Vector Calculus 4th Edition Chapter 1 Vectors
Page 59 Problem 1 Answer
Given: (1,2,3,…n)
To find the terms in standard basis.
Use the vector method and find the answer.
Therefore,
(1,2,3…,n)=1(1,0,0…,0)+2(0,1,0,0,…,0)+…+n(0,0,0,…0,1)
=e1+2e2+3e3+…+nen
Hence, e1+2e2+3e3+…+nen is the standard form of vector.
Page 59 Problem 2 Answer
Given: (1,0,−1,1,0,−1,…,1,0,−1)
To find the terms in standard basis.
Use vector method and find the answer.
Therefore,
(1,0,−1,1,0,−1,…,1,0,−1)=1(1,0,0…,0)+0(0,1,0,0,…,0)−1(0,0,1,0,..,0)…−n(0,0,0,….0,1)
=e1−e3+e4−e6+…+en−2−en
Hence, e1−e3+e4−e6+…+en−2−en is the standard form of vector.
Page 59 Problem 3 Answer
Given:e1−2e2+3e3−4e4+⋯+(−1)n+1 nen
To find the terms in standard basis.
Use the vector method and find the answer.
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Therefore,
e1−2e2+3e3−4e4+…+(−1)n+1nen
=(1,0,0,…,0)+(0,−2,0,0,…,0)+(0,0,3,0,…0)+(0,0,0,1,−4,0,…,0)+…+(0,0,0,…0,(−1)n+1n)
=(1,−2,3,−4…,(−1)n+1n)
Hence,(1,−2,3,−4…,(−1)n+1n) is the given vectors without recourse to standard basis notation.
Page 59 Problem 4 Answer
To find the given vectors without recourse to standard basis notation,
e1+en=(1,0,0,…,0)+(0,0,0,…0,1)
=(1,0,0,…,0,1)..
(1,0,0,…,0,1) is the given vectors without recourse to standard basis notation.
Page 59 Problem 5 Answer
Given: a=(1,3,5,…2n−1) and b=(2,−4,6,…,(−1)n+12n).To find: a+b.Evaluate and find the answer.
Calculating by using the given information, we get,
a+b=(1,3,5,7,…,2n−1)+(2,−4,6,−8,…,(−1)n+12n)
=(3,−1,11,−1,…,2n(1+(−1)n+1)−1)
Hence, the required solution is (3,−1,11,−1,…,2n(1+(−1)n+1)−1).
Page 59 Problem 6 Answer
Given: a=(1,3,5,…2n−1) and b=(2,−4,6,…,(−1)n+12n).To find: a−b.Evaluate and find the answer.
Calculating by using the given information, we get,
a−b=(1,3,5,7,…,2n−1)−(2,−4,6,−8,…,(−1)n+12n)
=(−1,7,−1,15,…,2n(1+(−1)n)−1)
Hence, the required solution is (−1,7,−1,15,…,2n(1+(−1)n)−1).
Page 59 Problem 7 Answer
Given: a=(1,3,5,…2n−1) and b=(2,−4,6,…,(−1)n+12n).To find: −3a.Evaluate and find the answer.
Calculating by using the given information, we get,
−3a=−3(1,3,5,7,…,2n−1)
=(−3,−9,−15,−21,…,−6n+3)
Hence, the required solution is (−3,−9,−15,−21,…,−6n+3).
Page 59 Problem 8 Answer
Given: a=(1,3,5,…2n−1) and b=(2,−4,6,…,(−1)n+12n).
To find: ∥a∥.
Evaluate and find the answer.
Calculating by using the given information, we get,
∥a∥=∥(1,3,5,7,…,2n−1)∥
=√12+32+52+⋯+(2n−1)2
Hence, the required solution is √12+32+52+⋯+(2n−1)2.
Page 59 Problem 9 Answer
Given: a=(1,3,5,….2n−1)
And b=(2,−4,6,…,(−1)n+12n)
To Find: Calculate a.b
Evaluate the given question to get the answer.
Since
a=(1,3,5,…2n−1)
And b=(2,−4,6,…,(−1)n+1/2n)
Therefore,
a⋅b=(1,3,5,7,…,2n−1)⋅(2,−4,6,−8,…,(−1)n+1/2n)
=2−12+30−56+⋯+(−1)n+1/2(2n−1)n
Hence, the required answer is a.b=2−12+30−56+⋯+(−1)n+1/2(2n−1)n.
