Page 29 - Engineering Electromagnetics, 8th Edition
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CHAPTER 1   Vector Analysis            11

                     Solution. Substituting the coordinates of point Q into the expression for G,wehave

                                           G(r Q ) = 5a x − 10a y + 3a z
                     Next we find the scalar component. Using the dot product, we have
                                                                   1
                                                  1
                         G · a N = (5a x − 10a y + 3a z ) · (2a x + a y − 2a z ) = (10 − 10 − 6) =−2
                                                  3
                                                                   3
                     The vector component is obtained by multiplying the scalar component by the unit
                     vector in the direction of a N ,
                                         1
                         (G · a N )a N =−(2) (2a x + a y − 2a z ) =−1.333a x − 0.667a y + 1.333a z
                                         3
                     The angle between G(r Q ) and a N is found from

                                         G · a N =|G| cos θ Ga
                                                 √
                                            −2 =   25 + 100 + 9 cos θ Ga
                     and
                                                       −2
                                           θ Ga = cos −1  √  = 99.9 ◦
                                                        134

                        D1.3. The three vertices of a triangle are located at A(6, −1, 2), B(−2, 3, −4),
                        and C(−3, 1, 5). Find: (a) R AB ;(b) R AC ;(c) the angle θ BAC at vertex A;(d) the
                        (vector) projection of R AB on R AC .

                                                              ◦
                        Ans. −8a x + 4a y − 6a z ; −9a x + 2a y + 3a z ;53.6 ; −5.94a x + 1.319a y + 1.979a z

                     1.7 THE CROSS PRODUCT
                     Given two vectors A and B,wenow define the cross product,or vector product,of A
                     and B, written with a cross between the two vectors as A × B and read “A cross B.”
                     The cross product A × B is a vector; the magnitude of A × B is equal to the product
                     of the magnitudes of A, B, and the sine of the smaller angle between A and B; the
                     direction of A×B is perpendicular to the plane containing A and B and is along one of
                     the two possible perpendiculars which is in the direction of advance of a right-handed
                     screw as A is turned into B. This direction is illustrated in Figure 1.5. Remember that
                     either vector may be moved about at will, maintaining its direction constant, until
                     the two vectors have a “common origin.” This determines the plane containing both.
                     However, in most of our applications we will be concerned with vectors defined at
                     the same point.
                         As an equation we can write


                                            A × B = a N |A||B| sin θ AB               (7)

                     where an additional statement, such as that given above, is required to explain the
                     direction of the unit vector a N . The subscript stands for “normal.”
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