Vector Calculus 4th Edition Chapter 1 Vectors
Page 26 Problem 1 Answer
Given: Two vectors with their coordinates.
To Find: The dot product of both vectors and magnitude of each of the given vectors
a=(1,5),and b=(−2,3)
The Dot Product of two vectors a and b is given by
a⋅b=a1b1+a2b2
The Magnitude of a vector is given by
∥a∥=√a⋅a
First, find out the dot product i.e., a⋅b
of the two vectors by substituting the respective coordinates in the formula.
a⋅b=(1)(−2)+(5)(3)
a⋅b=−2+15
∴a⋅b=13
Magnitude of vector a is,∥a∥=√12+52
∥a∥=√26
Magnitude of vector b is,∥b∥=√(−2)2+32
∥b∥=√13
Therefore, the required dot product i.e., a⋅b is 13 and the magnitude of the vectors are:
∥a∥=√26
∥b∥=√13
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Page 26 Problem 2 Answer
Given: Two vectors a and b with coordinates (4,−1) and (1/2,2) respectively.To Find: The dot product and magnitude of each vectors,
a=(4,−1),
b=(1/2,2)
The Dot Product of two vectors is given by,
a⋅b=a1b1+a2b2
The magnitude of a vector is given by,
∥a∥=√a⋅a
First, find out the dot product i.e., a⋅b of the two vectors by substituting the respective coordinates in the formula.
a⋅b=(4)(1/2)+(−1)(2)
a⋅b=2−2
∴a⋅b=0
Magnitude of vector a,
∥a∥=√42+(−1)2
∥a∥=√17
Magnitude of vector b,
∥b∥=√(1/2)2+22
∥b∥=√17/4
∥b∥=√17/2
Therefore, the required dot product i.e., a⋅b is 0 and the magnitude of the vectors are:
∥a∥=√17
∥b∥=√17/2
Page 26 Problem 3 Answer
Given: Two vectors a and b having coordinates (−1,0,7) and (2,4,−6) respectively.
To Find: We have to compute the dot product i.e., a⋅b of the vectors and magnitude of each vectors i.e.,
∥a∥ and ∥b∥.Substitute the coordinates in the respective formula and compute the result.
The dot product of two vectors is,
a⋅b=a1b1+a2b2
∴a⋅b=(−1)(2)+(0)(4)+(7)(−6)
a⋅b=−2+0−42
a⋅b=−44
The magnitude of a vector,
∥a∥=√a⋅a
∴∥a∥=√(−1)2+02+72
∥a∥=√50
∥a∥=5√2
And ∥b∥=√22+42+(−6)2
∥b∥=√4+16+36
∥b∥=√56
∥b∥=2√14
Therefore, the required dot product i.e., a⋅b is −44 and the magnitude of the vectors are:
∥a∥=5√2
∥b=2√14
Page 26 Problem 4 Answer
Given: Two vectors are given as follows:
a=(2,1,0)
b=(1,−2,3).To Find: The dot product and magnitude of each of the vectors a and b with coordinates (2,1,0) and (1,−2,3).
Substitute the coordinates in the respective formula for the dot product and magnitude of the vectors and compute the answer.
The dot product of the vectors a and b,
a⋅b=a1b1+a2b2
Substitute the coordinates of the respective vectors in the above formula.
a⋅b=(2)(1)+(1)(−2)+(0)(3)
a⋅b=2−2+0
∴a⋅b=0
The magnitude of a vector is,∥a∥=√a⋅a
Substitute the coordinates of vector a in the above formula.
∥a∥=√22+12+02
∴∥a∥=√5
Substitute the coordinates of vector b in the formula.
∥b∥=√12+(−2)2+32
∥b∥=√1+4+9
∴∥b∥=√14
Therefore, the required dot product of the given vectors is 0. The magnitude of the vectors are:
∥a∥=√5
∥b∥=√14
Page 26 Problem 5 Answer
Given: Two vectors,
a=4i−3j+k
b=i+j+k
To Find: To find the dot product i.e., a⋅b and the magnitude i.e., ∥a∥ of the vectors a and b.
