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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)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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(Electromagnetic field) Hello sir I want a detailed solution for this solution 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 IB-A| c) a unit vector directed from the origin to the midpoint of the line AB. аAB= aоM == 12 R12 470 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. This is the point we are looking for. (3,-1,3) √19 4- 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 be F₂= (8 × 10−6)(−5 × 10−6) (4ax +10ay) 47 €0 (116)1.5 (0.69,-0.23, 0.69) =(-1.15a, -2.88ay) mN b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: F3 = Q3 Q1R13 4TEOIR1313 + where R13 = (x - 2)a, + (y-5)ay and R23 = (x − 6)ax + (y-15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15-5)/(6-2)=2.5. Therefore, we look for P3 at 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]¹.3 X = Q₂R23 R233 where we require the term in large brackets to be zero. This leads to 8(x-2)[((2.5)² + 1)(x-6)²]¹.5-5(x-6)[((2.5)² + 1)(x - 2)²1¹5 = 0 which reduces to 5[(x-6)a, +2.5(x - 6)ay] [(x − 6)² + (2.5)²(x − 6)²]¹ 8(x-6)²-5(x - 2)² = 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) = 21.1
Transcribed Image Text:(Electromagnetic field) Hello sir I want a detailed solution for this solution 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 IB-A| c) a unit vector directed from the origin to the midpoint of the line AB. аAB= aоM == 12 R12 470 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. This is the point we are looking for. (3,-1,3) √19 4- 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 be F₂= (8 × 10−6)(−5 × 10−6) (4ax +10ay) 47 €0 (116)1.5 (0.69,-0.23, 0.69) =(-1.15a, -2.88ay) mN b) Find the coordinates of P3 if a charge Q3 experiences a total force F3 = 0 at P3: This force in general will be: F3 = Q3 Q1R13 4TEOIR1313 + where R13 = (x - 2)a, + (y-5)ay and R23 = (x − 6)ax + (y-15)ay. Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q₂. The slope of this vector is (15-5)/(6-2)=2.5. Therefore, we look for P3 at 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]¹.3 X = Q₂R23 R233 where we require the term in large brackets to be zero. This leads to 8(x-2)[((2.5)² + 1)(x-6)²]¹.5-5(x-6)[((2.5)² + 1)(x - 2)²1¹5 = 0 which reduces to 5[(x-6)a, +2.5(x - 6)ay] [(x − 6)² + (2.5)²(x − 6)²]¹ 8(x-6)²-5(x - 2)² = 6√8-2√5 √8-√5 The coordinates of P3 are thus P3 (21.1, 52.8, 8) = 21.1
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