ = _____ = ______ = ____ -12 -12 -6 V∙W 5 ║W║ 20 2 5

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

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