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  1. 1. Algebra > Vector Algebra >History and Terminology > Disciplinary Terminology > Aeronautical Terminology >Interactive Entries > Interactive Demonstrations >VectorA vector is formally defined as an element of a vector space. In the commonly encountered vector space (i.e.,Euclidean n-space), a vector is given by coordinates and can be specified as . Vectors aresometimes referred to by the number of coordinates they have, so a 2-dimensional vector is often called atwo-vector, an -dimensional vector is often called an n-vector, and so on.Vectors can be added together (vector addition), subtracted (vector subtraction) and multiplied by scalars (scalarmultiplication). Vector multiplication is not uniquely defined, but a number of different types of products, such asthedot product, cross product, and tensor direct product can be defined for pairs of vectors.A vector from a point to a point is denoted , and a vector may be denoted , or more commonly, . Thepoint is often called the "tail" of the vector, and is called the vectors "head." A vector with unit length is calledaunit vector and is denoted using a hat, .When written out componentwise, the notation generally refers to . On the other hand, whenwritten with a subscript, the notation (or ) generally refers to .An arbitrary vector may be converted to a unit vector by dividing by its norm (i.e., length; i.e., magnitude), (1)giving (2)A zero vector, denoted , is a vector of length 0, and thus has all components equal to zero.Since vectors remain unchanged under translation, it is often convenient to consider the tail as located at the originwhen, for example, defining vector addition and scalar multiplication.A vector may also be defined as a set of numbers , ..., that transform according to the rule (3)where Einstein summation notation has been used,
  2. 2. (4)are constants (corresponding to the direction cosines), with partial derivatives taken with respect to the original andtransformed coordinate axes, and , ..., (Arfken 1985, p. 10). This makes a vector a tensor of tensorrankone. A vector with components in called an -vector, and a scalar may therefore be thought of as a 1-vector(or a 0-tensor rank tensor). Vectors are invariant under translation, and they reverse sign upon inversion. Objects thatresemble vectors but do not reverse sign upon inversion are known as pseudovectors. To distinguish vectorsfrompseudovectors, the former are sometimes called polar vectors.A vector is represented in Mathematica as a list of numbers a1, a2, ..., an . Vector addition is then simply writtenusing a plus sign, e.g., a1, a2, ..., an + b1, b2, ..., bn , and scalar multiplication is indicated by placing a scalarnext to a vector (with or without an optional asterisk), s a1, a2, ..., an .Let be the unit vector defined in spherical coordinates by (5)Then the average value of the -component of the over the surface of the unit sphere is given by (6) (7) (8)More generally, (9)for , , or (indexed as 1, 2, 3), and (10) (11) (12)Given vectors , , , , the average values of a number of quantities over the unit sphere are given by (13) (14) (15) (16) (17)
  3. 3. and (18)where is the Kronecker delta, is a dot product, and Einstein summation has been used.A map that assigns each a vector function is called a vector field.Unit VectorA unit vector is a vector of length 1, sometimes also called a direction vector (Jeffreys and Jeffreys 1988). The unitvector having the same direction as a given (nonzero) vector is defined bywhere denotes the norm of , is the unit vector in the same direction as the (finite) vector . A unit vector inthe direction is given bywhere is the radius vector. CROSS PRODUCTThe cross product, also called the vector product, is anoperation on two vectors. The cross product of twovectors produces a third vector which is perpendicular tothe plane in which the first two lie. That is, for the crossof two vectors, A and B, we place A and B so that theirtails are at a common point. Then, their cross Figure 1 A x B = Cproduct, A x B, gives a third vector, say C, whose tail isalso at the same point as those of A and B. Thevector C points in a direction perpendicular (or normal)to both A and B. The direction of C depends on the RightHand Rule.
