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)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Question
![We can use Eq. 1.7 to find A
A = 7 and B = 4) and the angle
Then 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 know
their magnitudes and their dot product. The magnitudes of the vectors A and B are:
(−5)² + (−3)² + (2)² = 6.164
B =
and their dot product is:
A B = AB cos = (7)(4) cos 60° = 14
A = A² + A² + A² =
=
which then gives
A · B = AxBx+ AyBy + A₂B₂ = (−5)(0) + (−3)(−2) + (2)(−2) = 2
1.2. WORKED EXAMPLES
√B²+ B²+ B² = √(0)² + (−2)² + (−2)² = 2.828
B. We have the magnitudes of the two vectors (namely
between the two is
= 130° 70° = 60°.
Then from Eq. 1.7, if o is the angle between A and B, we have
2
(6.164) (2.828)
This gives us:
coso =
From which we get by:
A.B
AB
=
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 and
b. (b) Find the components of a vector c that is perpendicular to a, is in the xy
plane 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 the
angle 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 a
and b. First, find a b. Using Eq. 1.8 we find:
a b
=
=
cos =
a² + a²=
b²+ b²
COS =
= 83.4°.
Cx
=
axbx + ayby
(3.2) (0.50) + (1.6) (4.5)
8.8
a.b
ab
= 0.114
a.b
ab
V
(3.2)² + (1.6)² = 3.6
(0.50)² + (4.5)² = 4.5
1.6
3.2
23
8.8
= 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 tells
us:
a carCr
Aycy = (3.2) C+ (1.6)cy = 0
This equation doesn't allow us to solve for the components of c but it does give us:
Cy= -0.50cy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3556b677-568e-44d6-a204-9ef657dac686%2Fc531ad75-61d7-4478-8bed-0e4d5b875ad6%2Fz09acq9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:We can use Eq. 1.7 to find A
A = 7 and B = 4) and the angle
Then 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 know
their magnitudes and their dot product. The magnitudes of the vectors A and B are:
(−5)² + (−3)² + (2)² = 6.164
B =
and their dot product is:
A B = AB cos = (7)(4) cos 60° = 14
A = A² + A² + A² =
=
which then gives
A · B = AxBx+ AyBy + A₂B₂ = (−5)(0) + (−3)(−2) + (2)(−2) = 2
1.2. WORKED EXAMPLES
√B²+ B²+ B² = √(0)² + (−2)² + (−2)² = 2.828
B. We have the magnitudes of the two vectors (namely
between the two is
= 130° 70° = 60°.
Then from Eq. 1.7, if o is the angle between A and B, we have
2
(6.164) (2.828)
This gives us:
coso =
From which we get by:
A.B
AB
=
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 and
b. (b) Find the components of a vector c that is perpendicular to a, is in the xy
plane 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 the
angle 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 a
and b. First, find a b. Using Eq. 1.8 we find:
a b
=
=
cos =
a² + a²=
b²+ b²
COS =
= 83.4°.
Cx
=
axbx + ayby
(3.2) (0.50) + (1.6) (4.5)
8.8
a.b
ab
= 0.114
a.b
ab
V
(3.2)² + (1.6)² = 3.6
(0.50)² + (4.5)² = 4.5
1.6
3.2
23
8.8
= 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 tells
us:
a carCr
Aycy = (3.2) C+ (1.6)cy = 0
This equation doesn't allow us to solve for the components of c but it does give us:
Cy= -0.50cy
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