MATH 112 - TECHNICAL
MATHEMATICS - Vector operations
Question Bank - Set 2
Liberty University
Question 1
Question
Let a=
2
−3
1
,b=
4
5
−2
, and c=
1
2
−1
. Determine whether the vectors
band care orthogonal to each other.
Solution
To determine if two vectors are orthogonal to each other, we need to check if
their dot product is equal to zero.
Step 1: Calculate the dot product of band c.
b·c= (4)(1) + (5)(2) + (−2)(−1) = 4 + 10 + 2 = 16
Step 2: Determine if the dot product is zero. Since b·c= 16 = 0, the
vectors band care not orthogonal to each other.
Question 2
Question
Let a=⟨3,−2,4⟩and b=⟨−1,5,2⟩. Find the vector csuch that 3a−2b=c.
Solution
Step 1: Calculate 3aand 2b.
3a= 3⟨3,−2,4⟩=⟨9,−6,12⟩
2b= 2⟨−1,5,2⟩=⟨−2,10,4⟩
Step 2: Find 3a−2b.
3a−2b=⟨9,−6,12⟩ − ⟨−2,10,4⟩
=⟨9−(−2),−6−10,12 −4⟩
=⟨11,−16,8⟩
Therefore, the vector csuch that 3a−2b=cis c=⟨11,−16,8⟩.
Question 3
Question
Let a= 2i−3j+ 4kand b= 5i+j−2k.
Determine the angle between vectors aand b.
Solution
To find the angle between two vectors aand b, we can use the dot product
formula:
a·b=|a||b|cos θ
where θis the angle between aand b.
Step 1: Calculate the dot product of vectors a and b
The dot product of two vectors a=a1i+a2j+a3kand b=b1i+b2j+b3k
is given by:
a·b=a1b1+a2b2+a3b3
In this case, a·b= (2)(5) + (−3)(1) + (4)(−2) = 10 −3−8 = −1.
Step 2: Calculate the magnitudes of vectors a and b
The magnitude of a vector v=v1i+v2j+v3kis given by:
|v|=qv2
1+v2
2+v2
3
In this case, |a|=p22+ (−3)2+ 42=√4 + 9 + 16 = √29 and |b|=
p52+ 12+ (−2)2=√25 + 1 + 4 = √30.
Step 3: Calculate the angle θbetween vectors a and b
Substitute the dot product and magnitudes into the formula a·b=|a||b|cos θ:
2
−1 = √29 ·√30 ·cos θ
cos θ=−1
√29 ·√30 =−1
√870
θ= cos−1−1
√870
Therefore, the angle between vectors aand bis cos−1−1
√870 .
Question 4
Question
Let a=⟨3,−4,2⟩and b=⟨1,7,−5⟩be two vectors. Determine the angle
between aand bin radians.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= 3(1) + (−4)(7) + 2(−5) = 3 −28 −10 = −35
Step 2: Find the magnitudes of vectors aand b.
∥a∥=p32+ (−4)2+ 22=√9 + 16 + 4 = √29
∥b∥=p12+ 72+ (−5)2=√1 + 49 + 25 = √75
Step 3: Use the dot product formula to find the angle between the vectors.
cos(θ) = a
·b∥a∥∥b∥
cos(θ) = −35
√29 ·√75 =−35
√2175 =−35
√2175
Step 4: Calculate the angle between the vectors.
θ= arccos −35
√2175≈2.937 radians
Therefore, the angle between vectors aand bis approximately 2.937 radians.
Question 5
Question
Let u =
3
−1
2
and v =
5
4
−3
. Determine the vector projection of u onto v.
3
Solution
To find the vector projection of u onto v, we can use the formula
projvu =u ·v
∥v∥2v,
where u ·v is the dot product of vectors u and v, and ∥v∥is the magnitude of v.
Step 1: Calculate the dot product u ·v.
u ·v =
3
−1
2
·
5
4
−3
= (3)(5) + (−1)(4) + (2)(−3) = 15 −4−6=5.
Step 2: Calculate the magnitude of v,∥v∥.
∥v∥=p52+ 42+ (−3)2=√25 + 16 + 9 = √50 = 5√2.
Step 3: Substitute the values into the formula to find the vector projection.
projvu =5
(5√2)2
5
4
−3
=5
50
5
4
−3
=1
10
5
4
−3
=
0.5
0.4
−0.3
.
Therefore, the vector projection of u onto v is
0.5
0.4
−0.3
.
Question 6
Question
Let a= 3i−4j+ 5kand b=j−2k. Find the angle between vectors aand b.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (3)(0) + (−4)(1) + (5)(−2) = −4−10 = −14
Step 2: Calculate the magnitudes of vectors aand b.
|a|=p(3)2+ (−4)2+ (5)2=√9 + 16 + 25 = √50
|b|=p(0)2+ (1)2+ (−2)2=√0 + 1 + 4 = √5
Step 3: Use the formula cos θ=a·b
|a|·|b|to find the angle.
cos θ=−14
√50 ·√5=−14
√250 =−14
5√10 =−14√10
50
θ= arccos −14√10
50 !≈126.87◦
Therefore, the angle between vectors aand bis approximately 126.87◦.
4
Question 7
Question
Let a =
2
−3
4
and
b=
−1
5
−2
. Find the angle between vectors a and
b.
Solution
To find the angle between two vectors a and
b, we can use the formula
cos(θ) = a ·
b
∥a∥∥
b∥
where θis the angle between the two vectors, a ·
bis the dot product of a and
b, and ∥a∥,∥
b∥are the magnitudes of a and
b, respectively.
Step 1: Calculate the dot product of a and
bThe dot product a ·
bis
given by:
a ·
b= (2)(−1) + (−3)(5) + (4)(−2) = −2−15 −8 = −25
Step 2: Calculate the magnitudes of a and
bThe magnitude of a vector
v =
x
y
z
is given by:
∥v∥=px2+y2+z2
So, the magnitudes of vectors a and
bare:
∥a∥=p22+ (−3)2+ 42=√4 + 9 + 16 = √29
∥
b∥=p(−1)2+ 52+ (−2)2=√1 + 25 + 4 = √30
Step 3: Find the cosine of the angle between vectors a and
bNow,
we can substitute the values we found into the formula:
cos(θ) = a ·
b
∥a∥∥
b∥=−25
√29 ·√30 =−25
√29 ·30 =−25
√870
Step 4: Find the angle between vectors a and
bTo find the angle θ,
we take the arccosine of the value we found in step 3:
θ= arccos −25
√870≈116.56◦
Therefore, the angle between vectors a and
bis approximately 116.56◦.
5
Question 8
Question
Let a =3
−2and
b=−1
5. Find the scalar ksuch that c =ka −2
bis
orthogonal to a.
Solution
Step 1: Find the dot product of c and a.
c ·a = (ka −2
b)·a
=ka ·a −2
b·a
=k|a|2−2
b·a
=k|a|2−2|
b||a|cos θ
where θis the angle between a and
b.
Step 2: Since c is orthogonal to a, we know that c ·a = 0.
k|a|2−2|
b||a|cos θ= 0
k(32+ (−2)2)−2p(−1)2+ 52p32+ (−2)2cos θ= 0
k(9 + 4) −2√26√13 cos θ= 0
Step 3: Finally, since c is orthogonal to a, the dot product of c and a is 0.
Solve for the scalar k.
13k−26√26 cos θ= 0
k=26√26 cos θ
13
Question 9
Question
Let a=
2
−3
4
,b=
−1
2
1
, and c=
0
−2
7
. Find the scalar ksuch that
ka+b= 2c.
6
Solution
Step 1: First, we can expand the equation ka+b= 2c.
k
2
−3
4
+
−1
2
1
= 2
0
−2
7
Step 2: Simplifying further, we get
2k
−3k
4k
+
−1
2
1
=
0
−4
14
Step 3: Combining like terms, we have
2k−1
−3k+ 2
4k+ 1
=
0
−4
14
Step 4: Setting the corresponding components equal, we get the following
system of equations:
2k−1=0
−3k+ 2 = −4
4k+ 1 = 14
Step 5: Solving these equations, we find k= 1.
Therefore, the scalar ksuch that ka+b= 2cis k= 1.
Question 10
Question
Let u=
2
−1
3
and v=
−4
5
1
be two vectors in R3. Determine the vector
projection of uonto v.
Solution
Step 1: We can find the vector projection of uonto vusing the formula:
projv(u) = u·v
∥v∥2v
where: - u·vis the dot product of uand v, - ∥v∥is the magnitude of v.
7
Step 2: Calculate u·v:
u·v=
2
−1
3
·
−4
5
1
= (2)(−4) + (−1)(5) + (3)(1) = −8−5 + 3 = −10
Step 3: Calculate ∥v∥:
∥v∥=p(−4)2+ 52+ 12=√16 + 25 + 1 = √42
Step 4: Substitute the values into the formula to find the vector projection:
projv(u) = −10
42
−4
5
1
=
20
21
−25
21
−5
21
Therefore, the vector projection of uonto vis
20
21
−25
21
−5
21
.
Question 11
Question
Let a =
3
−4
5
and
b=
−2
1
3
be two vectors in R3. Find the angle θbetween
the two vectors (in radians) and the projection of a onto the direction of
b.
Solution
Step 1: To find the angle θbetween the two vectors, we will use the dot product
formula: a ·
b=||a||||
b||cos θ.
a ·
b=
3
−4
5
·
−2
1
3
= (3 · −2) + (−4·1) + (5 ·3)
=−6−4 + 15
= 5
Step 2: Calculate the magnitudes of a and
busing the formula ||v|| =
pv2
1+v2
2+v2
3.
||a|| =p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
8
||
b|| =p(−2)2+ 12+ 32=√4 + 1 + 9 = √14
Step 3: Substitute the values of the dot product and the magnitudes into
the dot product formula to solve for the angle θ.
5=5√2·√14 ·cos θ
= 5√28 ·cos θ
= 2√70 ·cos θ
cos θ=5
2√70 =5√70
140 =√70
28
θ= cos−1 √70
28 !≈0.6435 radians
Step 4: To find the projection of a onto
b, use the formula: proj
b(a) =
a·
b
||
b||2
b.
proj
b(a) = 5
14
−2
1
3
=
−5
7
5
14
15
14
Therefore, the angle between the two vectors is approximately 0.6435 radi-
ans, and the projection of a onto
bis
−5
7
5
14
15
14
.
