MATH 112 - TECHNICAL
MATHEMATICS - Vector operations
Question Bank - Set 5
Liberty University
Question 1
Question
Let a=
3
−1
2
and b=
−2
4
0
. Find the unit vector in the direction of a−3b.
Solution
Step 1: Find a−3b.
a−3b=
3
−1
2
−3
−2
4
0
=
3
−1
2
−
−6
12
0
=
9
−13
2
Step 2: Find the magnitude of a−3b.
|a−3b|=p92+ (−13)2+ 22=√246
Step 3: Find the unit vector in the direction of a−3b.
ˆu=a−3b
|a−3b|=1
√246
9
−13
2
=
9
√246
−13
√246
2
√246
Therefore, the unit vector in the direction of a−3bis
9
√246
−13
√246
2
√246
.
Question 2
Question
Let v=
3
−4
5
and w=
2
1
−3
be two vectors in R3. Determine the angle
between the vectors vand w.
Solution
To find the angle between two vectors vand win R3, we can use the dot product
formula:
v·w=|v||w|cos θ
where |v|and |w|are the magnitudes of vectors vand wrespectively, and θis
the angle between the two vectors.
Step 1: Calculate the dot product of vand w:
v·w= (3)(2) + (−4)(1) + (5)(−3) = 6 −4−15 = −13
Step 2: Calculate the magnitudes of vectors vand w:
|v|=p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
|w|=p22+ 12+ (−3)2=√4 + 1 + 9 = √14
Step 3: Substitute the dot product and magnitudes into the formula to find
the cosine of the angle θ:
−13 = (5√2)(√14) cos θ
cos θ=−13
5√28 =−13
10√2=−13√2
20
Step 4: Finally, find the angle θusing the inverse cosine function:
θ= cos−1 −13√2
20 !≈130.32◦
Therefore, the angle between the vectors vand wis approximately 130.32◦.
Question 3
Question
Let u=
2
−1
3
and v=
−4
0
1
. Find ∥u−2v∥.
2
Solution
Step 1: Find 2v.
2v= 2
−4
0
1
=
−8
0
2
Step 2: Find u−2v.
u−2v=
2
−1
3
−
−8
0
2
=
10
−1
1
Step 3: Find the norm of ∥u−2v∥.
∥u−2v∥=p102+ (−1)2+ 12=√100 + 1 + 1 = √102
Therefore, ∥u−2v∥=√102.
Question 4
Question
Let a= 3i−2j−kand b= 2i+ 4j+ 5kbe two vectors. Find the scalar
projection of aonto b.
Solution
Step 1: Find the dot product of aand b.
a·b= (3)(2) + (−2)(4) + (−1)(5) = 6 −8−5 = −7
Step 2: Find the magnitude of b.
|b|=p(2)2+ (4)2+ (5)2=√4 + 16 + 25 = √45
Step 3: Find the scalar projection of aonto busing the formula:
projba=a·b
|b|=−7
√45 =−7√45
45
Therefore, the scalar projection of aonto bis −7√45
45 .
Question 5
Question
Let u =
3
−1
2
and v =
1
4
−2
. Find the angle between vectors u and v to the
nearest degree.
3
Solution
Step 1: The dot product of two vectors u and v is given by:
u ·v =∥u∥∥v∥cos θ
where ∥u∥and ∥v∥are the magnitudes of vectors u and v respectively, and θis
the angle between the two vectors.
Step 2: Calculate the dot product of u and v:
u ·v = (3)(1) + (−1)(4) + (2)(−2) = 3 −4−4 = −5
Step 3: Calculate the magnitudes of vectors u and v:
∥u∥=p32+ (−1)2+ 22=√14
∥v∥=p12+ 42+ (−2)2=√21
Step 4: Substitute the dot product and magnitudes into the formula provided
in Step 1 to solve for the cosine of the angle θ:
−5 = √14 ·√21 ·cos θ
Step 5: Solve for cos θ:
cos θ=−5
√14 ·√21 =−5
√294 =−5√294
294
Step 6: Find the angle between the vectors in degrees using the inverse cosine
function:
θ= arccos −5√294
294 !≈129◦
Therefore, the angle between vectors u and v is approximately 129 degrees.
Question 6
Question
Let a= 2i−3j+kand b=i+ 4j−2k. Determine 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∥
4
where θis the angle between aand b,·denotes the dot product, and ∥ · ∥
denotes the magnitude.
