Addition and Subtraction of Vectors

Transcription

1 ddition and Subtraction of Vectors 1 ppendi ddition and Subtraction of Vectors In this appendi the basic elements of vector algebra are eplored. Vectors are treated as geometric entities represented b directed line segments. The was that the components of a vector can be written in Matlab will be introduced. Prerequisite knowledge: asic trigonometr and plane geometr lgebra including determinants.1 Scalars and Vectors Different tpes of phsical quantities can be distinguished b the number of pieces of information required to completel specif them. For eample, temperature and mass have onl a magnitude and thus a single number representing this magnitude is sufficient to completel specif them. Such phsical quantities are called scalars. Thus a scalar is a phsical quantit that is specified b giving onl a magnitude or single number. Certain quantities, however, cannot be completel specified b a magnitude onl. These are quantities such as velocit and electric field strength, which in addition to a magnitude also have a certain direction associated with them. Such quantities are called vectors. We shall see that in three-dimensional space vectors require three pieces of information to completel specif them. vector then is a phsical quantit that is specified b giving both a magnitude and a direction. There are other kinds of phsical quantities that require more than three pieces of information to completel specif them and are therefore neither scalars nor vectors. Such quantities are called tensors. Since a vector has both a magnitude and a direction, we can represent it b a directed line segment or an arrow in which the length of the line segment is proportional to the magnitude of the vector and the orientation of the line segment specifies the direction of the vector. Thus in Fig..1 the magnitude of the vector is equal to two times the magnitude of the vector and the directions of and differ b 90 degrees. Figure.1 Vectors have a magnitude and direction

2 2 ppendi Notice that we have written the vectors and in boldface tpe to distinguish them from scalars, which are written in italics (T,m). In writing vectors b hand ou cannot make a smbol boldface and so some convention must be used to indicate that the smbol refers to a vector quantit. useful convention is to write a vector quantit as with a wav line under the smbol. This is useful because if the material is printed, ~ the tpesetter will automaticall make such smbols boldface. nother common convention is to use an arrow over the smbol as in...2 ddition of Vectors Since vectors are specified b giving onl a magnitude and a direction, we can relocate the vectors and in Fig..1 provided we preserve their original orientation. The vectors are said to be invariant to translation. If we think of and as two successive displacement vectors that describe a person walking four units east and then two units north,, it is clear that the resultant vector = + can be found b placing such that its tail is at the same point as the head of. The resultant vector then has its tail at the tail of and its head at the head of as shown in Fig..2. Figure.2 The resultant vector = + Now move in Fig..2 so that its tail coincides with the tail of. The resultant is seen to lie along the diagonal of the parallelogram formed b and, with the tails of all three vectors coinciding. Finall shift in Fig..3 such that its tail coincides with the head of the shifted as shown in Fig..4. It is now obvious from Fig..4 that = + = +, showing that the commutative law holds for vector addition. Figure.3 Shift Figure.4 Shift The sum of more than two vectors can be found b continuing to place the tail of succeeding vectors at the head of the preceding vector, as shown in Fig..5. The resultant vector D = + + C is shown in Fig..6.

3 ddition and Subtraction of Vectors 3 C C D Figure C Figure.4 D = + + C Draw the vector sum ( + ) in Fig..4. The result is shown in Fig..5. Finall, draw the vector sum ( + C) in Fig..5. The result is shown in Fig..6. It is now clear from Fig..6 that D = + + C = ( + ) + C = + ( + C), showing that the associative law holds for vector addition. + C + C D D + C Figure.5 D = ( + ) + C Figure.6 D = + ( + C).3 Subtraction of Vectors Two vectors and are shown Fig..7. The vector - is a vector with the same magnitude as but with the opposite direction. Draw - in Fig..7. The result is shown in Fig Figure.7 Vectors and Figure 1.8 The vector - Draw the resultant vector = + in Fig..8. Since = +, verif that - = b showing that + (-) = in Frame.8. The result is shown in Fig..9.

4 4 ppendi - - Figure.9 - = Fig..10 shows two vectors and. Draw the resultant vector = + and the difference vector D = -. Note that the difference vector D can be drawn b connecting the head of with the head of and locating the head of D at the head of as shown in Fig..11. D = = D Figure.10 Vectors and Figure.11 D = -.4 Unit Vectors and Coordinate Sstems If a vector is multiplied b a scalar m, the resulting product m is a vector whose magnitude is equal to m times the magnitude of. The direction of m is the same as that of if m is positive and opposite to that of if m is negative. If λ is a vector having a magnitude of unit, then mλ is a vector whose magnitude is m. The magnitude of the vector is written as =. In Fig..12 the unit vector λ, which has a magnitude of unit, is in the same direction as. We can therefore write the vector as the magnitude of multiplied b the unit vector λ. That is, = λ. The unit vector λ in the direction of can then be written as λ =. λ Figure.12 = λ

