We can use Eq. 1.7 to find AA = 7 and B = 4) and the angleThen we get:26. Find the angle between A = -5i - 3j + 2k and B -2j - 2k. [Ser4 7-20]Eq. 1.7 allows us to find the cosine of the angle between two vectors as long as we knowtheir magnitudes and their dot product. The magnitudes of the vectors A and B are:(−5)² + (−3)² + (2)² = 6.164B =and their dot product is:A B = AB cos = (7)(4) cos 60° = 14A = A² + A² + A² ==which then givesA · B = AxBx+ AyBy + A₂B₂ = (−5)(0) + (−3)(−2) + (2)(−2) = 21.2. WORKED EXAMPLES√B²+ B²+ B² = √(0)² + (−2)² + (−2)² = 2.828B. We have the magnitudes of the two vectors (namelybetween the two is= 130° 70° = 60°.Then from Eq. 1.7, if o is the angle between A and B, we have2(6.164) (2.828)This gives us:coso =From which we get by:A.BAB=27. Two vectors a and b have the components, in arbitrary units, ax = 3.2,ay = 1.6, b = 0.50, by = 4.5. (a) Find the angle between the directions of a andb. (b) Find the components of a vector c that is perpendicular to a, is in the xyplane and has a magnitude of 5.0 units. [HRW5 3-51](a) The scalar product has something to do with the angle between two vectors... if theangle between a and b is o then from Eq. 1.7 we have:a =Now find the magnitudes of a and b:b =We can compute the right-hand-side of this equation since we know the components of aand b. First, find a b. Using Eq. 1.8 we find:a b==cos =a² + a²=b²+ b²COS == 83.4°.Cx=axbx + ayby(3.2) (0.50) + (1.6) (4.5)8.8a.bab= 0.114a.babV(3.2)² + (1.6)² = 3.6(0.50)² + (4.5)² = 4.51.63.2238.8= 0.54(3.6) (4.5)= cos ¹ (0.54) = 57°23/26(b) Let the components of the vector c be ca and cy (we are told that it lies in the xy plane).If c is perpendicular to a then the dot product of the two vectors must give zero. This tellsus:a carCrAycy = (3.2) C+ (1.6)cy = 0This equation doesn't allow us to solve for the components of c but it does give us:Cy= -0.50cy

Given two vectors A= -5i-3j+2kB=-2j-2k

26. Find the angle between A = -5i - 3j +2k and B = −2j - 2k. [Ser4 7-20] Eq. 1.7 allows us to find the cosine of the angle between two vectors as long as we know their magnitudes and their dot product. The magnitudes of the vectors A and B are: A = √√A²+ A²+ A² = √√(-5)² + (−3)² + (2)² = 6.164 B = B²+ B²+ B² = V (0)² + (−2)² + (-2)² = 2.828 and their dot product is: A B = A₂ Br + Ay By + A₂B₂ = (-5)(0) + (−3)(−2) + (2)(−2) = 2 1.2. WORKED EXAMPLES Then from Eq. 1.7, if o is the angle between A and B, we have 2 (6.164) (2.828) which then gives coso: = A. B AB = 83.4°. = 0.114 23

26. Find the angle between A = -5i - 3j +2k and B = −2j - 2k. [Ser4 7-20] Eq. 1.7 allows us to find the cosine of the angle between two vectors as long as we know their magnitudes and their dot product. The magnitudes of the vectors A and B are: A = √√A²+ A²+ A² = √√(-5)² + (−3)² + (2)² = 6.164 B = B²+ B²+ B² = V (0)² + (−2)² + (-2)² = 2.828 and their dot product is: A B = A₂ Br + Ay By + A₂B₂ = (-5)(0) + (−3)(−2) + (2)(−2) = 2 1.2. WORKED EXAMPLES Then from Eq. 1.7, if o is the angle between A and B, we have 2 (6.164) (2.828) which then gives coso: = A. B AB = 83.4°. = 0.114 23

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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