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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.**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**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.**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 = Vx+ Vy**Graphically the vector is broken into its components as**follows: Y VyV X Vx V = Vx+ Vy**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**The components of a vector add up to form the vector itself,**i.e. V = Vx+ Vyin 2 dimensions V Vy Vx**Or in three dimensions**V = Vx+ Vy+ Vz**Components in 3-d**Z V Y VxVz Vy X**When we resolve a vector into its components, e.g.**V = Vx+ Vy the magnitude of the two component vectors is given by the relations |Vx| = |V| cosϴ |Vy| = |V| sin ϴ**The Pythagorean theorem then gives a relation between the**magnitudes of the x and y components, i.e. |V|2 = |Vx |2 + |Vy |2 in 2-dimensions And |V|2 = |Vx |2 + |Vy |2 + |Vz |2 in 3-d**Use of Unit Vectors i j k**It is convenient to define three unit vectors iparallel 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. Vx = Vxiwhere Vx = |Vx|**Z**k j Y I X**Dot or Scalar Product**The dot or scalar product of two vectors A · B Is a scalar quantity. A = Axi + Ayj + Azk B = Bxi + Byj + Bzk A · B = |A||Bcosϴ A · B = AxBx + AyBy + AzBz**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| = [22 + 12 + (-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) cosϴ = 12.41 cosϴ = - 8 cos ϴ = - 8/12.41 = - 0.645 ϴ = cos-1 (- 0.645) = 130.2⁰**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 = Cxi + Cyj + Czk A·C = 2Cx + 3Cy – Cz = 0 You now have three unknowns and only one equation.**A·C = 2Cx + 3Cy – Cz = 0**Let Cy = 1 2Cx + 3 – Cz = 0 Let Cx = 1 2 + 3 – Cz = 0 Cz = 5 So the vector C = i+ j +5k is normal to A.**The order of the vectors in the dot product does not affect**the dot product itself, i.e. A · B = B · A**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.**Calculation of C = A x B**If the vectors A and B are A = Axi + Ayj + Azk B = Bxi + Byj + Bzk then i j k C = Ax AyAz Bx By Bz**To evaluate the determinant, it is convenient to write the**iand 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 = Ax AyAzAx Ay Bx By BzBxBy**C = AyBzi+ AzBxj + AxByk- AzByi– AxBzj - AyBxk**Or C = (AyBz – AzBy)i+ (AzBx– AxBz)j + (AxBy– AyBx)k**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**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**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.**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**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’|**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| = [22 + 12 + (-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**Useful Information**A · A = |A|2 i · i = 1 j · j = 1 k · k = 1 A x A = 0 i x i = 0 j x j = 0 k x k = 0 i x j = k j x k = i k x i = j j x i = -k k x j = -ii x k = -j