The vectors are represented by unit vector notation and we can get their coordinates from it.
After that, substitute the coordinates in the respective formulas to compute the dot product and their magnitudes.
The vectors are given in unit vector notation i.e.,
a=4i−3j+k
b=i+j+k
Converting it into their coordinate form we get,
a=(4,−3,1)
b=(1,1,1)
Now, the dot product of the vectors are,
a⋅b=a1b1+a2b2
Substitute the respective coordinates to compute it.
Dot Product:
a⋅b=(4)(1)+(−3)(1)+(1)(1)
a⋅b=4−3+1
∴a⋅b=2
Magnitude of a vector is given by:
∥a∥=√a⋅a
Magnitude of vector a,
∥a∥=√42+(−3)2+12
∥a∥=√16+9+1
∥a∥=√26
Magnitude of vector b,
∥b∥=√12+12+12
∥b∥=√3
Therefore, the required dot product i.e., a⋅b is 2 and the magnitude of the vectors are:
∥a∥=√26
∥b∥=√3
Page 26 Problem 6 Answer
Given: Two vectors a and b in their unit vector notation.To Find The dot product and magnitude of the vectors,
a=i+2j−k
b=−3j+2k.
We can find the coordinates from the unit vector representation easily.
After that, substitute the values in the formulas to find their dot product and the magnitude.
From the unit vector representation of the vectors, we found,
a=(1,2,−1)
b=(0,−3,2)
Now, the dot product of the vectors:
a⋅b=a1b1+a2b12
∴a⋅b=(1)(0)+(2)(−3)+(−1)(2)
a⋅b=0−6−2
a⋅b=−8
The magnitude of a vector is,
∥a∥=√a⋅a
∴∥a∥=√12+22+(−1)2
∥a∥=√1+4+1
∥a∥=√6
And ∥b∥=√02+(−3)2+22
Therefore, the required dot product i.e., a⋅b is −8 and the magnitude of the vectors are:
∥a∥=√6
∥b∥=√13
Page 26 Problem 7 Answer
Given: Two vectors a and b in their unit vector notation. To Find The angle between the vectors,
a=√3i+j, b=−√3i+j
First, find the dot product and the magnitude of the given vectors, then plug them into the formula,
θ=cos−1a⋅b
∥a∥∥b∥ and calculate the angle.
From the given vectors, we know can find out the coordinates. C
alculate the dot product and the magnitudes of the given vectors.
a⋅b=(√3)(−√3)+(1)(1)
a⋅b=−3+1
a⋅b=−2
And ∥a∥=√√3/2+12
∥a∥=√4
∥a∥=2,
∥b∥=√(−√3)2+12
∥b∥=√4
∥b∥=2
Plug them into the formula and find the angles between the vectors.
θ=cos−1−2/(2)(2)
θ=cos−1(−1/2)
θ=120∘
Therefore, the required angle between the given vectors is 120∘.
Page 26 Problem 8 Answer
Given: Two vectors a and b along with coordinates (−1,2) and (3,1) respectively.
To Find: The angle between the vectors,
a=(−1,2)
b=(3,1)
The angle between two vectors can be calculated as:
θ=cos−1a⋅b
∥a∥∥b∥
Calculate the dot product and the magnitude of the given vectors and plug them into the above formula to get the angle.
First, compute the dot product and magnitude of the vectors.
a⋅b=(−1)(3)+(2)(1)
a⋅b=−3+2
a⋅b=−1
And ∥a∥=√(−1)2+22
∥a∥=√5
∥b∥=√32+12
∥b∥=√10
Now, plug these values into the formula to calculate the angle between the two vectors.
θ=cos−1−1(√5)(√10)
θ=cos−1(−1/√50)
θ=98.13∘
Thus, the required angle between the given two vectors is 98.13∘.
Page 26 Problem 9 Answer
Given:a=i+j,
b=i+j+k
To find: Find the angle between each of the pairs of vectors. Evaluate to get the final answer.
