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CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2,11, −10) = (−580.5, 3193, −2902) 1.2. The three vertices of a triangle are located at A(−1, 2, 5), B(−4, −2, −3), and C(1, 3, −2). a) Find the length of the perimeter of the triangle: Begin with AB = (−3, −4, −8),√ = (5, 5, 1), BC and CA = (−2, √ 7). Then the perimeter will be = |AB| + |BC| + |CA| = 9 + 16 + 64 + −1, √ 25 + 25 + 1 + 4 + 1 + 49 = 23.9. b) Find a unit vector that is directed from themidpoint of the side AB to the midpoint of side BC: The vector from the origin to the midpoint of AB is MAB = 1 (A + B) = 1 (−5ax + 2az ). 2 2 The vector from the origin to the midpoint of BC is MBC = 1 (B + C) = 1 (−3ax + ay − 5az ). 2 2 The vector from midpoint to midpoint is now MAB − MBC = 1 (−2ax − ay + 7az ). The unit 2 vector is therefore aM M = (−2ax − ay + 7az ) MAB − MBC = = −0.27ax −0.14ay + 0.95az |MAB − MBC | 7.35 (26, 10, 4) = (0.92, 0.36, 0.14) |(26, 10, 4)|

where factors of 1/2 have cancelled. c) Show that this unit vector multiplied by a scalar is equal to the vector from A to C and that the unit vector is therefore parallel to AC. First we find AC = 2ax + ay − 7az , which we recognize as −7.35 aM M . The vectors are thus parallel (but oppositely-directed). 1.3. Thevector from the origin to the point A is given as (6, −2, −4), and the unit vector directed from the origin toward point B is (2, −2, 1)/3. If points A and B are ten units apart, find the coordinates of point B. With A = (6, −2, −4) and B = 1 B(2, −2, 1), we use the fact that |B − A| = 10, or 3 |(6 − 2 B)ax − (2 − 2 B)ay − (4 + 1 B)az | = 10 3 3 3 Expanding, obtain 36 − 8B + 4 B 2 + 4 − 8 B + 4 B 2 +16 + 8 B + 1 B 2 = 100 9 3 9 3 9 or B 2 − 8B − 44 = 0. Thus B = B=
√ 8± 64−176 2

= 11.75 (taking positive option) and so

2 2 1 (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az 3 3 3 1

1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit √ vector in rectangular components that lies in the xy plane, is tangent to the circleat ( 3, 1, 0), and is in the general direction of increasing values of y: A unit vector tangent to this circle in the general increasing y direction is t = √φ . Its x and y a a components are tx = aφ · ax = − sin φ, and ty = √φ · ay = cos φ. At the point ( 3, 1), φ = 30◦ , and so t = − sin 30◦ ax + cos 30◦ ay = 0.5(−ax + 3ay ). 1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z 2 az. Given two points, P (1, 2, −1) and Q(−2, 1, 3), find: a) G at P : G(1, 2, −1) = (48, 36, 18) b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so aG = (−48, 72, 162) = (−0.26, 0.39, 0.88) |(−48, 72, 162)|

c) a unit vector directed from Q toward P : aQP = (3, −1, 4) P−Q = √ = (0.59, 0.20, −0.78) |P − Q| 26

d) the equation of the surface on which |G| = 60: We write60 = |(24xy, 12(x2 + 2), 18z 2 )|, or 10 = |(4xy, 2x2 + 4, 3z 2 )|, so the equation is 100 = 16x2 y 2 + 4x4 + 16x2 + 16 + 9z 4

1.6. If a is a unit vector in a given direction, B is a scalar constant, and r = xax + yay + zaz , describe the surface r · a = B. What is the relation between the the unit vector a and the scalar B to this surface? (HINT: Consider first a simple example with a = ax and B= 1, and then consider any a and B.): We could consider a general unit vector, a = A1 ax + A2 ay + A3 az , where A2 + A2 + A2 = 1. 1 2 3 Then r · a = A1 x + A2 y + A3 z = f (x, y, z) = B. This is the equation of a planar surface, where f = B. The relation of a to the surface becomes clear in the special case in which a = ax . We obtain r · a = f (x) = x = B, where it is evident that a is a unit...