Question 12
Question
Let u =
3
−1
2
,v =
−2
4
1
, and w =
5
0
−3
. Find (2u −3v)·w.
Solution
Step 1: Compute 2u and 3v.
2u = 2
3
−1
2
=
6
−2
4
9
3v = 3
−2
4
1
=
−6
12
3
Step 2: Calculate 2u −3v.
2u −3v =
6
−2
4
−
−6
12
3
=
12
−14
1
Step 3: Compute the dot product of (2u −3v) and w.
(2u −3v)·w =
12
−14
1
·
5
0
−3
= (12)(5) + (−14)(0) + (1)(−3) = 60 −3 = 57
Therefore, (2u −3v)·w = 57 .
Question 13
Question
Let u =
2
−1
3
and v =
−4
0
1
. Find the angle between the vectors u and v.
Solution
Step 1: Find the dot product of u and v:
u ·v = (2)(−4) + (−1)(0) + (3)(1) = −8 + 0 + 3 = −5
Step 2: Find the magnitude of u and v:
|u|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|v|=p(−4)2+ 02+ 12=√16 + 0 + 1 = √17
Step 3: Use the formula cos(θ) = u·v
|u|·|v|to find the cosine of the angle
between u and v.
cos(θ) = −5
√14 ·√17 =−5
√238
Step 4: Find the angle θusing the inverse cosine function:
θ= arccos −5
√238≈123.8◦
Therefore, the angle between the vectors u and v is approximately 123.8◦.
10
Question 14
Question
Let a= 3i−2j+ 4kand b=i+ 5j−2k. Find the angle between vectors aand
b.
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= (3)(1) + (−2)(5) + (4)(−2) = 3 −10 −8 = −15
Step 2: Calculate the magnitudes of vectors aand b:
|a|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|b|=p12+ 52+ (−2)2=√1 + 25 + 4 = √30
Step 3: Use the formula for the angle between two vectors:
cos θ=a
·b|a||b|=−15
√29 ·√30 =−15
√870 =−15
√870
Step 4: Find the angle θ:
θ= arccos −15
√870
Therefore, the angle between vectors aand bis arccos −15
√870 .
Question 15
Question
Let a=2
1and b=−1
3. Calculate the angle between vectors aand b.
Solution
To find the angle between two vectors aand b, we can use the dot product
formula:
a·b=|a|·|b| · cos(θ)
where |a|and |b|are the magnitudes of vectors aand b, respectively, and θ
is the angle between the two vectors.
Step 1: Calculate the magnitudes of vectors aand b.
|a|=p22+ 12=√5
11
|b|=p(−1)2+ 32=√10
Step 2: Calculate the dot product of vectors aand b.
a·b= 2 ·(−1) + 1 ·3 = −2 + 3 = 1
Step 3: Substitute the values into the dot product formula and solve for θ.
1 = √5·√10 ·cos(θ)
Step 4: Solve for θ.
cos(θ) = 1
√50 =1
5√2
Step 5: Finally, calculate the angle θ.
θ= cos−11
5√2≈74.74◦
Therefore, the angle between vectors aand bis approximately 74.74◦.
Question 16
Question
Let u=3
−2and v=−4
5. Determine whether the vectors uand vare
orthogonal (perpendicular) to each other.
Solution
Step 1: To determine if two vectors are orthogonal, we need to calculate their
dot product. If the dot product is equal to zero, then the vectors are orthogonal.
Step 2: The dot product of two vectors uand vis given by the formula:
u·v=3
−2·−4
5= (3)(−4) + (−2)(5)
Step 3: Calculating the dot product, we get:
u·v= (−12) + (−10) = −22
Step 4: Since the dot product u·v=−22 = 0, the vectors uand vare not
orthogonal to each other.
Question 17
Question
Let a=
2
−1
3
and b=
4
−2
1
. Determine the angle between vectors aand b.
12
Solution
To find the angle between two vectors aand b, we can use the dot product
formula:
a·b=|a|·|b| · cos θ
where θis the angle between the two vectors.
Step 1: Calculate the dot product of vectors aand b.
a·b= (2)(4) + (−1)(−2) + (3)(1) = 8 + 2 + 3 = 13
Step 2: Calculate the magnitudes of vectors aand b.
|a|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|b|=p42+ (−2)2+ 12=√16 + 4 + 1 = √21
Step 3: Substitute the values into the dot product formula. Solve for θ.
13 = (√14) ·(√21) ·cos θ
cos θ=13
√14 ·√21 =13
√294 =13
√294 ·√294
√294 =13√294
294
θ= cos−1 13√294
294 !
Therefore, the angle between vectors aand bis θ= cos−113√294
294 .
Question 18
Question
Let u =4
−3and v =−2
5. Determine 3u −2v.
Solution
Step 1: Calculate 3u.
3
−3·4
−3=3(4)
3(−3)=12
−9
Step 2: Calculate 2v.
2
5·−2
5=2(−2)
2(5) =−4
10
Step 3: Subtract 2v from 3u to find 3u −2v.
12
−9−−4
10 =12 + 4
−9−10=16
−19
Therefore, 3u −2v =16
−19.
13
Question 19
Question
Given two vectors a=
2
−3
1
and b=
−1
4
2
, find the angle between them in
degrees.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (2)(−1) + (−3)(4) + (1)(2) = −2−12 + 2 = −12
Step 2: Find the magnitudes of vectors aand b.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p(−1)2+ 42+ 22=√1 + 16 + 4 = √21
Step 3: Use the formula for the angle between two vectors: θ= cos−1a
·b∥a∥∥b∥.
θ= cos−1−12
√14 ·√21
Step 4: Calculate θin degrees.
θ≈cos−1−12
√294
θ≈cos−1−12
17.1464
θ≈cos−1(−0.7001)
θ≈135.52◦
Therefore, the angle between vectors aand bis approximately 135.52◦.
Question 20
Question
Given vectors a=3
−1and b=−2
4, find a vector csuch that c·a= 2c·b.
14
Solution
To find the vector c, we use the property of the dot product that states x·y=
x1y1+x2y2for vectors x=x1
x2and y=y1
y2.
Step 1: Compute c·aand 2c·b.
c·a=c1
c2·3
−1= 3c1−c2
2c·b= 2 c1
c2·−2
4= 2(−2c1+ 4c2) = −4c1+ 8c2
Step 2: Set up the equation given by the problem and solve for c. The
equation given is c·a= 2c·b, so we have:
3c1−c2=−4c1+ 8c2
Simplifying, we get:
7c1= 9c2
Step 3: Find a solution for c. To find a solution, we can let c1= 9 and
c2= 7. Therefore, the vector cis 9
7.
Thus, a vector cthat satisfies the given condition is 9
7.
Question 21
Question
Let a= 2i−3j+kand b=i+ 4j−2k. Determine ∥a−b∥.
Solution
Step 1: Compute a−b:
a−b= (2i−3j+k)−(i+ 4j−2k)
= 2i−3j+k−i−4j+ 2k
=i−7j+ 3k
Step 2: Compute the norm of a−b:
∥a−b∥=p(1)2+ (−7)2+ (3)2
=√1 + 49 + 9
=√59
Therefore, the magnitude of a−bis √59.
15
Question 22
Question
Let a = 2ˆ
i−3ˆ
j+ 4ˆ
kand
b=ˆ
i+ 5ˆ
j−2ˆ
k. Find the angle between vectors a and
b.
Solution
To find the angle between the vectors a and
b, we can use the dot product
formula:
a ·
b=|a|·|
b| · cos θ
where θis the angle between the vectors.
Step 1: Calculate the dot product of a and
b:
a ·
b= (2)(1) + (−3)(5) + (4)(−2)
a ·
b= 2 −15 −8
a ·
b=−21
Step 2: Calculate the magnitudes of a and
b:
|a|=p22+ (−3)2+ 42
|a|=√4 + 9 + 16
|a|=√29
|
b|=p12+ 52+ (−2)2
|
b|=√1 + 25 + 4
|
b|=√30
Step 3: Substitute into the dot product formula to find the angle:
−21 = √29 ·√30 ·cos θ
−21 = √870 ·cos θ
cos θ=−21
√870
θ= arccos −21
√870
θ≈2.04 radians
Therefore, the angle between vectors a and
bis approximately 2.04 radians.
16
Question 23
Question
Let a= 3i−2j+ 4kand b= 2i+ 5j−3k. Determine the scalar projection of
aonto b.
Solution
Step 1: To find the scalar projection of aonto b, we use the formula:
Scalar projection of aonto b=a·b
∥b∥
where a·bis the dot product of aand b, and ∥b∥is the magnitude of b.
Step 2: First, let’s find the dot product a·b:
a·b= (3)(2) + (−2)(5) + (4)(−3)
a·b= 6 −10 −12 = −16
Step 3: Next, calculate the magnitude of b:
∥b∥=p(2)2+ (5)2+ (−3)2
∥b∥=√4 + 25 + 9 = √38
Step 4: Now, substitute the dot product and the magnitude into the formula
to find the scalar projection:
Scalar projection of aonto b=−16
√38 =−16√38
38 =−8√38
19
Therefore, the scalar projection of aonto bis −8√38
19 .
Question 24
Question
Let a = 3
i−2
j+ 4
kand
b= 2
i+
j−2
k. Find the angle between vectors a and
b.
Solution
Step 1: To find the angle between two vectors a and
b, we can use the formula:
cos(θ) = a ·
b
∥a∥∥
b∥
17
where θis the angle between the two vectors, a ·
bis the dot product of a and
b, and ∥a∥and ∥
b∥are the magnitudes of a and
b, respectively.
Step 2: First, calculate the dot product of a and
b:
a ·
b= (3
i−2
j+ 4
k)·(2
i+
j−2
k)
= 3(2) + (−2)(1) + 4(−2) = 6 −2−8 = −4
Step 3: Next, calculate the magnitudes of a and
b:
∥a∥=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
∥
b∥=p22+ 12+ (−2)2=√4 + 1 + 4 = √9
Step 4: Substitute the dot product and magnitudes into the formula to find
cos(θ):
cos(θ) = −4
√29 ·√9=−4
√261
Step 5: Finally, to find the angle θ, take the arccosine of −4
√261 :
θ= arccos −4
√261
Question 25
Question
Let a=2
−1and b=3
4. Find the angle between vectors aand b.
Solution
To find the angle between two vectors aand b, we can use the formula:
cos(θ) = a·b
∥a∥∥b∥
where θis the angle between the vectors, a·bis the dot product of aand b,
and ∥a∥and ∥b∥are the magnitudes of vectors aand b, respectively.
