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...

See similar textbooks

Similar questions

版 EEE204 JLY 2021 (Campatibildy Mode) Word (Product Activation Failed) DESIGN PAGE LAYOUT REFERENCES MAILINGS REVIEW VIEW Ci. Show that the following vectors are orthogonal: A = 4a, +6a, - 2a and B= -2a, + 4a, +8a. ii If p = 21 +)-k and q=-3/+ 2k determine the norm of the addition of p and q, and the magnitude.
Find the cosine of the angle between ?⃗ and ?⃗⃗ .?⃗ = −2?⃗? − 3?⃗? and ?⃗⃗ = −6?⃗? + 4?⃗?
Question 6: [Basic: Review of Complex Number] Consider the following complex numbers: e*2, 4 + 5j, 4 – 5j, , j², ne-3Um+) 1) Is the complex number in rectangular form or polar form? Specify for each. 2) Write down the complex number in the other form. If a complex number is given in rectangular form, write down its polar form, vice versa. 3) Specify the amplitude and phase for each given complex number. Show each of the numbers in complex plane. Remark: p. 71 in the textbook contains some very important discussions about the polar form
What is the Cartesian representation of z = 460/135°? O a. z = -325.27-j325.27 O b. z = -325.27 +j325.27 O c. z = 325.27 +j325.27 O d. z = 325.27-j325.27
Problem 42: Perform the following subtractions in rectangular form: a. (9.8 + j6.2) – (4.6 + j4.6) b. (167 + j243) –(-42.3 – j68) c. (-36.0 +j78) - (-4 – j6) + (10.8 – j72)
The dimensional formula for frequency is: (a) MOLOT-1 (c) MºLºT-² (b) MºL¹T-1 (d) M°L°T-3
Given: A 2a,- 4.3a, + a, and B = a, + 2a, - 3a,. Find the z-component of (2A + B) x (A-2B). a. 35.00 b. 76.93 OC. -54.50 d. 41.50 Your answer is incorrect. The correct answer is: 41.50
Prove the CFL condition is the stability required when the Lax-Wendroff method is applied to solve the simple 1-D wave equation. The difference equation is of the form
Refer to Table 4.3 and find the Fourier coefficient Ck for the waveform shown: x.(1) 1/2 cycle of sinusoid -2 2 4 Select one: O -12e-jkr /T(4k² – 1) O -12/T(4k2 – 1) O -12/T(2k2 – 1) O 12/T(4k2 – 1)
Consider the complex numbers Z3 and Z4 defined below. Z 2490 Z = 42- 90° Compute Z4/Z3 and specify its magnitude in the box below.
But the phase angle plot does exist. I'm not sure why you are saying it doesn't. Its in my textbook.
Refer to Table 4.3 and find the Fourier coefficient C for the waveform shown: M. 0 0.2 -0.2 Select one: Oj8/nk O 4/nk Oj4/xk O-j4/nk