Solve last part Solution is provided to 3 parts

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Solve last part 

Solution is provided to  3 parts

 

 

=r₁
the z axis component of the r 1
(₁)₂=r₁cos40=6cos40 = 4.596
the x axis component of the ₁
(1₁)x
X
the y axis component of the ₁
(1₁), = -r₁sin40sin30=-6sin40sin30 =
Step 3: solution
for vector 2
₂=
where
a= 60°
B = 45°
Y = 120⁰
1₂=9
6m
the vector form of the ₁
₁ = 3.34i-1.928j + 4.596k
r₂(cosai+cos³j+cosyk)
cose=
48
cose=
sin40cos30=6sin40cos30=3.
substitute all the values
₂=9x (cos60i+cos45j +cos120k)
7₂ = 4.5i +6.364j - 4.5k
cose =
Z
The angle between ₁ and ₂
1
cose=
Solution
1.2
1₁.1₂
0 = 109.4°
(3.34i-1.928j+4.596k).(4.5i+
-17.922
54
15.03-12.27-20.682
54
6x9
Answer
part a-₁ =3.34i-1.928j+4.596k
part b-72₂=4.5i +6.364j-4.5k
part c- 0=109.4°
Transcribed Image Text:=r₁ the z axis component of the r 1 (₁)₂=r₁cos40=6cos40 = 4.596 the x axis component of the ₁ (1₁)x X the y axis component of the ₁ (1₁), = -r₁sin40sin30=-6sin40sin30 = Step 3: solution for vector 2 ₂= where a= 60° B = 45° Y = 120⁰ 1₂=9 6m the vector form of the ₁ ₁ = 3.34i-1.928j + 4.596k r₂(cosai+cos³j+cosyk) cose= 48 cose= sin40cos30=6sin40cos30=3. substitute all the values ₂=9x (cos60i+cos45j +cos120k) 7₂ = 4.5i +6.364j - 4.5k cose = Z The angle between ₁ and ₂ 1 cose= Solution 1.2 1₁.1₂ 0 = 109.4° (3.34i-1.928j+4.596k).(4.5i+ -17.922 54 15.03-12.27-20.682 54 6x9 Answer part a-₁ =3.34i-1.928j+4.596k part b-72₂=4.5i +6.364j-4.5k part c- 0=109.4°
1. Find the components of .
2.
A
Find the components of T2-
(Reminder: cos(a) =)
3. Find the angle between and 2.
4. Find the magnitude of the projection of r2 along/parallel to
T = 6m
30°
40°
Z
8 60°
r₁=
r₂ =
√2
120°
45⁰
a
0 =
Y=
2 = 9 m
optional
Transcribed Image Text:1. Find the components of . 2. A Find the components of T2- (Reminder: cos(a) =) 3. Find the angle between and 2. 4. Find the magnitude of the projection of r2 along/parallel to T = 6m 30° 40° Z 8 60° r₁= r₂ = √2 120° 45⁰ a 0 = Y= 2 = 9 m optional
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