Step 1: Calculate the dot product of aand b.
a·b= (2)(3) + (−1)(4) = 6 −4=2
Step 2: Calculate the magnitudes of vectors aand b.
∥a∥=p22+ (−1)2=√4 + 1 = √5
∥b∥=p32+ 42=√9 + 16 = √25 = 5
18
Step 3: Substitute the dot product and magnitudes into the formula for
the cosine of the angle between vectors.
cos(θ) = 2
√5·5=2
5√5
Step 4: Solve for the angle θ.
θ= arccos 2
5√5≈71.57◦
Therefore, the angle between vectors aand bis approximately 71.57◦.
Question 26
Question
Given two vectors u=⟨3,−2,5⟩and v=⟨−1,4,7⟩, find:
a) ∥u+v∥(magnitude of the sum)
b) ∥u−v∥(magnitude of the difference)
Solution
a) To find the magnitude of the sum ∥u+v∥: Step 1: Calculate the sum of the
two vectors:
u+v=⟨3,−2,5⟩+⟨−1,4,7⟩=⟨3+(−1),−2+4,5+7⟩=⟨2,2,12⟩
Step 2: Find the magnitude of the sum:
∥u+v∥=p22+ 22+ 122=√4 + 4 + 144 = √152
b) To find the magnitude of the difference ∥u−v∥: Step 1: Calculate the
difference of the two vectors:
u−v=⟨3,−2,5⟩ − ⟨−1,4,7⟩=⟨3−(−1),−2−4,5−7⟩=⟨4,−6,−2⟩
Step 2: Find the magnitude of the difference:
∥u−v∥=p42+ (−6)2+ (−2)2=√16 + 36 + 4 = √56
Question 27
Question
Let a=3
−2and b=−1
4. Determine the angle between vectors aand b,
rounded to the nearest degree.
19
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= 3(−1) + (−2)(4) = −3−8 = −11
Step 2: Calculate the magnitude of vector a:
|a|=p32+ (−2)2=√9 + 4 = √13
Step 3: Calculate the magnitude of vector b:
|b|=p(−1)2+ 42=√1 + 16 = √17
Step 4: Calculate the angle θbetween vectors aand busing the formula:
cos(θ) = a·b
|a|·|b|
cos(θ) = −11
√13 ·√17
θ= arccos −11
√13 ·√17
Step 5: Use a calculator to find the angle θ. Rounding to the nearest degree,
the angle between vectors aand bis approximately 123◦.
Question 28
Question
Let u=
3
−1
2
and v=
2
4
−3
be two vectors. Find the angle between the
vectors uand v.
Solution
Step 1: Find the dot product of vectors uand v,
u·v= (3)(2) + (−1)(4) + (2)(−3) = 6 −4−6 = −4.
Step 2: Find the magnitudes of vectors uand v,
|u|=p32+ (−1)2+ 22=√9 + 1 + 4 = √14,
|v|=p22+ 42+ (−3)2=√4 + 16 + 9 = √29.
20
Step 3: Use the formula for finding the angle between two vectors,
cos θ=u·v
|u||v|.
Step 4: Substitute the values of the dot product and magnitudes,
cos θ=−4
√14 ·√29 =−4
√406.
Step 5: Solve for θby taking the arccosine of both sides,
θ= arccos −4
√406≈109.76◦.
Therefore, the angle between vectors uand vis approximately 109.76◦.
Question 29
Question
Let a=
3
−2
5
and b=
−1
4
2
. Find the vector csuch that c·(a×b) = |a×b|.
Solution
Step 1: Find the cross product of vectors aand b, denoted by a×b.
a×b=
3
−2
5
×
−1
4
2
=
−22
−13
10
Step 2: Find the magnitude of the cross product |a×b|.
|a×b|=p(−22)2+ (−13)2+ 102=√484 + 169 + 100 = √753 = 27√3
Step 3: Find the unit vector in the direction of a×b.
ˆ
u=a×b
|a×b|=1
27√3
−22
−13
10
=
−22
27√3
−13
27√3
10
27√3
Step 4: Find the vector cby multiplying the unit vector with the magnitude
of a×b.
c=|a×b| · ˆ
u= 27√3·
−22
27√3
−13
27√3
10
27√3
=
−22
−13
10
Therefore, c=
−22
−13
10
.
21
Question 30
Question
Let a=
2
−1
3
,b=
−1
3
5
, and c=
4
2
−1
. Determine if a,b, and care linearly
independent.
Solution
To determine if a,b, and care linearly independent, we need to check if the
equation
k1a+k2b+k3c=0
has only the trivial solution k1=k2=k3= 0.
Let’s set up the equation and solve for k1,k2, and k3.
Step 1: Set up the equation.
Consider the equation
k1
2
−1
3
+k2
−1
3
5
+k3
4
2
−1
=
0
0
0
Step 2: Solve the equation.
This gives the system of equations
2k1−k2+ 4k3= 0
−k1+ 3k2+ 2k3= 0
3k1+ 5k2−k3= 0
We can use Gaussian elimination to solve this system of equations. After
reducing the augmented matrix, we find that the only solution is k1=k2=
k3= 0.
Since the equation has only the trivial solution, a,b, and care linearly
independent.
Question 31
Question
Let u =⟨3,−1,2⟩and v =⟨−2,5,4⟩. Find a vector w such that w ·u = 8 and
w ·v =−3.
22
Solution
Step 1: Recall that the dot product of two vectors a =⟨a1, a2, a3⟩and
b=
⟨b1, b2, b3⟩is given by a ·
b=a1b1+a2b2+a3b3.
Step 2: Given that w ·u = 8, we have w · ⟨3,−1,2⟩= 8, which gives us
3w1−w2+ 2w3= 8.
Step 3: Similarly, from w ·v =−3, we have w · ⟨−2,5,4⟩=−3, which gives
us −2w1+ 5w2+ 4w3=−3.
Step 4: To find a vector w that satisfies both equations, we set up a system
of linear equations:
(3w1−w2+ 2w3= 8
−2w1+ 5w2+ 4w3=−3
Step 5: Solving the system of equations, we find w1=−2, w2=−1, and
w3= 3.
Therefore, the vector w =⟨−2,−1,3⟩satisfies the given conditions.
Question 32
Question
Let a= 3i−4j+ 2kand b= 2i+j−3k. Find the angle between vectors aand
b.
Solution
Step 1: Calculate the dot product of aand b.
a·b= (3)(2) + (−4)(1) + (2)(−3) = 6 −4−6 = −4
Step 2: Calculate the magnitudes of aand b.
|a|=p32+ (−4)2+ 22=√9 + 16 + 4 = √29
|b|=p22+ 12+ (−3)2=√4 + 1 + 9 = √14
Step 3: Calculate the cosine of the angle between aand busing the dot
product formula.
cos θ=a
·b|a|·|b|=−4
√29√14 =−4
√406
Step 4: Find the angle between aand busing inverse cosine.
θ= arccos −4
√406
Therefore, the angle between vectors aand bis θ.
23
Question 33
Question
Let a= 3i+ 7j−2kand b= 5i−4j+ 6kbe two vectors in R3. Determine the
vector projection of aonto b.
Solution
Step 1: Calculate the dot product of aand b.
The dot product of two vectors a=a1i+a2j+a3kand b=b1i+b2j+b3k
is given by:
a·b=a1b1+a2b2+a3b3
For our vectors aand b, we have:
a·b= (3)(5) + (7)(−4) + (−2)(6) = 15 −28 −12 = −25
Step 2: Calculate the magnitude of vector b.
The magnitude of a vector b=b1i+b2j+b3kis given by:
|b|=qb2
1+b2
2+b2
3
For our vector b, we have:
|b|=p52+ (−4)2+ 62=√25 + 16 + 36 = √77
Step 3: Calculate the vector projection of aonto b.
The vector projection of aonto bis given by:
projba=a·b
|b|2b
Substitute the calculated values to get:
projba=−25
77 ·1
77(5i−4j+ 6k)
So, the vector projection of aonto bis:
−25
77 (5i−4j+ 6k)
Question 34
Question
Let a = (−3,4,2) and
b= (1,−2,0). Find the angle between vectors a and
b.
24
Solution
Step 1: To find the angle θbetween vectors a and
b, we can use the dot product
formula:
a ·
b=|a||
b|cos θ.
Step 2: Calculate the dot product of a and
b:
a ·
b= (−3)(1) + (4)(−2) + (2)(0) = −3−8 + 0 = −11.
Step 3: Calculate the magnitudes of vectors a and
b:
|a|=p(−3)2+ 42+ 22=√9 + 16 + 4 = √29,
|
b|=p12+ (−2)2+ 02=√1 + 4 + 0 = √5.
Step 4: Substitute the dot product and magnitudes into the dot product
formula:
−11 = √29 ·√5·cos θ.
Step 5: Solve for θby isolating cos θ:
cos θ=−11
√29 ·√5=−11
√145.
Step 6: Finally, find the angle θin radians by taking the inverse cosine:
θ= cos−1−11
√145≈2.62 radians.
Therefore, the angle between vectors a and
bis approximately 2.62 radians.
Question 35
Question
Let a=
3
−2
5
,b=
−4
1
2
, and c=
1
3
−4
. Find a vector dsuch that a+ 2b−
3c=d.
Solution
Step 1: Perform scalar multiplication on vectors a,b, and c.
2b= 2
−4
1
2
=
−8
2
4
−3c=−3
1
3
−4
=
−3
−9
12
25
Step 2: Add the scaled vectors a, 2b, and −3ctogether to find d.
d=a+ 2b−3c
=
3
−2
5
+
−8
2
4
+
−3
−9
12
=
3
−2
5
+
−8
2
4
+
−3
−9
12
=
(3) + (−8) + (−3)
(−2) + 2 + (−9)
5 + 4 + 12
=
−8
−9
21
Therefore, the vector dis
−8
−9
21
.