Step 1: Find the dot product a ·b.
a·b= (2i−3j+k)·(i+ 4j−2k)
a·b= 2 ·1+(−3) ·4+1·(−2)
a·b= 2 −12 −2
a·b=−12
Step 2: Find the magnitudes ∥a∥and ∥b∥.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p12+ 42+ (−2)2=√1 + 16 + 4 = √21
Step 3: Find the angle θ.
cos θ=−12
√14 ·√21 =−12
√294
θ= cos−1−12
√294≈104.04◦
Therefore, the angle between vectors aand bis approximately 104.04◦.
Question 7
Question
Let a = 3ˆ
i+ 2ˆ
j+ˆ
kand
b= 2ˆ
i−ˆ
j+ 4ˆ
kbe two vectors. Find the projection of
a onto
b.
Solution
Step 1: The projection of a onto
bcan be calculated using the formula:
proj
b(a) = a ·
b
|
b|2·
b
Step 2: First, we calculate the dot product of a and
b:
a ·
b= (3ˆ
i+ 2ˆ
j+ˆ
k)·(2ˆ
i−ˆ
j+ 4ˆ
k)
= 3(2) + 2(−1) + 1(4) = 6 −2 + 4 = 8
Step 3: Next, we calculate the magnitude of
b:
|
b|=p22+ (−1)2+ 42=√4 + 1 + 16 = √21
5
Step 4: Now, substitute the values into the projection formula:
proj
b(a) = 8
|
b|2·(2ˆ
i−ˆ
j+ 4ˆ
k)
=8
21 ·(2ˆ
i−ˆ
j+ 4ˆ
k) = 16
21ˆ
i−8
21ˆ
j+32
21ˆ
k
Therefore, the projection of a onto
bis 16
21 ˆ
i−8
21 ˆ
j+32
21 ˆ
k.
Question 8
Question
Let u=3
−2and v=−1
4. Find the unit vector in the direction of u+ 2v.
Solution
Step 1: Find u+ 2v.
u+ 2v=3
−2+ 2 −1
4=3
−2+−2
8=1
6
Step 2: Find the magnitude of u+ 2v.
∥u+ 2v∥=p12+ 62=√1 + 36 = √37
Step 3: Find the unit vector in the direction of u+ 2v.
ˆ
u+ 2v=u+ 2v
∥u+ 2v∥=1
√37 1
6="1
√37
6
√37 #
Question 9
Question
Let a = 4ˆ
i−3ˆ
j+ 2ˆ
kand
b= 2ˆ
i+ 5ˆ
j−ˆ
k. Find the angle between vectors a and
b.
Solution
Step 1: Calculate the dot product of vectors a and
b
a ·
b= (4)(2) + (−3)(5) + (2)(−1) = 8 −15 −2 = −9
Step 2: Calculate the magnitudes of vectors a and
b
|a|=p42+ (−3)2+ 22=√16 + 9 + 4 = √29
6
|
b|=p22+ 52+ (−1)2=√4 + 25 + 1 = √30
Step 3: Using the dot product formula, find the angle θbetween vectors a
and
b.
cos θ=a ·
b
|a||
b|=−9
√29√30 =−9
√870
Step 4: Solve for θby taking the arccosine of −9
√870 .
θ= arccos −9
√870
Question 10
Question
Let a=3
−2and b=1
5. Determine the magnitude of the vector a−2b.
Solution
Step 1: Find 2b.
Given b=1
5,we have 2b= 2 1
5=2
10.
Step 2: Calculate a−2b.
a−2b=3
−2−2
10=3−2
−2−10=1
−12.
Step 3: Find the magnitude of a−2b. The magnitude of a vector v=x
y
is given by ∥v∥=px2+y2. So, for a−2b=1
−12:
∥a−2b∥=p12+ (−12)2=√1 + 144 = √145.
Therefore, the magnitude of the vector a−2bis √145 .
Question 11
Question
Let v =−3
4and w =1
−2. Find the angle between vectors v and w to the
nearest degree.
7
Solution
Step 1: Calculate the dot product of vectors v and w:
v ·w = (−3)(1) + (4)(−2) = −3−8 = −11
Step 2: Calculate the magnitudes of vectors v and w:
|v|=p(−3)2+ 42=√9 + 16 = √25 = 5
|w|=p12+ (−2)2=√1 + 4 = √5
Step 3: Use the formula cos θ=v·w
|v||w|to find the cosine of the angle between
the vectors:
cos θ=−11
5·√5=−11
5√5
Step 4: Find the angle θby taking the arccosine of −11
5√5:
θ= arccos −11
5√5≈131◦
Therefore, the angle between vectors v and w is approximately 131◦.
Question 12
Question
Let u =
3
−1
4
and v =
−2
5
1
. Determine the projection of v onto u.