Page 59 Problem 10 Answer
To Prove: The triangle inequality in Rn
for a=(1,0,1,0,…,0)
And b=(0,1,0,1,…,1).
Evaluate the given question to get the answer.
Let, ∥a+b∥≤∥a∥+∥b∥ where a=(1,0,1,0…,0)
And b=(0,1,0,1,…,0,1)
Now, a+b=(1,1,1….1,1)
Since, there are exactly n/2I′
s in the a vector
Therefore, ∥a∥=√12+12+…+12=√n/2
Similarly for b we have:
∥b∥=√n/2
Thus, ∥a∥+∥b∥=2√n/2
Now, ∥a+b∥=√12+12+…+12
=√n
=√(2)√n/2≤2√n/2
=∥a∥+∥b∥
Hence, the required answer is ∣∣a+b∣∣≤∥a∣∣+∥b∣
Page 59 Problem 11 Answer
Given: The vectors are: a=(1,2,…,n)
And b=(1,1,…,1)
To Prove: The Cauchy–Schwarz inequality holds for the given vectorsEvaluate the given question to get the answer.
Let,∣a⋅b∣=1+2+⋯+n=n(n+1)/2
∥a∥=√12+22+⋯+n2
=√n(n+1)(2n+1)/6
∥b∥=√12+12+⋯+12=√n
And, 2n+1/6≥n+1/4.
Now, ∥a∥2∥b∥2=n2(n+1)(2n+1)/6
=n2(n+1)2n+1/6≥n2(n+1)n+1/4
=n2(n+1)2/4
=∣a⋅b∣2
Now, taking square root on both sides we h get,
∥a∥∥b∥=∣a⋅b∣
Hence, the Cauchy–Schwarz inequality holds for the given vectors is verified.
Page 59 Problem 12 Answer
Given: a=(1,−1,7,3,2)
And b=(2,5,0,9,−1).
To Find: The projection proj a/b.
Evaluate the given question to get the answer.
Let, a=(1,−1,7,3,2), And b=(2,5,0,9,−1)
Now, projb=2−5+27−2
1+1+49+9+4aˉ
=11/32aˉ
=11/32(1,−1,7,3,2)
=(11/32,−11/32,77/32,33/32,11/16)
Hence the projection is given by: proja
b=(11/32,−11/32,77/32,33/32,11/16)
Page 59 Problem 13 Answer
Given: The vectors, a,b,c∈Rn,
To Prove: ∣∣a−b∥≤∥a−c∣+∣c−b∣
Evaluate the given question to get the answer.
Let, the triangle inequality be
∥a+b∥≤∥a∥+∥b∣
Now ∥a−b∥=∥(a−c)+(c−b)∥
Thus, from the triangle inequality we get,
∣(a−c)+(c−b)∥≤∥(a−c)∥+∥(c−b)∥
Hence the requited answer is proved ∣∣a−b∥≤∥a−c∥+∥c−b∥.
Page 59 Problem 14 Answer
Given: The Pythagorean theorem where,a,b,and c are vectors in Rn
such that a+b=c
And a⋅b=0, To Prove: ∣a∥2+∥b∣2=∥c∣2 and why is this called the Pythagorean theorem Evaluate the given question to get the answer.
Here, a+b=c
Since, If the vectors are equal then there magnitude must also be equal
Therefore,
∥c∥2=∥a+b∥2
∥c∥2=(a+b)⋅(a+b)
∥c∥2=a⋅a+a⋅b+b⋅a+b⋅b
Now, since a⋅b=0
Therefore, ∥c∥2=a⋅a+b⋅b
∥c∣2=∥a∥2+∥b∥2
Since, a and b are like the pair of orthogonal sides of a right angled triangle also called the base and height and the third side of the triangle is a+b=c.
According to theorem, the sum of the squares of the lengths of the ⊥r sides is the square of the length of the “hypotenuse”, which is c in our case.
Thus this is called the Pythagorean theorem.
Hence, ∥c∣2=∥a∥2+∥b∥2 is proved and it is also found why it is called the Pythagorean theorem.
Page 59 Problem 15 Answer
Given: a and b be vectors in Rn .To Prove: If ∣a+b∥=∥a−b∣, then a and b are orthogonal Evaluate the given question to get the answer.