26
Question 7
Question
Let a =
2
−3
4
and
b=
−1
5
−2
. Find the angle between vectors a and
b.
Solution
To find the angle between two vectors a and
b, we can use the formula
cos(θ) = a ·
b
∥a∥∥
b∥
where θis the angle between the two vectors, a ·
bis the dot product of a and
b, and ∥a∥,∥
b∥are the magnitudes of a and
b, respectively.
Step 1: Calculate the dot product of a and
bThe dot product a ·
bis
given by:
a ·
b= (2)(−1) + (−3)(5) + (4)(−2) = −2−15 −8 = −25
Step 2: Calculate the magnitudes of a and
bThe magnitude of a vector
v =
x
y
z
is given by:
∥v∥=px2+y2+z2
So, the magnitudes of vectors a and
bare:
∥a∥=p22+ (−3)2+ 42=√4 + 9 + 16 = √29
∥
b∥=p(−1)2+ 52+ (−2)2=√1 + 25 + 4 = √30
Step 3: Find the cosine of the angle between vectors a and
bNow,
we can substitute the values we found into the formula:
cos(θ) = a ·
b
∥a∥∥
b∥=−25
√29 ·√30 =−25
√29 ·30 =−25
√870
Step 4: Find the angle between vectors a and
bTo find the angle θ,
we take the arccosine of the value we found in step 3:
θ= arccos −25
√870≈116.56◦
Therefore, the angle between vectors a and
bis approximately 116.56◦.
5
Question 8
Question
Let a =3
−2and
b=−1
5. Find the scalar ksuch that c =ka −2
bis
orthogonal to a.
Solution
Step 1: Find the dot product of c and a.
c ·a = (ka −2
b)·a
=ka ·a −2
b·a
=k|a|2−2
b·a
=k|a|2−2|
b||a|cos θ
where θis the angle between a and
b.
Step 2: Since c is orthogonal to a, we know that c ·a = 0.
k|a|2−2|
b||a|cos θ= 0
k(32+ (−2)2)−2p(−1)2+ 52p32+ (−2)2cos θ= 0
k(9 + 4) −2√26√13 cos θ= 0
Step 3: Finally, since c is orthogonal to a, the dot product of c and a is 0.
Solve for the scalar k.
13k−26√26 cos θ= 0
k=26√26 cos θ
13
Question 9
Question
Let a=
2
−3
4
,b=
−1
2
1
, and c=
0
−2
7
. Find the scalar ksuch that
ka+b= 2c.
6
Solution
Step 1: First, we can expand the equation ka+b= 2c.
k
2
−3
4
+
−1
2
1
= 2
0
−2
7
Step 2: Simplifying further, we get
2k
−3k
4k
+
−1
2
1
=
0
−4
14
Step 3: Combining like terms, we have
2k−1
−3k+ 2
4k+ 1
=
0
−4
14
Step 4: Setting the corresponding components equal, we get the following
system of equations:
2k−1=0
−3k+ 2 = −4
4k+ 1 = 14
Step 5: Solving these equations, we find k= 1.
Therefore, the scalar ksuch that ka+b= 2cis k= 1.
Question 10
Question
Let u=
2
−1
3
and v=
−4
5
1
be two vectors in R3. Determine the vector
projection of uonto v.
Solution
Step 1: We can find the vector projection of uonto vusing the formula:
projv(u) = u·v
∥v∥2v
where: - u·vis the dot product of uand v, - ∥v∥is the magnitude of v.
7
Step 2: Calculate u·v:
u·v=
2
−1
3
·
−4
5
1
= (2)(−4) + (−1)(5) + (3)(1) = −8−5 + 3 = −10
Step 3: Calculate ∥v∥:
∥v∥=p(−4)2+ 52+ 12=√16 + 25 + 1 = √42
Step 4: Substitute the values into the formula to find the vector projection:
projv(u) = −10
42
−4
5
1
=
20
21
−25
21
−5
21
Therefore, the vector projection of uonto vis
20
21
−25
21
−5
21
.
Question 11
Question
Let a =
3
−4
5
and
b=
−2
1
3
be two vectors in R3. Find the angle θbetween
the two vectors (in radians) and the projection of a onto the direction of
b.
Solution
Step 1: To find the angle θbetween the two vectors, we will use the dot product
formula: a ·
b=||a||||
b||cos θ.
a ·
b=
3
−4
5
·
−2
1
3
= (3 · −2) + (−4·1) + (5 ·3)
=−6−4 + 15
= 5
Step 2: Calculate the magnitudes of a and
busing the formula ||v|| =
pv2
1+v2
2+v2
3.
||a|| =p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
8
||
b|| =p(−2)2+ 12+ 32=√4 + 1 + 9 = √14
Step 3: Substitute the values of the dot product and the magnitudes into
the dot product formula to solve for the angle θ.
5=5√2·√14 ·cos θ
= 5√28 ·cos θ
= 2√70 ·cos θ
cos θ=5
2√70 =5√70
140 =√70
28
θ= cos−1 √70
28 !≈0.6435 radians
Step 4: To find the projection of a onto
b, use the formula: proj
b(a) =
a·
b
||
b||2
b.
proj
b(a) = 5
14
−2
1
3
=
−5
7
5
14
15
14
Therefore, the angle between the two vectors is approximately 0.6435 radi-
ans, and the projection of a onto
bis
−5
7
5
14
15
14
.
Question 12
Question
Let u =
3
−1
2
,v =
−2
4
1
, and w =
5
0
−3
. Find (2u −3v)·w.
Solution
Step 1: Compute 2u and 3v.
2u = 2
3
−1
2
=
6
−2
4
9
3v = 3
−2
4
1
=
−6
12
3
Step 2: Calculate 2u −3v.
2u −3v =
6
−2
4
−
−6
12
3
=
12
−14
1
Step 3: Compute the dot product of (2u −3v) and w.
(2u −3v)·w =
12
−14
1
·
5
0
−3
= (12)(5) + (−14)(0) + (1)(−3) = 60 −3 = 57
Therefore, (2u −3v)·w = 57 .
Question 13
Question
Let u =
2
−1
3
and v =
−4
0
1
. Find the angle between the vectors u and v.
Solution
Step 1: Find the dot product of u and v:
u ·v = (2)(−4) + (−1)(0) + (3)(1) = −8 + 0 + 3 = −5
Step 2: Find the magnitude of u and v:
|u|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|v|=p(−4)2+ 02+ 12=√16 + 0 + 1 = √17
Step 3: Use the formula cos(θ) = u·v
|u|·|v|to find the cosine of the angle
between u and v.
cos(θ) = −5
√14 ·√17 =−5
√238
Step 4: Find the angle θusing the inverse cosine function:
θ= arccos −5
√238≈123.8◦
Therefore, the angle between the vectors u and v is approximately 123.8◦.
10
Question 14
Question
Let a= 3i−2j+ 4kand b=i+ 5j−2k. Find the angle between vectors aand
b.
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= (3)(1) + (−2)(5) + (4)(−2) = 3 −10 −8 = −15
Step 2: Calculate the magnitudes of vectors aand b:
|a|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|b|=p12+ 52+ (−2)2=√1 + 25 + 4 = √30
Step 3: Use the formula for the angle between two vectors:
cos θ=a
·b|a||b|=−15
√29 ·√30 =−15
√870 =−15
√870
Step 4: Find the angle θ:
θ= arccos −15
√870
Therefore, the angle between vectors aand bis arccos −15
√870 .
Question 15
Question
Let a=2
1and b=−1
3. Calculate the angle between vectors aand b.
Solution
To find the angle between two vectors aand b, we can use the dot product
formula:
a·b=|a|·|b| · cos(θ)
where |a|and |b|are the magnitudes of vectors aand b, respectively, and θ
is the angle between the two vectors.
Step 1: Calculate the magnitudes of vectors aand b.
|a|=p22+ 12=√5
11
|b|=p(−1)2+ 32=√10
Step 2: Calculate the dot product of vectors aand b.
a·b= 2 ·(−1) + 1 ·3 = −2 + 3 = 1
Step 3: Substitute the values into the dot product formula and solve for θ.
1 = √5·√10 ·cos(θ)
Step 4: Solve for θ.
cos(θ) = 1
√50 =1
5√2
Step 5: Finally, calculate the angle θ.
θ= cos−11
5√2≈74.74◦
Therefore, the angle between vectors aand bis approximately 74.74◦.
Question 16
Question
Let u=3
−2and v=−4
5. Determine whether the vectors uand vare
orthogonal (perpendicular) to each other.
Solution
Step 1: To determine if two vectors are orthogonal, we need to calculate their
dot product. If the dot product is equal to zero, then the vectors are orthogonal.
Step 2: The dot product of two vectors uand vis given by the formula:
u·v=3
−2·−4
5= (3)(−4) + (−2)(5)
Step 3: Calculating the dot product, we get:
u·v= (−12) + (−10) = −22
Step 4: Since the dot product u·v=−22 = 0, the vectors uand vare not
orthogonal to each other.
Question 17
Question
Let a=
2
−1
3
and b=
4
−2
1
. Determine the angle between vectors aand b.
12
Solution
To find the angle between two vectors aand b, we can use the dot product
formula:
a·b=|a|·|b| · cos θ
where θis the angle between the two vectors.
Step 1: Calculate the dot product of vectors aand b.
a·b= (2)(4) + (−1)(−2) + (3)(1) = 8 + 2 + 3 = 13
Step 2: Calculate the magnitudes of vectors aand b.
|a|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|b|=p42+ (−2)2+ 12=√16 + 4 + 1 = √21
Step 3: Substitute the values into the dot product formula. Solve for θ.
13 = (√14) ·(√21) ·cos θ
cos θ=13
√14 ·√21 =13
√294 =13
√294 ·√294
√294 =13√294
294
θ= cos−1 13√294
294 !
Therefore, the angle between vectors aand bis θ= cos−113√294
294 .
Question 18
Question
Let u =4
−3and v =−2
5. Determine 3u −2v.
Solution
Step 1: Calculate 3u.
3
−3·4
−3=3(4)
3(−3)=12
−9
Step 2: Calculate 2v.
2
5·−2
5=2(−2)
2(5) =−4
10
Step 3: Subtract 2v from 3u to find 3u −2v.
12
−9−−4
10 =12 + 4
−9−10=16
−19
Therefore, 3u −2v =16
−19.
