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)
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53e85b40-3118-4666-aced-ba645696f424%2F497a708d-ae08-48cf-bd4d-9978fd8d0f38%2Fa0zyxe_processed.jpeg&w=3840&q=75)
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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