Here, ∥a+b∥=∥a−b∣
Since, equal vectors have the same magnitude
Therefore,
∥a+b∥2=∥a−b∥2
(a+b)⋅(a+b)=(a−b)⋅(a−b)
a⋅a+a⋅b+b⋅a+b⋅b=a⋅a−a⋅b−b⋅a+b⋅b
Since a⋅b=b⋅a
Thus, 4a⋅b=0
Since a⋅b=0,
Hence, the final answer is a and b are orthogonal vectors.
Page 59 Problem 16 Answer
Given: a and b be vectors in Rn.
To Prove: If ∣a−bˉ∣>∣a+b∣ then the angle between a and b is obtuse Evaluate the given question to get the answer.
Here, ∥a−b∣>∣∣a+b∣∣
Now, squaring both sides we get,
∣∣a−b∥2 >∥∥a+b∥2
(a−b)⋅(a−b)>(a+b)⋅(a+b)
a⋅a−a⋅b−b⋅a+b⋅b>a⋅a+a⋅b+b⋅a+b⋅b
Since, a⋅b=b⋅a
Therefore,4a⋅b<0
Now, since a⋅b<0 we get:cosθ=a⋅
∥a∣∥b∥<0
Since, cosine is negative for angles between (π/2,π)
Hence, the angle between a and b, that is θ is obtuse.
Page 59 Problem 17 Answer
Given: The points in R5 satisfying the equation 2(x1−1)+3(x2+2)−7×3+x4−4−5(x5+1)=0.
To Find: Describe geometrically the set of the given points.
Evaluate the given question to get the answer.
Since it passes through the point (1,−2,0,4,−1) and is parallel to every vector which is perpendicular to (2,3,−7,1,−5).
It is a 4-dimensional affine space
Thus this is a hyperplane in R5.
Hence, the set of the given points is a hyperplane in R5.
Page 59 Problem 18 Answer
Given: An inventory of 20 shirts that can sell for$8 each, 30 shirts that sell for $10 each, 24 shirts that sell for $12 each and20 shirts that you sell for $15each and in friend’s inventory it has 30 caps that can be sold for$10 each, 16 caps that can be sold for $10 each, 20 caps that can be sold for $12 each, and 28 caps that can be sold for $15 each.
To Find: If suggested swapping half the inventory of each type of T-shirt for half his inventory of each type of baseball cap. Is our friend likely to accept your offer? Why or why not? Evaluate the given question to get the answer.
Let the number of each t-shirt in my inventory be
n1=(20,30,24,20)
And their respective costs (per unit) be,
c1=(8,10,12,15)
Also, n2=(30,16,20,28)
And c2=(10,10,12,15) be the friend’s inventory.
Now, the value of half of my inventory is
1/2 n1⋅c1=1/2(160+300+288+300)=$524
And the value of half of friend’s inventory is
1/2 n1⋅c1=1/2(300+160+240+420)
=$560
Hence, the friend may not be so happy with the deal as he’d be losing $36 worth of assets.
Page 59 Problem 19 Answer
Given: Prices of the six type of grains are $200,$250,$300,$375,$450,$500
per ton and the commodity bundle vector is x=(x1,…,x6)
To Find: Express the total cost of the commodity bundle as a dot product of two vectors in R6
Evaluate the given question to get the answer.
Now,
Total cost = cost per unit × number of units.
Since, while buying more than one product, it needs to sum each subtotal.
Therefore,
Total cost=(200,250,300,375,450,500)⋅(x1,x2,x3,x4,x5,x6)
=200×1+250×2+300×3+375×4+450×5+500×6
Hence, the total cost of the commodity bundle as a dot product of two vectors in R6 is 200×1+250×2+300×3+375×4+450×5+500×6.
Page 59 Problem 20 Answer
Given: The prices of six type of grains are $(200,250,300,375,450,500)
per ton and a customer has $100,000
to be to purchase the grain.
To Find: The set of possible commodity bundle vectors that the customer can afford.
Also describe the relevant budget hyperplane in R6
Evaluate the given question to get the answer.
Let,100000≥(200,250,300,375,450,500)⋅x
According to set-builder notation, the set is,
{x∈R6∣(200,250,300,375,450,500)⋅x≤100000}.
Geometrically, this describes the region in R6 between the hyperplanes
Now,
(200,250,300,375,450,500)⋅x=100000
And,(200,250,300,375,450,500)⋅x=−100000
Hence, the set of possible commodity bundle vectors that the customer can afford.
Also describe the relevant budget hyperplane in R6 is found.