13
Question 19
Question
Given two vectors a=
2
−3
1
and b=
−1
4
2
, find the angle between them in
degrees.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (2)(−1) + (−3)(4) + (1)(2) = −2−12 + 2 = −12
Step 2: Find the magnitudes of vectors aand b.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p(−1)2+ 42+ 22=√1 + 16 + 4 = √21
Step 3: Use the formula for the angle between two vectors: θ= cos−1a
·b∥a∥∥b∥.
θ= cos−1−12
√14 ·√21
Step 4: Calculate θin degrees.
θ≈cos−1−12
√294
θ≈cos−1−12
17.1464
θ≈cos−1(−0.7001)
θ≈135.52◦
Therefore, the angle between vectors aand bis approximately 135.52◦.
Question 20
Question
Given vectors a=3
−1and b=−2
4, find a vector csuch that c·a= 2c·b.
14
Solution
To find the vector c, we use the property of the dot product that states x·y=
x1y1+x2y2for vectors x=x1
x2and y=y1
y2.
Step 1: Compute c·aand 2c·b.
c·a=c1
c2·3
−1= 3c1−c2
2c·b= 2 c1
c2·−2
4= 2(−2c1+ 4c2) = −4c1+ 8c2
Step 2: Set up the equation given by the problem and solve for c. The
equation given is c·a= 2c·b, so we have:
3c1−c2=−4c1+ 8c2
Simplifying, we get:
7c1= 9c2
Step 3: Find a solution for c. To find a solution, we can let c1= 9 and
c2= 7. Therefore, the vector cis 9
7.
Thus, a vector cthat satisfies the given condition is 9
7.
Question 21
Question
Let a= 2i−3j+kand b=i+ 4j−2k. Determine ∥a−b∥.
Solution
Step 1: Compute a−b:
a−b= (2i−3j+k)−(i+ 4j−2k)
= 2i−3j+k−i−4j+ 2k
=i−7j+ 3k
Step 2: Compute the norm of a−b:
∥a−b∥=p(1)2+ (−7)2+ (3)2
=√1 + 49 + 9
=√59
Therefore, the magnitude of a−bis √59.
15
Question 22
Question
Let a = 2ˆ
i−3ˆ
j+ 4ˆ
kand
b=ˆ
i+ 5ˆ
j−2ˆ
k. Find the angle between vectors a and
b.
Solution
To find the angle between the vectors a and
b, we can use the dot product
formula:
a ·
b=|a|·|
b| · cos θ
where θis the angle between the vectors.
Step 1: Calculate the dot product of a and
b:
a ·
b= (2)(1) + (−3)(5) + (4)(−2)
a ·
b= 2 −15 −8
a ·
b=−21
Step 2: Calculate the magnitudes of a and
b:
|a|=p22+ (−3)2+ 42
|a|=√4 + 9 + 16
|a|=√29
|
b|=p12+ 52+ (−2)2
|
b|=√1 + 25 + 4
|
b|=√30
Step 3: Substitute into the dot product formula to find the angle:
−21 = √29 ·√30 ·cos θ
−21 = √870 ·cos θ
cos θ=−21
√870
θ= arccos −21
√870
θ≈2.04 radians
Therefore, the angle between vectors a and
bis approximately 2.04 radians.
16
Question 23
Question
Let a= 3i−2j+ 4kand b= 2i+ 5j−3k. Determine the scalar projection of
aonto b.
Solution
Step 1: To find the scalar projection of aonto b, we use the formula:
Scalar projection of aonto b=a·b
∥b∥
where a·bis the dot product of aand b, and ∥b∥is the magnitude of b.
Step 2: First, let’s find the dot product a·b:
a·b= (3)(2) + (−2)(5) + (4)(−3)
a·b= 6 −10 −12 = −16
Step 3: Next, calculate the magnitude of b:
∥b∥=p(2)2+ (5)2+ (−3)2
∥b∥=√4 + 25 + 9 = √38
Step 4: Now, substitute the dot product and the magnitude into the formula
to find the scalar projection:
Scalar projection of aonto b=−16
√38 =−16√38
38 =−8√38
19
Therefore, the scalar projection of aonto bis −8√38
19 .
Question 24
Question
Let a = 3
i−2
j+ 4
kand
b= 2
i+
j−2
k. Find the angle between vectors a and
b.
Solution
Step 1: To find the angle between two vectors a and
b, we can use the formula:
cos(θ) = a ·
b
∥a∥∥
b∥
17
where θis the angle between the two vectors, a ·
bis the dot product of a and
b, and ∥a∥and ∥
b∥are the magnitudes of a and
b, respectively.
Step 2: First, calculate the dot product of a and
b:
a ·
b= (3
i−2
j+ 4
k)·(2
i+
j−2
k)
= 3(2) + (−2)(1) + 4(−2) = 6 −2−8 = −4
Step 3: Next, calculate the magnitudes of a and
b:
∥a∥=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
∥
b∥=p22+ 12+ (−2)2=√4 + 1 + 4 = √9
Step 4: Substitute the dot product and magnitudes into the formula to find
cos(θ):
cos(θ) = −4
√29 ·√9=−4
√261
Step 5: Finally, to find the angle θ, take the arccosine of −4
√261 :
θ= arccos −4
√261
Question 25
Question
Let a=2
−1and b=3
4. Find the angle between vectors aand b.
Solution
To find the angle between two vectors aand b, we can use the formula:
cos(θ) = a·b
∥a∥∥b∥
where θis the angle between the vectors, a·bis the dot product of aand b,
and ∥a∥and ∥b∥are the magnitudes of vectors aand b, respectively.
Step 1: Calculate the dot product of aand b.
a·b= (2)(3) + (−1)(4) = 6 −4=2
Step 2: Calculate the magnitudes of vectors aand b.
∥a∥=p22+ (−1)2=√4 + 1 = √5
∥b∥=p32+ 42=√9 + 16 = √25 = 5
18
Step 3: Substitute the dot product and magnitudes into the formula for
the cosine of the angle between vectors.
cos(θ) = 2
√5·5=2
5√5
Step 4: Solve for the angle θ.
θ= arccos 2
5√5≈71.57◦
Therefore, the angle between vectors aand bis approximately 71.57◦.
Question 26
Question
Given two vectors u=⟨3,−2,5⟩and v=⟨−1,4,7⟩, find:
a) ∥u+v∥(magnitude of the sum)
b) ∥u−v∥(magnitude of the difference)
Solution
a) To find the magnitude of the sum ∥u+v∥: Step 1: Calculate the sum of the
two vectors:
u+v=⟨3,−2,5⟩+⟨−1,4,7⟩=⟨3+(−1),−2+4,5+7⟩=⟨2,2,12⟩
Step 2: Find the magnitude of the sum:
∥u+v∥=p22+ 22+ 122=√4 + 4 + 144 = √152
b) To find the magnitude of the difference ∥u−v∥: Step 1: Calculate the
difference of the two vectors:
u−v=⟨3,−2,5⟩ − ⟨−1,4,7⟩=⟨3−(−1),−2−4,5−7⟩=⟨4,−6,−2⟩
Step 2: Find the magnitude of the difference:
∥u−v∥=p42+ (−6)2+ (−2)2=√16 + 36 + 4 = √56
Question 27
Question
Let a=3
−2and b=−1
4. Determine the angle between vectors aand b,
rounded to the nearest degree.
19
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= 3(−1) + (−2)(4) = −3−8 = −11
Step 2: Calculate the magnitude of vector a:
|a|=p32+ (−2)2=√9 + 4 = √13
Step 3: Calculate the magnitude of vector b:
|b|=p(−1)2+ 42=√1 + 16 = √17
Step 4: Calculate the angle θbetween vectors aand busing the formula:
cos(θ) = a·b
|a|·|b|
cos(θ) = −11
√13 ·√17
θ= arccos −11
√13 ·√17
Step 5: Use a calculator to find the angle θ. Rounding to the nearest degree,
the angle between vectors aand bis approximately 123◦.
Question 28
Question
Let u=
3
−1
2
and v=
2
4
−3
be two vectors. Find the angle between the
vectors uand v.
Solution
Step 1: Find the dot product of vectors uand v,
u·v= (3)(2) + (−1)(4) + (2)(−3) = 6 −4−6 = −4.
Step 2: Find the magnitudes of vectors uand v,
|u|=p32+ (−1)2+ 22=√9 + 1 + 4 = √14,
|v|=p22+ 42+ (−3)2=√4 + 16 + 9 = √29.
20
Step 3: Use the formula for finding the angle between two vectors,
cos θ=u·v
|u||v|.
Step 4: Substitute the values of the dot product and magnitudes,
cos θ=−4
√14 ·√29 =−4
√406.
Step 5: Solve for θby taking the arccosine of both sides,
θ= arccos −4
√406≈109.76◦.
Therefore, the angle between vectors uand vis approximately 109.76◦.
Question 29
Question
Let a=
3
−2
5
and b=
−1
4
2
. Find the vector csuch that c·(a×b) = |a×b|.
Solution
Step 1: Find the cross product of vectors aand b, denoted by a×b.
a×b=
3
−2
5
×
−1
4
2
=
−22
−13
10
Step 2: Find the magnitude of the cross product |a×b|.
|a×b|=p(−22)2+ (−13)2+ 102=√484 + 169 + 100 = √753 = 27√3
Step 3: Find the unit vector in the direction of a×b.
ˆ
u=a×b
|a×b|=1
27√3
−22
−13
10
=
−22
27√3
−13
27√3
10
27√3
Step 4: Find the vector cby multiplying the unit vector with the magnitude
of a×b.
c=|a×b| · ˆ
u= 27√3·
−22
27√3
−13
27√3
10
27√3
=
−22
−13
10
Therefore, c=
−22
−13
10
.
21
Question 30
Question
Let a=
2
−1
3
,b=
−1
3
5
, and c=
4
2
−1
. Determine if a,b, and care linearly
independent.
Solution
To determine if a,b, and care linearly independent, we need to check if the
equation
k1a+k2b+k3c=0
has only the trivial solution k1=k2=k3= 0.
Let’s set up the equation and solve for k1,k2, and k3.
Step 1: Set up the equation.
Consider the equation
k1
2
−1
3
+k2
−1
3
5
+k3
4
2
−1
=
0
0
0
Step 2: Solve the equation.