By using this formula:θ=cos−1/a⋅b∥a∣∣∣b∣∣
First we finda⋅b,∥a∥,∥b∥:
a⋅b=(1)(1)+(1)(1)+(0)(1)
=1+1+0
=2∥a∥=√(1)2+12+02
=√2
∣∣b∥=√12+12+12
=√3
Now find angle between vectors:
θ=cos−1/2(√2)(√3)
⇒cos−1(2√6)
θ=33.26∘
Hence, the angle between the given the pair of vectors isθ=33.26∘
Page 26 Problem 10 Answer
Given:a=i+j−k,
b=−i+2j+2k
To find: Find the angle between each of the pairs of vectors. Evaluate to get the solution.
By using this formula:θ=cos−1/a⋅b
∥a∥∣∣b∣∣
First we find a⋅b,∥a∥,∥b∥:
a⋅b=(1)(−1)+(1)(2)+(−1)(2)
⇒−1+2−2=−1
∣a∥=√12+12+(−1)2=√3
∥b∥=√(−1)2+22+22
⇒√9
=3
Now angle between vectors are:
θ=cos−1−1(√3)(3)
θ=101.10∘
Hence, the angle between the given pair of vectors isθ=101.10∘
Page 26 Problem 11 Answer
Given:a=1,−2,3,
b=3,−6,−5
To find: Find the angle between each of the pairs of vectors. Evaluate to get the final angle.
By using this formula:θ=cos−1/a⋅b
∥a∥∣∣b∥
First we find a⋅b,∥a∥,∥b∥:
a⋅b=(1)(3)+(−2)(−6)+(3)(−5)
=3+12−15
=0
∥a∥=√12+(−2)2+32
=√14
∥b∥=√32+(−6)2+(−5)2
=√70
Now find angle between vectors are:
θ=cos−10(√14)(√70)
=cos−10
θ=90∘
Hence, the angle between the given pair of vectors isθ=90∘
Page 26 Problem 12 Answer
Given:a=i+j,
b=2i+3j−k
To find: Calculate the projection ab.
Evaluate to get the final answer.
By using this formula:proj ab=(a⋅b/a⋅a)a
First we find a⋅b,a⋅a:
a⋅b=(1)(2)+(1)(3)+(0)(−1)
⇒2+3=5
a⋅a=12+12+02
=2
Now, we find projection:
projb=(5/2)(1,1,0)
=(5/2,5/2,0)
Hence, the required projection is:(5/2,5/2,0)
Page 26 Problem 13 Answer
Given:a=(i+j)/√2,´
b=2i+3j−k
To find: Calculate the projection a b
Evaluate to get the projection.
By using this formula we get:proj ab=(a⋅b/a⋅a)a
Then first we find a⋅b,a⋅a:
a⋅b=(1/√2)(2)+(1/√2)(3)+(0)(−1)
=5/√2a⋅a=(1/√2)2+(1/√2)2+02
=1/2+1/2 =1
Now find projection:
projb=(5/√21)(1/√2,1/√2,0)
=5/√2(1/√2,1/√2,0)=(5/2,5/2,0)
Hence, the projection a b are(5/2,5/2,0)
Page 26 Problem 14 Answer
Given:a=5k,
b=i−j+2k
To find: Calculate projection of ab.
Evaluate to get the final answer.
By using this formula:proj ab=(a⋅b/a⋅a)a
Then first find a⋅b,a⋅a:
a⋅b=(0)(1)+(0)(−1)+(5)(2)
=0+0+10
=10
a⋅a =02+02+52=25
Now find projection:
Proj b=(10/25)(0,0,5)
=(0,0,2)
Hence, the projection are:(0,0,2)
Page 26 Problem 15 Answer
Given:a=−3k,
b=i−j+2k
To find: Calculate the projection ab.
Evaluate to get the final answer.
By using this formula:proj ab=(a⋅b/a⋅a)a
Then first we find a⋅b,a⋅a:
a⋅b=(0)(1)+(0)(−1)+(−3)(2)
=0+0−6
=−6
a⋅a=02+02+(−3)2
=9
Now find projection:
Proj b=(−6/9)(0,0,−3)
=(0,0,2)
Hence, the projection are :(0,0,2)