Solution
Step 1: Calculate the dot product of u and v.
u ·v = (3 · −2) + (−1·5) + (4 ·1)
u ·v =−6−5+4
u ·v =−7
Step 2: Calculate the magnitude of u.
|u|=p32+ (−1)2+ 42
|u|=√9 + 1 + 16
|u|=√26
8
Step 3: Calculate the projection of v onto u using the formula:
projuv =u ·v
|u|2u
projuv =−7
26
3
−1
4
projuv =
−21
26
7
26
−28
26
Therefore, the projection of v onto u is
−21
26
7
26
−28
26
.
Question 13
Question
A force of 120 N is acting along a direction of 30 degrees east of north. Resolve
this force into two components, one along the north direction and the other
along the east direction.
Solution
Step 1: Draw a diagram to represent the force vector.
Step 2: Resolve the force into two components: Let Fbe the magnitude
of the force and θbe the angle it makes with the north direction. Then, the
force components along the north and east directions can be represented as:
Fn=F·cos(θ) and Fe=F·sin(θ)
Step 3: Calculate the components. Given F= 120 N and θ= 30◦, we have:
Fn= 120 ·cos(30◦)Fe= 120 ·sin(30◦)
Step 4: Solve for the components. Fn= 120 ·cos(30◦) = 120 ·√3
2= 60√3≈
103.92 N Fe= 120 ·sin(30◦) = 120 ·1
2= 60 N
Step 5: Write the final answer. Therefore, the force component along the
north direction is approximately 103.92 N and along the east direction is 60 N.
Question 14
Question
Let a= 3i−2j+kand b= 2i+ 5j−4k. Find the angle between vectors aand
b.
9
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (3i−2j+k)·(2i+ 5j−4k)
= 3(2) + (−2)(5) + 1(−4)
= 6 −10 −4
=−8
Step 2: Calculate the magnitudes of vectors aand b.
|a|=p32+ (−2)2+ 12=√9 + 4 + 1 = √14
|b|=p22+ 52+ (−4)2=√4 + 25 + 16 = √45
Step 3: Use the formula for the angle between two vectors θ= cos−1a·b
|a||b|.
θ= cos−1−8
√14√45
θ= cos−1−8
√630
θ≈cos−1(−0.5131)
θ≈120.47◦
Therefore, the angle between vectors aand bis approximately 120.47◦.
Question 15
Question
Let u =
3
−1
4
and v =
2
5
−2
. Determine a vector w such that w ·u = 2v ·u.
Solution
Step 1: Find w ·u.
w ·u = 2v ·u
w ·u = 2(v ·u)
w ·u = 2(3(2) + (−1)(5) + 4(−2))
w ·u = 2(6 −5−8)
w ·u = 2(−7)
10
w ·u =−14
Step 2: Solve for w. Let w =
x
y
z
. Setting up the dot product equation:
x
y
z
·
3
−1
4
=−14
3x−y+ 4z=−14
3x−y+ 4z=−14
To satisfy the above equation, we have infinitely many solutions. One such
solution is w =
0
−14
0
.
Therefore, w =
0
−14
0
is a vector such that w ·u = 2v ·u.
Question 16
Question
Let u=
3
−2
4
and v=
1
5
2
.
Find u·v(dot product) and ∥u×v∥(magnitude of cross product).
Solution
Step 1: Calculate the dot product u·vusing the formula u·v=u1v1+u2v2+
u3v3.
u·v= (3)(1) + (−2)(5) + (4)(2) = 3 −10 + 8 = 1
Step 2: Calculate the cross product u×vusing the formula
u×v=
u2v3−u3v2
u3v1−u1v3
u1v2−u2v1
u×v=
(−2)(2) −(4)(5)
(4)(1) −(3)(2)
(3)(5) −(−2)(1)
=
−4−20
4−6
15 + 2
=
−24
−2
17
Step 3: Calculate the magnitude of the cross product ∥u×v∥using the
formula ∥u×v∥=p(u2v3−u3v2)2+ (u3v1−u1v3)2+ (u1v2−u2v1)2.
∥u×v∥=p(−24)2+ (−2)2+ 172=√576 + 4 + 289 = √869
11
Question 17
Question
Let a=⟨2,−3,4⟩and b=⟨−1,0,2⟩. Find ∥a+ 2b∥.
Solution
Step 1: Calculate a+ 2bby multiplying bby 2 and adding the result to a.
a+2b=⟨2,−3,4⟩+2⟨−1,0,2⟩=⟨2,−3,4⟩+⟨−2,0,4⟩=⟨2−2,−3+0,4+4⟩=⟨0,−3,8⟩
Step 2: Calculate the magnitude of a+ 2busing the formula for a vector in
3D space ∥⟨x, y, z⟩∥ =px2+y2+z2.
