Dr. Fritz Wilhelm, DVC,8/30/2004;4:25 PM E:\Excel files\ch 03 Vector calculations.doc Last printed 8/30/2004 4:25:00 PM

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1 E:\Ecel files\ch 03 Vector calculations.doc Last printed 8/30/2004 4:25:00 PM Vector calculations 1 of 6 Vectors are ordered sequences of numbers. In three dimensions we write vectors in an of the following forms: (1.1) = i + j + k =,, The quantities i are called the components of the vector ; the magnitude of the vector is called (no arrow) and is equal to = = magnitude = + + = i i i= 1 The dot indicates the scalar or dot product. The direction of the vector requires three angles in three dimensions, but fortunatel onl one angle in two dimensions. In the case of two dimensional vectors we have: θ =tan -1 -ais j u θ i -ais n location vector r can be written in terms of a directional unit-vector ur and the magnitude of the vector which is r. unit vector is an vector of unit sie. We can obtain it b taking an regular vector and dividing it b its magnitude. (1.2) u = =

2 E:\Ecel files\ch 03 Vector calculations.doc Last printed 8/30/2004 4:25:00 PM 2 of 6 We c hoose an orthogonal set of unit vectors, which together form the basis for our vector space. This means that an vector in this vector-space can be written in terms of a linear combination of the unit-vectors. (1.3) = i + j + k =,, The Cartesian basis consists of the fied vectors i, j, k in the,, and direction respectivel. These are all unit vectors, which are perpendicular to each other. The components of a vector on this basis of unit vectors are the projections of the vector on the unit vectors. In the case of the Cartesian unit vectors: = cos θ, = cos β, = cosγ (1.4) the angles θβγ,, are the angles between the vector and the unit vectors. nother set of unit vectors is used often in conjunction with circular motion. We use a radial and a tangential unit vector ur and u θ. We define the radial unit vector as the vector pointing from the center of the circle to the point moving on the circle. These unit vectors move with the particle. Their time-derivative is therefore not equal to 0. So (1.5) r =ru r or for our vector = magnitude times unit vector 2 2 = u = + uwith u = = If we add, subtract or multipl vectors we can do so using an coordinate sstem. ddition of vectors: (1.6) + = +, + which corresponds to graphicall forming a parallelogram out of the two vectors in such a wa that the tail of one meets the head of the other. Head to tail, for short. If we move the two vectors around parallel to themselves in such a wa that their tails start at the same point, we create the difference between the two vectors. - is the vector which starts at the end of and goes to the end of. -=,

3 E:\Ecel files\ch 03 Vector calculations.doc Last printed 8/30/2004 4:25:00 PM 3 of 6 + -ais u θ j i -ais The location of a point at the time t is described b the location vector rt () = t (), t (), t (). Its derivative is the velocit vector. moving point then describes a continuous motion associated with a velocit vector r r( t+ t) r( t) (1.7) v(t)= lim = lim t 0 t t 0 t Each component of the velocit vector is obtained b the same limiting process shown below, which we used for a function (t) of a single variable t: d() t ( t + t) () t (1.8) = v() t = () t = lim = lim ; ; t dt t 0 t t 0 t We can of course form the time-derivative of an vector: If we use the Cartesian coordinates with constant unit vectors, we get d d d d d d d (1.9) = i + j + k =,, dt dt dt dt dt dt dt

4 E:\Ecel files\ch 03 Vector calculations.doc Last printed 8/30/2004 4:25:00 PM 4 of 6 Scalar Product: (see also chapter 7) (1.10) = cos θ = + + where the angle θ is the angle between the two vectors, which, b definition, is the smallest angle ou get b putting the tails of the two vectors together. Measure angles alwas going counter clock wise. s ou can see, the scalar product is just a number. Now that we have defined the scalar product between two vectors we see that the components of a vector are the scalar products between the vector and the unit vector. For eample: (1.11) = i i = 1 cos θ : projection of the vector on the -direction. We can appl the rule for scalar products convenientl to determine the law of cosines: + = C C = 2 2 C γ ( C) = ( ) = α β 2iC+ C = = 2iCcosα + C This is a nice proof of the law of cosines. Find the angles for =10, =8, C=4 C cosα = 2C i Vector Product or cross Product: (See also chapter 11) Vector product: C = = The magnitude of the vector C is equal to sin θ where θ is the (1.12) angl e between and. The direction of the vector C is perpendicular to the plane spanned b and, using the right hand rule. The thumb of our right hand points in the direction of, our inde finger is, our middle finger, perpendicular to the other two fingers points in the direction of C. 2

5 E:\Ecel files\ch 03 Vector calculations.doc Last printed 8/30/2004 4:25:00 PM 5 of 6 C α The area spanned b the two vectors and is equal to sinα, which means that ½ sinα is equal to the triangle between the two vectors (1.13) C = C = C = sinθ One convenient wa to calculate the cross product is b means of a determinant: i j k = C, C, C = i j + k = (1.14) i ( ) j( ) + k( ) C = ; C = ; C = (1.15) ssume that =,, =,, where is a scalar. i j k i j k = = = C, C, C Here is the rule: common factor in a row or a column of a determinant can be pulled in front of the determinant. (1.16) The double cross-product is another vector. Double vector product: C = C C ( ) ( ) ( )

6 E:\Ecel files\ch 03 Vector calculations.doc Last printed 8/30/2004 4:25:00 PM 6 of 6 Mied product: Cclical permutation (1.17) ( C) = C ( ) = ( C ) = C C C The mied product is a scalar, its absolute value is equal to the volume of the parallelepiped spanned b the three vectors. Derivatives of products: Let c(t) be a scalar function of t, let () t and () t be vector functions of t: d dc d ( c) = + c dt dt dt d d d (1.18) ( ) = + dt dt dt d d d ( ) = + dt dt dt The time derivative of an vector with constant magnitude is perpendicular to the original vector: 2 2 dc = = = c; = 0 dt (1.19) d 2 d d d d ( ) = + = 2 = 0= 2 cosθ cosθ = 0 dt dt dt dt dt