5 ddition and Subtraction of Vectors 5 Fig..13 shows to be the vector sum of and. That is, = +. The vectors and lie along the and aes; therefore, we sa that the vector has been resolved into its and components. The unit vectors i and j are directed along the and aes as shown in Fig..13. Using the technique of Fig..12, we can therefore write = i and = j. We can then write in terms of the unit vectors as the vector sum = i + j.. j i θ Figure.13 = i + j In the previous frame we saw that a vector ling in the - plane can be written as = i + j. From the figure we see that the magnitudes are related b from which the ratio = cos θ = sin θ sinθ = = tanθ cosθ Square and and add the results to obtain ( cos θ sin θ) from which = = + = ddition of Vectors b Components To illustrate the addition of vectors b components, consider the vector sum = + shown in Fig..14. resolving and into and components, we can write = + = i + j + i + j from which = ( + )i + ( + )j

6 6 ppendi j i Figure.14 = ( + )i + ( + )j ut in Fig..14 can be written as = i + j. Therefore = + = + These results are summarized in Fig..15. Figure.15 = i + j = ( + )i + ( + )j.6 3-Dimensional Vectors The results above can readil be etended to three dimensions. From Fig..16 the vector is the vector sum = + + z or, in terms of the unit vectors i, j, and k, = i + j + z k The magnitudes associated with this vector sum are shown in Fig In terms of θ and φ we can write

7 ddition and Subtraction of Vectors 7 = cos φ = sin φ z = cos θ z z i k z j + z θ φ Figure.16 = i + j + z k Figure.17 Magnitudes of the components of Since = sin θ, the components in Fig..17 can be written as = sin θ cos φ = sin θ sin φ z = cos θ Square,, and z and add the results to obtain ( ) z = sin θ cos φ+ sin φ + cos θ = sin θ + cos θ 2 =.7 Direction Cosines n alternate wa of describing the vector in three dimensions is b projecting the vector directl onto the,, and z coordinates through the angles α, β, and γ, respectivel, as shown in Fig..18. Thus = cos α = cos β z = cos γ

8 8 ppendi z z α γ β Figure.18 Definition of direction cosines The cosines of the angles α, β, and γ in Fig..18 are called the direction cosines and are designated b l, m, and n, respectivel. Thus, in terms of,,, and z l = cosα = m = cos β = z n = cosγ = Note that l + m + n = cos α + cos β + cos γ z = Since from Eample 1e, = + +, it follows that l 2 + m 2 + n 2 =1.8 Eample 1 z Given the vectors = i 2j + 4k and = 3i + j - 2k, find = +. nswer 1: The components of the given vectors are = 1, = -2, z = 4, = 3, = 1, z = -2. Thus, = + = 4, = + = -1, z = z + z = 2, so that = + = 4i - j + 2k

9 ddition and Subtraction of Vectors 9 nswer 2: In Matlab the vector can be written as the following row matri containing the components of. = [1-2 4] Note that the components of a row vector are separated b spaces. Similarl, the vector can be written as the following row matri containing the components of. = [3 1-2] The sum of these two vectors,, can be found b writing = + in Matlab as shown in Matlab Eample 1a. Matlab Eample 1a >> = [1-2 4] = >> = [3 1-2] = >> = + = >> nswer 3: In Matlab the vector can be written as the following column matri containing the components of. = [1; -2; 4] Note that the components of a column vector are separated b semicolons. Similarl, the vector can be written as the following column matri containing the components of. = [3; 1; -2] The sum of these two vectors,, can be found b writing = + in Matlab as shown in Matlab Eample 1b.

10 10 ppendi Matlab Eample 1b >> = [1; -2; 4] = >> = [3; 1; -2] = >> = + = >> Eample 2 Find = and = for the vectors in Eample 1. nswer 1: = = = 21 = z = + + = = 14 = z nswer 2: In Matlab the magnitude of the vector can be written as norm() as shown in Matlab Eample 2. Matlab Eample 2 >> = [1-2 4] = >> mag = norm() mag = >> = [3 1-2] = >> mag = norm() mag = >>

11 ddition and Subtraction of Vectors Eample 3 Find the unit vector λ in the direction of the vector given in Eample 1. nswer 1: λ = = i j+ k λ = i j k nswer 2: In Matlab the unit vector in the direction of can be found as shown in Matlab Eample 3. Matlab Eample 3 >> = [1-2 4] = >> lambda = /norm() lambda = >>.11 Eample 4 Find the direction cosines of the vector given in Eample 1. nswer 1: 1 l = cosα = = = cos m = β = = = n = cosγ = z = = nswer 2: In Matlab the direction cosines of can be found as shown in Matlab Eample 4.

12 12 ppendi Matlab Eample 4 >> = [1-2 4] = >> l = (1)/norm() l = >> m = (2)/norm() m = >> n = (3)/norm() n = >> Problems Use Matlab to find the answers to the following problems. -1 Given the vectors = 2i + 6j - 3k and = 3i - 3j + 2k, find (a) and (c) 3-4 (b) + (d) - -2 epeat E. -1 for = 5i + 2j - 7k and = -2i - 3j + 4k -3 Given the vectors = 2i + 3j - k, = 4i - 3j + 2k, and C = i + 2j - 3k, find (a) + + C (c) (b) + - C (d) + + C -4 Find the direction cosines and the direction angles α, β, and γ of the vector = 2i + 5j - 3k. -5 epeat E. -4 for = 6i - 5k. -6 Find the unit vector λ in the direction of the vector = 5i - 5j + 10k. Epress λ in terms of i, j, and k.