This gives the system of equations
2k1−k2+ 4k3= 0
−k1+ 3k2+ 2k3= 0
3k1+ 5k2−k3= 0
We can use Gaussian elimination to solve this system of equations. After
reducing the augmented matrix, we find that the only solution is k1=k2=
k3= 0.
Since the equation has only the trivial solution, a,b, and care linearly
independent.
Question 31
Question
Let u =⟨3,−1,2⟩and v =⟨−2,5,4⟩. Find a vector w such that w ·u = 8 and
w ·v =−3.
22
Solution
Step 1: Recall that the dot product of two vectors a =⟨a1, a2, a3⟩and
b=
⟨b1, b2, b3⟩is given by a ·
b=a1b1+a2b2+a3b3.
Step 2: Given that w ·u = 8, we have w · ⟨3,−1,2⟩= 8, which gives us
3w1−w2+ 2w3= 8.
Step 3: Similarly, from w ·v =−3, we have w · ⟨−2,5,4⟩=−3, which gives
us −2w1+ 5w2+ 4w3=−3.
Step 4: To find a vector w that satisfies both equations, we set up a system
of linear equations:
(3w1−w2+ 2w3= 8
−2w1+ 5w2+ 4w3=−3
Step 5: Solving the system of equations, we find w1=−2, w2=−1, and
w3= 3.
Therefore, the vector w =⟨−2,−1,3⟩satisfies the given conditions.
Question 32
Question
Let a= 3i−4j+ 2kand b= 2i+j−3k. Find the angle between vectors aand
b.
Solution
Step 1: Calculate the dot product of aand b.
a·b= (3)(2) + (−4)(1) + (2)(−3) = 6 −4−6 = −4
Step 2: Calculate the magnitudes of aand b.
|a|=p32+ (−4)2+ 22=√9 + 16 + 4 = √29
|b|=p22+ 12+ (−3)2=√4 + 1 + 9 = √14
Step 3: Calculate the cosine of the angle between aand busing the dot
product formula.
cos θ=a
·b|a|·|b|=−4
√29√14 =−4
√406
Step 4: Find the angle between aand busing inverse cosine.
θ= arccos −4
√406
Therefore, the angle between vectors aand bis θ.
23
Question 33
Question
Let a= 3i+ 7j−2kand b= 5i−4j+ 6kbe two vectors in R3. Determine the
vector projection of aonto b.
Solution
Step 1: Calculate the dot product of aand b.
The dot product of two vectors a=a1i+a2j+a3kand b=b1i+b2j+b3k
is given by:
a·b=a1b1+a2b2+a3b3
For our vectors aand b, we have:
a·b= (3)(5) + (7)(−4) + (−2)(6) = 15 −28 −12 = −25
Step 2: Calculate the magnitude of vector b.
The magnitude of a vector b=b1i+b2j+b3kis given by:
|b|=qb2
1+b2
2+b2
3
For our vector b, we have:
|b|=p52+ (−4)2+ 62=√25 + 16 + 36 = √77
Step 3: Calculate the vector projection of aonto b.
The vector projection of aonto bis given by:
projba=a·b
|b|2b
Substitute the calculated values to get:
projba=−25
77 ·1
77(5i−4j+ 6k)
So, the vector projection of aonto bis:
−25
77 (5i−4j+ 6k)
Question 34
Question
Let a = (−3,4,2) and
b= (1,−2,0). Find the angle between vectors a and
b.
24
Solution
Step 1: To find the angle θbetween vectors a and
b, we can use the dot product
formula:
a ·
b=|a||
b|cos θ.
Step 2: Calculate the dot product of a and
b:
a ·
b= (−3)(1) + (4)(−2) + (2)(0) = −3−8 + 0 = −11.
Step 3: Calculate the magnitudes of vectors a and
b:
|a|=p(−3)2+ 42+ 22=√9 + 16 + 4 = √29,
|
b|=p12+ (−2)2+ 02=√1 + 4 + 0 = √5.
Step 4: Substitute the dot product and magnitudes into the dot product
formula:
−11 = √29 ·√5·cos θ.
Step 5: Solve for θby isolating cos θ:
cos θ=−11
√29 ·√5=−11
√145.
Step 6: Finally, find the angle θin radians by taking the inverse cosine:
θ= cos−1−11
√145≈2.62 radians.
Therefore, the angle between vectors a and
bis approximately 2.62 radians.
Question 35
Question
Let a=
3
−2
5
,b=
−4
1
2
, and c=
1
3
−4
. Find a vector dsuch that a+ 2b−
3c=d.
Solution
Step 1: Perform scalar multiplication on vectors a,b, and c.
2b= 2
−4
1
2
=
−8
2
4
−3c=−3
1
3
−4
=
−3
−9
12
25
Step 2: Add the scaled vectors a, 2b, and −3ctogether to find d.
d=a+ 2b−3c
=
3
−2
5
+
−8
2
4
+
−3
−9
12
=
3
−2
5
+
−8
2
4
+
−3
−9
12
=
(3) + (−8) + (−3)
(−2) + 2 + (−9)
5 + 4 + 12
=
−8
−9
21
Therefore, the vector dis
−8
−9
21
.
26
Question 7
Question
Let a =
2
−3
4
and
b=
−1
5
−2
. Find the angle between vectors a and
b.
Solution
To find the angle between two vectors a and
b, we can use the formula
cos(θ) = a ·
b
∥a∥∥
b∥
where θis the angle between the two vectors, a ·
bis the dot product of a and
b, and ∥a∥,∥
b∥are the magnitudes of a and
b, respectively.
Step 1: Calculate the dot product of a and
bThe dot product a ·
bis
given by:
a ·
b= (2)(−1) + (−3)(5) + (4)(−2) = −2−15 −8 = −25
Step 2: Calculate the magnitudes of a and
bThe magnitude of a vector
v =
x
y
z
is given by:
∥v∥=px2+y2+z2
So, the magnitudes of vectors a and
bare:
∥a∥=p22+ (−3)2+ 42=√4 + 9 + 16 = √29
∥
b∥=p(−1)2+ 52+ (−2)2=√1 + 25 + 4 = √30
Step 3: Find the cosine of the angle between vectors a and
bNow,
we can substitute the values we found into the formula:
cos(θ) = a ·
b
∥a∥∥
b∥=−25
√29 ·√30 =−25
√29 ·30 =−25
√870
Step 4: Find the angle between vectors a and
bTo find the angle θ,
we take the arccosine of the value we found in step 3:
θ= arccos −25
√870≈116.56◦
Therefore, the angle between vectors a and
bis approximately 116.56◦.
5
Question 8
Question
Let a =3
−2and
b=−1
5. Find the scalar ksuch that c =ka −2
bis
orthogonal to a.
Solution
Step 1: Find the dot product of c and a.
c ·a = (ka −2
b)·a
=ka ·a −2
b·a
=k|a|2−2
b·a
=k|a|2−2|
b||a|cos θ
where θis the angle between a and
b.
Step 2: Since c is orthogonal to a, we know that c ·a = 0.
k|a|2−2|
b||a|cos θ= 0
k(32+ (−2)2)−2p(−1)2+ 52p32+ (−2)2cos θ= 0
k(9 + 4) −2√26√13 cos θ= 0
Step 3: Finally, since c is orthogonal to a, the dot product of c and a is 0.
Solve for the scalar k.
13k−26√26 cos θ= 0
k=26√26 cos θ
13
Question 9
Question
Let a=
2
−3
4
,b=
−1
2
1
, and c=
0
−2
7
. Find the scalar ksuch that
ka+b= 2c.
6
Solution
Step 1: First, we can expand the equation ka+b= 2c.
k
2
−3
4
+
−1
2
1
= 2
0
−2
7
Step 2: Simplifying further, we get
2k
−3k
4k
+
−1
2
1
=
0
−4
14
Step 3: Combining like terms, we have
2k−1
−3k+ 2
4k+ 1
=
0
−4
14
Step 4: Setting the corresponding components equal, we get the following
system of equations:
2k−1=0
−3k+ 2 = −4
4k+ 1 = 14
Step 5: Solving these equations, we find k= 1.
Therefore, the scalar ksuch that ka+b= 2cis k= 1.
Question 10
Question
Let u=
2
−1
3
and v=
−4
5
1
be two vectors in R3. Determine the vector
projection of uonto v.
Solution
Step 1: We can find the vector projection of uonto vusing the formula:
projv(u) = u·v
∥v∥2v
where: - u·vis the dot product of uand v, - ∥v∥is the magnitude of v.
7
Step 2: Calculate u·v:
u·v=
2
−1
3
·
−4
5
1
= (2)(−4) + (−1)(5) + (3)(1) = −8−5 + 3 = −10
Step 3: Calculate ∥v∥:
∥v∥=p(−4)2+ 52+ 12=√16 + 25 + 1 = √42
Step 4: Substitute the values into the formula to find the vector projection:
projv(u) = −10
42
−4
5
1
=
20
21
−25
21
−5
21
Therefore, the vector projection of uonto vis
20
21
−25
21
−5
21
.
Question 11
Question
Let a =
3
−4
5
and
b=
−2
1
3
be two vectors in R3. Find the angle θbetween
the two vectors (in radians) and the projection of a onto the direction of
b.
Solution
Step 1: To find the angle θbetween the two vectors, we will use the dot product
formula: a ·
b=||a||||
b||cos θ.
a ·
b=
3
−4
5
·
−2
1
3
= (3 · −2) + (−4·1) + (5 ·3)
=−6−4 + 15
= 5
Step 2: Calculate the magnitudes of a and
busing the formula ||v|| =
pv2
1+v2
2+v2
3.
||a|| =p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
8
||
b|| =p(−2)2+ 12+ 32=√4 + 1 + 9 = √14
Step 3: Substitute the values of the dot product and the magnitudes into
the dot product formula to solve for the angle θ.
5=5√2·√14 ·cos θ
= 5√28 ·cos θ
= 2√70 ·cos θ
cos θ=5
2√70 =5√70
140 =√70
28
θ= cos−1 √70
28 !≈0.6435 radians
Step 4: To find the projection of a onto
b, use the formula: proj
b(a) =
a·
b
||
b||2
b.
proj
b(a) = 5
14
−2
1
3
=
−5
7
5
14
15
14
Therefore, the angle between the two vectors is approximately 0.6435 radi-
ans, and the projection of a onto
bis
−5
7
5
14
15
14
.
