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Vectors

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Definitions Scalar – magnitude only Vector – magnitude and direction I am traveling at 65 mph – speed is a scalar. It has magnitude but no direction. I am traveling north at 65 mph – speed is a vector. It has both magnitude and direction.

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Graphical Representation The vector V is denoted graphically by an arrow. The length of the Arrow represents the magnitude of the vector. The direction of the arrow represents the direction of the vector. V

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Vector Representation A vector may be represented by a letter with an arrow over it, e.g. V A vector may be represented by a letter in bold faced type, e.g. V For ease of typing or word processing, vectors will be represented by the bold faced type.

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Components of a Vector In two dimensions, a vector will have an x- component (parallel to the X-axis) and a y- component (parallel to the Y-axis). In vector terms, V = V x + V y

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Graphically the vector is broken into its components as follows: Y V y V X V x V = V x + V y

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Vector Addition Vectors may be added graphically by placing the tail of the second vector at the head of the first vector and then drawing a new vector from the origin to the head of the second vector. B C C = A + B A

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The components of a vector add up to form the vector itself, i.e. V = V x + V y in 2 dimensions V V y V x

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Or in three dimensions V = V x + V y + V z

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Components in 3-d Z V Y V x V z V y X

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When we resolve a vector into its components, e.g. V = V x + V y the magnitude of the two component vectors is given by the relations |V x | = |V| cos ϴ | V y | = |V| sin ϴ

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The Pythagorean theorem then gives a relation between the magnitudes of the x and y components, i.e. |V| 2 = |V x | 2 + |V y | 2 in 2-dimensions And |V| 2 = |V x | 2 + |V y | 2 + |V z | 2 in 3-d

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Use of Unit Vectors i j k It is convenient to define three unit vectors i parallel to the X axis j parallel to the Y axis k parallel to the Z axis And to express the components of the vector in terms of a scalar times the unit vector along that axis. V x = V x iwhere V x = | V x |

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Z k j Y I X

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Dot or Scalar Product The dot or scalar product of two vectors A · B Is a scalar quantity. A = A x i + A y j + A z k B = B x i + B y j + B z k A · B = |A||B cos ϴ A · B = A x B x + A y B y + A z B z

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Example of Dot Product Consider A = 2i + j – 3k B = -i - 3j + k A·B = (2)(-1) + (1)(-3) + (-3)(1) = -2-3-3 = -8 |A| = [2 2 + 1 2 + (-3) 2 ] 1/2 = [14] 1/2 = 3.74 |B| = [(-1) 2 + (-3) 2 + (1) 2 ] 1/2 = [11] 1/2 = 3.32 A·B = (3.74)(3.32) cos ϴ = 12.41 cos ϴ = - 8 cos ϴ = - 8/12.41 = - 0.645 ϴ = cos -1 (- 0.645) = 130.2⁰

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Given a vector A = 2i + 3j – k, we can find a vector C that is normal to A by using the fact that the dot product A·C = 0 if A is normal to C. C = C x i + C y j + C z k A·C = 2C x + 3C y – C z = 0 You now have three unknowns and only one equation.

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How many equations do you need to solve for three unknowns?

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I can solve for three unknowns with only this one equation!

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A·C = 2C x + 3C y – C z = 0 Let C y = 1 2C x + 3 – C z = 0 Let C x = 1 2 + 3 – C z = 0 C z = 5 So the vector C = i + j +5k is normal to A.

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The order of the vectors in the dot product does not affect the dot product itself, i.e. A · B = B · A

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Cross (Vector) Product The cross product of two vectors produces a third vector which is normal to the first two vectors, i.e. C = A x B So vector C is normal to both A and B.

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Calculation of C = A x B If the vectors A and B are A = A x i + A y j + A z k B = B x i + B y j + B z k then i j k C = A x A y A z B x B y B z

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To evaluate the determinant, it is convenient to write the i and j columns to the right and multiply along each of the diagonals 1, 2, and 3 and add them, then multiply along 4, 5, and 6 and subtract them. 1 2 3 4 5 6 i j k i j C = A x A y A z A x A y B x B y B z B x B y

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C = A y B z i + A z B x j + A x B y k - A z B y i – A x B z j - A y B x k Or C = (A y B z – A z B y )i + (A z B x – A x B z ) j + (A x B y – A y B x )k

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Example of cross product Calculate C = A x B where A = i + 2j - 3k B = 2i - 3j + k i j k C = 1 2 -3 2 -3 1

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C = (2)(1)i + (-3)(2)j + (1)(-3)k - (-3)(-3)i – (1)(1)j – (2)(2)k C = (2 – 9)i + (- 6 – 1)j + (- 3 -4)k C = -7i -7j -7k

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To check that we have not made any mistakes in calculating the cross product, we can calculate the dot product C·A which should be equal to 0 since vector C is normal to vectors A and B. C·A = (-7)(1) + (-7)(2) + (-7)(-3) = -7 – 14 +21 = 0 So we have not made any mistakes in calculating C.

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The cross product is different if the order is reversed, i.e. A x B = C But B x A = - C B x A = - A x B

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When we look at the vector C = -7i – 7j – 7k It has the same direction as the vector C’ = - i – j – k But a different magnitude. |C| = 7 |C’|

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Unit Vectors To create a unit vector g in the same direction as G, simply divide the vector by its own magnitude, e.g. g = G/|G| If G = 2i + j – 3k Then |G| = [2 2 + 1 2 + (-3) 2 ] 1/2 = [4 + 1 + 9] 1/2 = [14] 1/2 = 3.74 g = (2/3.74)i + (1/3.74)j – (3/3.74)k

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Useful Information A · A = |A| 2 i · i = 1 j · j = 1 k · k = 1 A x A = 0 i x i = 0j x j = 0k x k = 0 i x j = kj x k = i k x i = j j x i = -kk x j = -i i x k = -j

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