(Electromagnetic field)Hello sir I want a detailed solution for this solution3-given points A(8,-5, 4) and B(-2, 3, 2), find:a) the distance from A to B.|BA| = |(-10, 8, -2)] = 12.96b) a unit vector directed from A towards B. This is found throughB-AIB-A|c) a unit vector directed from the origin to the midpoint of the line AB.аAB=aоM ==12 R12470 R12³=(-0.77, 0.62,-0.15)(A+B)/2|(A + B)/2]or=d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3.Note that the midpoint, (3,-1, 3), as determined from part c happens to have z coordinate of 3. Thisis the point we are looking for.(3,-1,3)√194-Let Q1 = 8 μC be located at P1 (2, 5, 8) while Q2 = -5 μC is at P2(6, 15, 8). Let e = €0.a) Find F2, the force on Q2: This force will beF₂=(8 × 10−6)(−5 × 10−6) (4ax +10ay)47 €0(116)1.5(0.69,-0.23, 0.69)=(-1.15a, -2.88ay) mNb) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force ingeneral will be:F3 =Q3 Q1R134TEOIR1313+where R13 = (x - 2)a, + (y-5)ay and R23 = (x − 6)ax + (y-15)ay. Note, however, thatall three charges must lie in a straight line, and the location of Q3 will be along the vector R12extended past Q₂. The slope of this vector is (15-5)/(6-2)=2.5. Therefore, we look for P3at coordinates (x, 2.5.x, 8). With this restriction, the force becomes:Q38[(x-2)a, +2.5(x - 2)ay]F3 = 47 €0 [(x - 2)² + (2.5)²(x - 2)²]¹.3X =Q₂R23R233where we require the term in large brackets to be zero. This leads to8(x-2)[((2.5)² + 1)(x-6)²]¹.5-5(x-6)[((2.5)² + 1)(x - 2)²1¹5 = 0which reduces to5[(x-6)a, +2.5(x - 6)ay][(x − 6)² + (2.5)²(x − 6)²]¹8(x-6)²-5(x - 2)² =6√8-2√5√8-√5The coordinates of P3 are thus P3 (21.1, 52.8, 8)= 21.1

Note: As per Bartleby guidelines we are supposed to provide solution of one question and first 3…

given points A(8,-5, 4) and B(-2, 3, 2), find: a) the distance from A to B. |BA| = |(-10, 8, -2)] = 12.96 b) a unit vector directed from A towards B. This is found through B-A |B-A| c) a unit vector directed from the origin to the midpoint of the line AB. aAB= =(-0.77, 0.62,-0.15)

given points A(8,-5, 4) and B(-2, 3, 2), find: a) the distance from A to B. |BA| = |(-10, 8, -2)] = 12.96 b) a unit vector directed from A towards B. This is found through B-A |B-A| c) a unit vector directed from the origin to the midpoint of the line AB. aAB= =(-0.77, 0.62,-0.15)

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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