Vectors have both magnitude and direction, represented by arrows. The sum of two vectors is obtained by placing the tail of one vector at the head of the other. If the vectors are at right angles, their dot product is zero, while their cross product is maximum. Scalar multiplication scales the magnitude but not the direction of a vector.

5. A x = A cos  A y = A sin  │ A │ =√ ( A x 2 + A y 2 ) The magnitude (length) of A is found by using the Pythagorean Theorem The length of a vector clearly does not depend on its direction. y x A A x A y A 

6. The direction of A can be stated as tan  = Ay / Ax  =tan -1 (Ay / Ax) y x A A x A y A 

7. Some Properties of Vectors Equality of Two Vectors Two vectors A and B may be defined to be equal if they have the same magnitude and point in the same directions. i.e. A = B A B A A B B

8. Negative of a Vector The negative of vector A is defined as giving the vector sum of zero value when added to A . That is, A + (- A) = 0 . The vector A and –A have the same magnitude but are in opposite directions. A -A

10. B =  A If  = 0, therefore B =  A = 0, which is also known as a zero vector  (  A) =  A =  (  A) (  +  )A =  A +  A Example

12. More than two vectors? x 1 x 5 x 4 x 3 x 2  x i  x i = x 1 + x 2 + x 3 + x 4 + x 5 Example

13. Vector Subtraction Equivalent to adding the negative vector Example A -B A - B B A B C = A + (-B) C =

17. Parallelogram method of addition (tailtotail) The magnitude of the resultant depends on the relative directions of the vectors A B A + B

25. Dot product (scalar) of two vectors The definition: θ B A A · B = │A││B │cos θ

31. Measurement and Error

32. THE END

33. Vectors are represented by an arrow A - B B A A θ

34. Conceptual Example If B is added to A, under what condition does the resultant vector A + B have the magnitude equal to A + B ? Under what conditions is the resultant vector equal to zero? *

35. Example (1Dimension) x 1 = 5 x 2 = 3  x = x 2 - x 1 = 2 x 1 + x 2 x 1 + x 2 = 8 MORE EXAMPLE x 1 x 2 x 1 x 2  x = x 2 - x 1

38. Example 2 (A Vacation Trip) A car travels 20.0 km due north and then 35.0 km in a direction 60 0 west of north. Find the magnitude and direction of the car’s resultant displacement. SOLUTION

39. Solution The magnitude of R can be obtained using the law of cosines as in figure Since θ =180 0 – 60 0 = 120 0 and C 2 = A 2 + B 2 – 2AB cos θ , we find that C = 48.2 km C A B 60 θ β Continue C = √A 2 + B 2 – 2AB cos θ C = √20 2 + 35 2 – 2(20)(35) cos 120 0

40. The direction of C measured from the northerly direction can be obtained from the sines law β = 38.9 0 Therefore, the resultant displacement of the car is 48.2 km in direction 38.9 0 west of north

41. Conceptual Example If one component of a vector is not zero, can its magnitude be zero? Explain. * MORE EXAMPLE

42. Conceptual Example If A + B = 0, what can you say about the components of the two vectors? *

43. Example 1 Find the sum of two vectors A and B lying in the xy plane and given by A = 2.0i + 2.0j and B = 2.0i – 4.0j SOLUTION

44. Solution Comparing the above expression for A with the general relation A = A x i + A y j , we see that A x = 2.0 and A y = 2.0. Likewise, B x = 2.0, and B y = -4.0 Therefore, the resultant vector C is obtained by using Equation C = A + B + (2.0 + 2.0)i + (2.0 - 4.0)j = 4.0i -2.0j or C x = 4.0 C y = -2.0 The magnitude of C given by equation * Find the angle θ that C makes with the positive x axis Exercise C = √C x 2 + C y 2 = √20 = 4.5

45. Example A particle undergoes three consecutive displacements d 1 = (1.5i + 3.0j – 1.2k) cm, d 2 = (2.3i – 1.4j – 3.6k) cm d 3 = (-1.3i + 1.5j) cm. Find the component and its magnitude.

46. Solution R = d 1 + d 2 + d 3 = (1.5 + 2.3 – 1.3)i + (3.0 – 1.4 + 1.5)j + (-1.2 – 3.6 + 0)k = (2.5i + 3.1j – 4.8k) cm That is, the resultant displacement has component R x = 2.5 cm R y = 3.1 cm and R z = -4.8 cm Its magnitude is R = √ R x 2 + R y 2 + R z 2 = 6. 2 cm

47. Example - 2D [headtotail] x 1 + x 2 (1, 0) (2, 2) x 1 + x 2 = (1, 0) + (2, 2) = (3, 2) x 1 x 2

48. Example - 2D [tailtotail] x 1 - x 2 ? (1, 0) (2, 2) x 1 + x 2 = (1, 0) + (2, 2) = (3, 2) (x 2 ) x 1 x 1 + x 2 x 2

49. Example of 2D (subtraction) (1, 0) (2, 2) x 1 x 2 x 1 + x 2

50. Example -2D for subtraction x 1 -x 2 x 1 - x 2 (1, 0) (2, 2) x 1 - x 2 = (1, 0) - (2, 2) = (-1, -2) x 1 - x 2 = x 1 + (-x 2 )

51. Not given the components? 1 m 2  2 m 45 o X 1 = (1, 0) X 2 = (x 2E , x 2N ) = (2  2cos(45 o ), 2  2sin(45 o )) = (2, 2) x 1 -x 2 x 1 - x 2 2  2 m 1 m 45 o Cosine rule: a 2 =b 2 + c 2 - 2bccosA = 1 + 8 - 2  2(1/  2) a =  5 m