Question 12
Question
Let u =
3
−1
2
,v =
−2
4
1
, and w =
5
0
−3
. Find (2u −3v)·w.
Solution
Step 1: Compute 2u and 3v.
2u = 2
3
−1
2
=
6
−2
4
9
3v = 3
−2
4
1
=
−6
12
3
Step 2: Calculate 2u −3v.
2u −3v =
6
−2
4
−
−6
12
3
=
12
−14
1
Step 3: Compute the dot product of (2u −3v) and w.
(2u −3v)·w =
12
−14
1
·
5
0
−3
= (12)(5) + (−14)(0) + (1)(−3) = 60 −3 = 57
Therefore, (2u −3v)·w = 57 .
Question 13
Question
Let u =
2
−1
3
and v =
−4
0
1
. Find the angle between the vectors u and v.
Solution
Step 1: Find the dot product of u and v:
u ·v = (2)(−4) + (−1)(0) + (3)(1) = −8 + 0 + 3 = −5
Step 2: Find the magnitude of u and v:
|u|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|v|=p(−4)2+ 02+ 12=√16 + 0 + 1 = √17
Step 3: Use the formula cos(θ) = u·v
|u|·|v|to find the cosine of the angle
between u and v.
cos(θ) = −5
√14 ·√17 =−5
√238
Step 4: Find the angle θusing the inverse cosine function:
θ= arccos −5
√238≈123.8◦
Therefore, the angle between the vectors u and v is approximately 123.8◦.
10
Question 14
Question
Let a= 3i−2j+ 4kand b=i+ 5j−2k. Find the angle between vectors aand
b.
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= (3)(1) + (−2)(5) + (4)(−2) = 3 −10 −8 = −15
Step 2: Calculate the magnitudes of vectors aand b:
|a|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|b|=p12+ 52+ (−2)2=√1 + 25 + 4 = √30
Step 3: Use the formula for the angle between two vectors:
cos θ=a
·b|a||b|=−15
√29 ·√30 =−15
√870 =−15
√870
Step 4: Find the angle θ:
θ= arccos −15
√870
Therefore, the angle between vectors aand bis arccos −15
√870 .
Question 15
Question
Let a=2
1and b=−1
3. Calculate the angle between vectors aand b.
Solution
To find the angle between two vectors aand b, we can use the dot product
formula:
a·b=|a|·|b| · cos(θ)
where |a|and |b|are the magnitudes of vectors aand b, respectively, and θ
is the angle between the two vectors.
Step 1: Calculate the magnitudes of vectors aand b.
|a|=p22+ 12=√5
11
|b|=p(−1)2+ 32=√10
Step 2: Calculate the dot product of vectors aand b.
a·b= 2 ·(−1) + 1 ·3 = −2 + 3 = 1
Step 3: Substitute the values into the dot product formula and solve for θ.
1 = √5·√10 ·cos(θ)
Step 4: Solve for θ.
cos(θ) = 1
√50 =1
5√2
Step 5: Finally, calculate the angle θ.
θ= cos−11
5√2≈74.74◦
Therefore, the angle between vectors aand bis approximately 74.74◦.
Question 16
Question
Let u=3
−2and v=−4
5. Determine whether the vectors uand vare
orthogonal (perpendicular) to each other.
Solution
Step 1: To determine if two vectors are orthogonal, we need to calculate their
dot product. If the dot product is equal to zero, then the vectors are orthogonal.
Step 2: The dot product of two vectors uand vis given by the formula:
u·v=3
−2·−4
5= (3)(−4) + (−2)(5)
Step 3: Calculating the dot product, we get:
u·v= (−12) + (−10) = −22
Step 4: Since the dot product u·v=−22 = 0, the vectors uand vare not
orthogonal to each other.
Question 17
Question
Let a=
2
−1
3
and b=
4
−2
1
. Determine the angle between vectors aand b.
12
Solution
To find the angle between two vectors aand b, we can use the dot product
formula:
a·b=|a|·|b| · cos θ
where θis the angle between the two vectors.
Step 1: Calculate the dot product of vectors aand b.
a·b= (2)(4) + (−1)(−2) + (3)(1) = 8 + 2 + 3 = 13
Step 2: Calculate the magnitudes of vectors aand b.
|a|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|b|=p42+ (−2)2+ 12=√16 + 4 + 1 = √21
Step 3: Substitute the values into the dot product formula. Solve for θ.
13 = (√14) ·(√21) ·cos θ
cos θ=13
√14 ·√21 =13
√294 =13
√294 ·√294
√294 =13√294
294
θ= cos−1 13√294
294 !
Therefore, the angle between vectors aand bis θ= cos−113√294
294 .
Question 18
Question
Let u =4
−3and v =−2
5. Determine 3u −2v.
Solution
Step 1: Calculate 3u.
3
−3·4
−3=3(4)
3(−3)=12
−9
Step 2: Calculate 2v.
2
5·−2
5=2(−2)
2(5) =−4
10
Step 3: Subtract 2v from 3u to find 3u −2v.
12
−9−−4
10 =12 + 4
−9−10=16
−19
Therefore, 3u −2v =16
−19.
13
Question 19
Question
Given two vectors a=
2
−3
1
and b=
−1
4
2
, find the angle between them in
degrees.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (2)(−1) + (−3)(4) + (1)(2) = −2−12 + 2 = −12
Step 2: Find the magnitudes of vectors aand b.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p(−1)2+ 42+ 22=√1 + 16 + 4 = √21
Step 3: Use the formula for the angle between two vectors: θ= cos−1a
·b∥a∥∥b∥.
θ= cos−1−12
√14 ·√21
Step 4: Calculate θin degrees.
θ≈cos−1−12
√294
θ≈cos−1−12
17.1464
θ≈cos−1(−0.7001)
θ≈135.52◦
Therefore, the angle between vectors aand bis approximately 135.52◦.
Question 20
Question
Given vectors a=3
−1and b=−2
4, find a vector csuch that c·a= 2c·b.
14
Solution
To find the vector c, we use the property of the dot product that states x·y=
x1y1+x2y2for vectors x=x1
x2and y=y1
y2.
Step 1: Compute c·aand 2c·b.
c·a=c1
c2·3
−1= 3c1−c2
2c·b= 2 c1
c2·−2
4= 2(−2c1+ 4c2) = −4c1+ 8c2
Step 2: Set up the equation given by the problem and solve for c. The
equation given is c·a= 2c·b, so we have:
3c1−c2=−4c1+ 8c2
Simplifying, we get:
7c1= 9c2
Step 3: Find a solution for c. To find a solution, we can let c1= 9 and
c2= 7. Therefore, the vector cis 9
7.
Thus, a vector cthat satisfies the given condition is 9
7.
Question 21
Question
Let a= 2i−3j+kand b=i+ 4j−2k. Determine ∥a−b∥.
Solution
Step 1: Compute a−b:
a−b= (2i−3j+k)−(i+ 4j−2k)
= 2i−3j+k−i−4j+ 2k
=i−7j+ 3k
Step 2: Compute the norm of a−b:
∥a−b∥=p(1)2+ (−7)2+ (3)2
=√1 + 49 + 9
=√59
Therefore, the magnitude of a−bis √59.
15
Question 22
Question
Let a = 2ˆ
i−3ˆ
j+ 4ˆ
kand
b=ˆ
i+ 5ˆ
j−2ˆ
k. Find the angle between vectors a and
b.
Solution
To find the angle between the vectors a and
b, we can use the dot product
formula:
a ·
b=|a|·|
b| · cos θ
where θis the angle between the vectors.
Step 1: Calculate the dot product of a and
b:
a ·
b= (2)(1) + (−3)(5) + (4)(−2)
a ·
b= 2 −15 −8
a ·
b=−21
Step 2: Calculate the magnitudes of a and
b:
|a|=p22+ (−3)2+ 42
|a|=√4 + 9 + 16
|a|=√29
|
b|=p12+ 52+ (−2)2
|
b|=√1 + 25 + 4
|
b|=√30
Step 3: Substitute into the dot product formula to find the angle:
−21 = √29 ·√30 ·cos θ
−21 = √870 ·cos θ
cos θ=−21
√870
θ= arccos −21
√870
θ≈2.04 radians
Therefore, the angle between vectors a and
bis approximately 2.04 radians.
16
Question 23
Question
Let a= 3i−2j+ 4kand b= 2i+ 5j−3k. Determine the scalar projection of
aonto b.
Solution
Step 1: To find the scalar projection of aonto b, we use the formula:
Scalar projection of aonto b=a·b
∥b∥
where a·bis the dot product of aand b, and ∥b∥is the magnitude of b.
Step 2: First, let’s find the dot product a·b:
a·b= (3)(2) + (−2)(5) + (4)(−3)
a·b= 6 −10 −12 = −16
Step 3: Next, calculate the magnitude of b:
∥b∥=p(2)2+ (5)2+ (−3)2
∥b∥=√4 + 25 + 9 = √38
Step 4: Now, substitute the dot product and the magnitude into the formula
to find the scalar projection:
Scalar projection of aonto b=−16
√38 =−16√38
38 =−8√38
19
Therefore, the scalar projection of aonto bis −8√38
19 .
Question 24
Question
Let a = 3
i−2
j+ 4
kand
b= 2
i+
j−2
k. Find the angle between vectors a and
b.
Solution
Step 1: To find the angle between two vectors a and
b, we can use the formula:
cos(θ) = a ·
b
∥a∥∥
b∥
17
where θis the angle between the two vectors, a ·
bis the dot product of a and
b, and ∥a∥and ∥
b∥are the magnitudes of a and
b, respectively.
Step 2: First, calculate the dot product of a and
b:
a ·
b= (3
i−2
j+ 4
k)·(2
i+
j−2
k)
= 3(2) + (−2)(1) + 4(−2) = 6 −2−8 = −4
Step 3: Next, calculate the magnitudes of a and
b:
∥a∥=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
∥
b∥=p22+ 12+ (−2)2=√4 + 1 + 4 = √9
Step 4: Substitute the dot product and magnitudes into the formula to find
cos(θ):
cos(θ) = −4
√29 ·√9=−4
√261
Step 5: Finally, to find the angle θ, take the arccosine of −4
√261 :
θ= arccos −4
√261
Question 25
Question
Let a=2
−1and b=3
4. Find the angle between vectors aand b.
Solution
To find the angle between two vectors aand b, we can use the formula:
cos(θ) = a·b
∥a∥∥b∥
where θis the angle between the vectors, a·bis the dot product of aand b,
and ∥a∥and ∥b∥are the magnitudes of vectors aand b, respectively.