  4. 4. If we let the angle between A and B be , then the crossproduct of A and B can be expressed as A x B = A B sin( )If the components for vectors A and B are known, thenwe can express the components of their crossproduct, C = Ax B in the following way Cx = AyBz - AzBy Cy = AzBx - AxBz Figure 2 B x A = D Cz = AxBy - AyBxFurther, if you are familiar with determinants, A x B, isComparing Figures 1 and 2, we notice that AxB=-BxA A very nice simulation which allows you to investigate the properties of the cross product is available by clickingHERE. Use the "back" button to return to this place.
  5. 5. Lecture given by Subrata Mukherjee at Cornell University in 1995 Cross ProductWe covered the scalar dot product of two vectors in the last lecture and now move onto the second vector product that can be performed, the Cross Product. The CrossProduct involves taking two vectors and getting as a result another vector which isperpindicular to both vectors. We use the formula below.The direction of n follows the right hand rule, which is easy to learn. Tkae the picturebelow. Point your thumb in the direction of A, the first vector in the cross product.Now point your fingers in the direction of B. Your palm will face in the direction of n,out of the screen.By this method we can do the Cross Products of i,j, and k.
  6. 6. ixj = kjxk = ikxi = jWe can also note that AxB = -BxA, since when we switch the position of our thumband fingers we have to flip our hand over and our palm then faces in the oppositedirection.NOTE: MOST RIGHT HANDED PEOPLE TEND TO LOOK AT THEIR LEFT HANDDURING TESTS, AS THEYRE BUSY WRITING WITH THE RIGHT HAND. MAKESURE TO USE THE RIGHT HAND, BECAUSE THE LEFT HAND WILL GIVE THEWRONG ANSWER. Motivations for the Cross Product. o Mechanics- Torque To find the torque of a force, you multiply the magnitude of the force times the magnitude of the torque arm times the sine of the angle between them. The you define a vector in the direction a right handed screw would turn if twisted in the direction the force is being applied. That is exactly the same as the mathematics of the Cross Product. o Area of a Parrallelogram Computing the Cross Product o By Distributive Property
  7. 7. Example: A=i-j B=i+k AxB = (i - j) x (i + k) = ixi + ixk - jxi - jxk = (0 + (-j ) - (-k) - i) = -i - j + k o By Determinant FormulaNote that both ways return the same answer. (This is very good.)Applications 1. Find the Area of the shaded triangle.
  8. 8. The area can be found using the fact that the area of the corresponding parrallelogram is |PAxPB|. So, the area of the triangle is one half of the area of the paralleogram, or .5*|PAxPB|. PA = -i + j PB = -i + j + k Area = .5*|PAxPB| = .5*|i + j| = sqrt(2)/2 2. Find a vector perpendicular to triangle PAB. PAxPB =i+j3D: Lines, Planes and Segments 0. Lines v = Ai + Bj + Ck (x-x0)i + (y-y0)j + (z-z0)k = t(Ai + Bj + Ck) Taking each component seperately
  9. 9. x = x0 + tA y= y0 + tB z = z0 + tC Example: Find a vector parallel to the line described by the following equations x=2-t y= 3 + t z = 1 - 2t The Components of the vector are just the coefficients on t, so v = -i + j - 2k Lets return to the cube with the triangle inside and find the equation of the line segment PB. Base Point: P(1,0,0) v = PB = -i + j + k x=1-t y=t z=t Now we must place limits on t, since we have a line segment not a full line. At P, t=0, and at B(0,1,1) t=1. So 0<=t<=1.1. Equation of a Pane
  10. 10. n is normal to any segment P0P in the plane. Thus the dot productof n and P0P is 0.[(x-x0)i + (y-y0)j + (z-z0)k] (dot) (Ai + Bj + Ck) = 0A(x-x0) + B(y-y0) + C(z-z0) = 0Ax + By + Cz = Ax0 + By0 + Cz0 = DExample:A Plane is described by the equation 2x - 3y = 5Vector Normal to Plane: 2i - 3j
  11. 11. Einstein Cartesian Vector Notation: BAC-CAB RuleThe Einstein notation for Cartesian vectors is a very useful way of dealingwith many complex vector expressions. Mostly one has to recognize anduse a few basic index patterns. This dialogue provides some practice.