Step 1: Calculate the dot product of aand b.
a·b= (2)(3) + (−1)(4) = 6 −4=2
Step 2: Calculate the magnitudes of vectors aand b.
∥a∥=p22+ (−1)2=√4 + 1 = √5
∥b∥=p32+ 42=√9 + 16 = √25 = 5
18
Step 3: Substitute the dot product and magnitudes into the formula for
the cosine of the angle between vectors.
cos(θ) = 2
√5·5=2
5√5
Step 4: Solve for the angle θ.
θ= arccos 2
5√5≈71.57◦
Therefore, the angle between vectors aand bis approximately 71.57◦.
Question 26
Question
Given two vectors u=⟨3,−2,5⟩and v=⟨−1,4,7⟩, find:
a) ∥u+v∥(magnitude of the sum)
b) ∥u−v∥(magnitude of the difference)
Solution
a) To find the magnitude of the sum ∥u+v∥: Step 1: Calculate the sum of the
two vectors:
u+v=⟨3,−2,5⟩+⟨−1,4,7⟩=⟨3+(−1),−2+4,5+7⟩=⟨2,2,12⟩
Step 2: Find the magnitude of the sum:
∥u+v∥=p22+ 22+ 122=√4 + 4 + 144 = √152
b) To find the magnitude of the difference ∥u−v∥: Step 1: Calculate the
difference of the two vectors:
u−v=⟨3,−2,5⟩ − ⟨−1,4,7⟩=⟨3−(−1),−2−4,5−7⟩=⟨4,−6,−2⟩
Step 2: Find the magnitude of the difference:
∥u−v∥=p42+ (−6)2+ (−2)2=√16 + 36 + 4 = √56
Question 27
Question
Let a=3
−2and b=−1
4. Determine the angle between vectors aand b,
rounded to the nearest degree.
19
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= 3(−1) + (−2)(4) = −3−8 = −11
Step 2: Calculate the magnitude of vector a:
|a|=p32+ (−2)2=√9 + 4 = √13
Step 3: Calculate the magnitude of vector b:
|b|=p(−1)2+ 42=√1 + 16 = √17
Step 4: Calculate the angle θbetween vectors aand busing the formula:
cos(θ) = a·b
|a|·|b|
cos(θ) = −11
√13 ·√17
θ= arccos −11
√13 ·√17
Step 5: Use a calculator to find the angle θ. Rounding to the nearest degree,
the angle between vectors aand bis approximately 123◦.
Question 28
Question
Let u=
3
−1
2
and v=
2
4
−3
be two vectors. Find the angle between the
vectors uand v.
Solution
Step 1: Find the dot product of vectors uand v,
u·v= (3)(2) + (−1)(4) + (2)(−3) = 6 −4−6 = −4.
Step 2: Find the magnitudes of vectors uand v,
|u|=p32+ (−1)2+ 22=√9 + 1 + 4 = √14,
|v|=p22+ 42+ (−3)2=√4 + 16 + 9 = √29.
20
Step 3: Use the formula for finding the angle between two vectors,
cos θ=u·v
|u||v|.
Step 4: Substitute the values of the dot product and magnitudes,
cos θ=−4
√14 ·√29 =−4
√406.
Step 5: Solve for θby taking the arccosine of both sides,
θ= arccos −4
√406≈109.76◦.
Therefore, the angle between vectors uand vis approximately 109.76◦.
Question 29
Question
Let a=
3
−2
5
and b=
−1
4
2
. Find the vector csuch that c·(a×b) = |a×b|.
Solution
Step 1: Find the cross product of vectors aand b, denoted by a×b.
a×b=
3
−2
5
×
−1
4
2
=
−22
−13
10
Step 2: Find the magnitude of the cross product |a×b|.
|a×b|=p(−22)2+ (−13)2+ 102=√484 + 169 + 100 = √753 = 27√3
Step 3: Find the unit vector in the direction of a×b.
ˆ
u=a×b
|a×b|=1
27√3
−22
−13
10
=
−22
27√3
−13
27√3
10
27√3
Step 4: Find the vector cby multiplying the unit vector with the magnitude
of a×b.
c=|a×b| · ˆ
u= 27√3·
−22
27√3
−13
27√3
10
27√3
=
−22
−13
10
Therefore, c=
−22
−13
10
.
21
Question 30
Question
Let a=
2
−1
3
,b=
−1
3
5
, and c=
4
2
−1
. Determine if a,b, and care linearly
independent.
Solution
To determine if a,b, and care linearly independent, we need to check if the
equation
k1a+k2b+k3c=0
has only the trivial solution k1=k2=k3= 0.
Let’s set up the equation and solve for k1,k2, and k3.
Step 1: Set up the equation.
Consider the equation
k1
2
−1
3
+k2
−1
3
5
+k3
4
2
−1
=
0
0
0
Step 2: Solve the equation.
This gives the system of equations
2k1−k2+ 4k3= 0
−k1+ 3k2+ 2k3= 0
3k1+ 5k2−k3= 0
We can use Gaussian elimination to solve this system of equations. After
reducing the augmented matrix, we find that the only solution is k1=k2=
k3= 0.
Since the equation has only the trivial solution, a,b, and care linearly
independent.
Question 31
Question
Let u =⟨3,−1,2⟩and v =⟨−2,5,4⟩. Find a vector w such that w ·u = 8 and
w ·v =−3.
22
Solution
Step 1: Recall that the dot product of two vectors a =⟨a1, a2, a3⟩and
b=
⟨b1, b2, b3⟩is given by a ·
b=a1b1+a2b2+a3b3.
Step 2: Given that w ·u = 8, we have w · ⟨3,−1,2⟩= 8, which gives us
3w1−w2+ 2w3= 8.
Step 3: Similarly, from w ·v =−3, we have w · ⟨−2,5,4⟩=−3, which gives
us −2w1+ 5w2+ 4w3=−3.
Step 4: To find a vector w that satisfies both equations, we set up a system
of linear equations:
(3w1−w2+ 2w3= 8
−2w1+ 5w2+ 4w3=−3
Step 5: Solving the system of equations, we find w1=−2, w2=−1, and
w3= 3.
Therefore, the vector w =⟨−2,−1,3⟩satisfies the given conditions.
Question 32
Question
Let a= 3i−4j+ 2kand b= 2i+j−3k. Find the angle between vectors aand
b.
Solution
Step 1: Calculate the dot product of aand b.
a·b= (3)(2) + (−4)(1) + (2)(−3) = 6 −4−6 = −4
Step 2: Calculate the magnitudes of aand b.
|a|=p32+ (−4)2+ 22=√9 + 16 + 4 = √29
|b|=p22+ 12+ (−3)2=√4 + 1 + 9 = √14
Step 3: Calculate the cosine of the angle between aand busing the dot
product formula.
cos θ=a
·b|a|·|b|=−4
√29√14 =−4
√406
Step 4: Find the angle between aand busing inverse cosine.
θ= arccos −4
√406
Therefore, the angle between vectors aand bis θ.
23
Question 33
Question
Let a= 3i+ 7j−2kand b= 5i−4j+ 6kbe two vectors in R3. Determine the
vector projection of aonto b.
Solution
Step 1: Calculate the dot product of aand b.
The dot product of two vectors a=a1i+a2j+a3kand b=b1i+b2j+b3k
is given by:
a·b=a1b1+a2b2+a3b3
For our vectors aand b, we have:
a·b= (3)(5) + (7)(−4) + (−2)(6) = 15 −28 −12 = −25
Step 2: Calculate the magnitude of vector b.
The magnitude of a vector b=b1i+b2j+b3kis given by:
|b|=qb2
1+b2
2+b2
3
For our vector b, we have:
|b|=p52+ (−4)2+ 62=√25 + 16 + 36 = √77
Step 3: Calculate the vector projection of aonto b.
The vector projection of aonto bis given by:
projba=a·b
|b|2b
Substitute the calculated values to get:
projba=−25
77 ·1
77(5i−4j+ 6k)
So, the vector projection of aonto bis:
−25
77 (5i−4j+ 6k)
Question 34
Question
Let a = (−3,4,2) and
b= (1,−2,0). Find the angle between vectors a and
b.
24
Solution
Step 1: To find the angle θbetween vectors a and
b, we can use the dot product
formula:
a ·
b=|a||
b|cos θ.
Step 2: Calculate the dot product of a and
b:
a ·
b= (−3)(1) + (4)(−2) + (2)(0) = −3−8 + 0 = −11.
Step 3: Calculate the magnitudes of vectors a and
b:
|a|=p(−3)2+ 42+ 22=√9 + 16 + 4 = √29,
|
b|=p12+ (−2)2+ 02=√1 + 4 + 0 = √5.
Step 4: Substitute the dot product and magnitudes into the dot product
formula:
−11 = √29 ·√5·cos θ.
Step 5: Solve for θby isolating cos θ:
cos θ=−11
√29 ·√5=−11
√145.
Step 6: Finally, find the angle θin radians by taking the inverse cosine:
θ= cos−1−11
√145≈2.62 radians.
Therefore, the angle between vectors a and
bis approximately 2.62 radians.
Question 35
Question
Let a=
3
−2
5
,b=
−4
1
2
, and c=
1
3
−4
. Find a vector dsuch that a+ 2b−
3c=d.
Solution
Step 1: Perform scalar multiplication on vectors a,b, and c.
2b= 2
−4
1
2
=
−8
2
4
−3c=−3
1
3
−4
=
−3
−9
12
25
Step 2: Add the scaled vectors a, 2b, and −3ctogether to find d.
d=a+ 2b−3c
=
3
−2
5
+
−8
2
4
+
−3
−9
12
=
3
−2
5
+
−8
2
4
+
−3
−9
12
=
(3) + (−8) + (−3)
(−2) + 2 + (−9)
5 + 4 + 12
=
−8
−9
21
Therefore, the vector dis
−8
−9
21
.
26