∥a+ 2b∥=p02+ (−3)2+ 82=√0 + 9 + 64 = √73
Therefore, ∥a+ 2b∥=√73.
Question 18
Question
Consider the vectors u =
3
−1
2
,v =
−2
4
−1
, and w =
1
−3
5
. Given that
a = 2u −v and
b=v ·w, determine the value of a ·
b.
Solution
Step 1: Find vector a = 2u −v.
a = 2u −v = 2
3
−1
2
−
−2
4
−1
=
6
−2
4
−
−2
4
−1
=
8
−6
5
Step 2: Find the dot product of
b=v ·w.
b=v ·w =
−2
4
−1
·
1
−3
5
= (−2)(1) + (4)(−3) + (−1)(5) = −2−12 −5 = −19
Step 3: Find the dot product of a ·
b.
a ·
b=
8
−6
5
·−19= (8)(−19) + (−6)(0) + (5)(0) = −152
Therefore, the value of a ·
bis −152 .
12
Question 19
Question
Let a=
2
−3
4
,b=
1
5
−2
, and c=
−3
0
6
. Determine if the vectors a,b, and c
are linearly independent or linearly dependent. If they are linearly dependent,
express one of the vectors as a linear combination of the other two.
Solution
Step 1: To determine if the vectors a,b, and care linearly independent or
dependent, we set up the equation k1a+k2b+k3c=0, where k1,k2, and k3
are scalars.
Step 2: Substituting the given vectors into the equation, we have: k1
2
−3
4
+
k2
1
5
−2
+k3
−3
0
6
=
0
0
0
.
Step 3: This gives us the system of equations: 2k1+k2−3k3= 0, −3k1+
5k2= 0, 4k1−2k2+ 6k3= 0.
Step 4: To determine if the vectors are linearly independent, we solve this
system of equations. One possible way to proceed is by row reducing the aug-
mented matrix [A — 0]:
2 1 −3|0
−3 5 0 |0
4−2 6 |0
Step 5: After performing row operations, the matrix is row reduced to:
102|0
011|0
000|0
Step 6: The system has an infinite number of solutions, indicating that the
vectors a,b, and care linearly dependent.
Step 7: To express one of the vectors as a linear combination of the other
two, we can express cas a linear combination of aand b. From the row-reduced
matrix, we have k1=−2k3and k2=−k3. Thus, we can write c= 2a+b.
Question 20
Question
Let a = 3ˆ
i−2ˆ
j+ 4ˆ
kand
b=ˆ
i+ 3ˆ
j−2ˆ
k. Find the angle between a and
bin
degrees.
13
Solution
Step 1: Calculate the dot product of a and
b.
a ·
b= (3)(1) + (−2)(3) + (4)(−2) = 3 −6−8 = −11
Step 2: Calculate the magnitudes of a and
b.
|a|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|
b|=p12+ 32+ (−2)2=√1 + 9 + 4 = √14
Step 3: Calculate the angle between a and
busing the formula cos(θ) = a·
b
|a||
b|.
cos(θ) = −11
√29√14
Step 4: Find the angle between a and
b.
θ= cos−1−11
√29√14≈128.6◦
Therefore, the angle between a and
bis approximately 128.6◦.
Question 21
Question
Given two vectors u = 2ˆ
i−3ˆ
j+ 4ˆ
kand v =−ˆ
i+ 2ˆ
j−5ˆ
k, find the angle between
the two vectors.
Solution
To find the angle between two vectors, we can use the dot product formula:
u ·v =|u||v|cos(θ), where θis the angle between the vectors.
Step 1: Calculate the dot product
u ·v = (2)(−1) + (−3)(2) + (4)(−5)
=−2−6−20
=−28
Step 2: Calculate the magnitudes of the vectors
|u|=p22+ (−3)2+ 42
=√4 + 9 + 16
=√29
|v|=p(−1)2+ 22+ (−5)2
=√1 + 4 + 25
=√30
14
Step 3: Find the angle between the vectors
cos(θ) = u ·v
|u||v|
cos(θ) = −28
√29 ·√30
θ= cos−1−28
√29 ·√30
Therefore, the angle between the vectors u and v is θ≈109.46◦.
Question 22
Question
Let a= 3i−2j+ 4kand b=i+ 5j−2k. Determine the angle between vectors
aand b.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (3)(1) + (−2)(5) + (4)(−2) = 3 −10 −8 = −15
Step 2: Find the magnitudes of vectors aand b.
|a|=p(3)2+ (−2)2+ (4)2=√9 + 4 + 16 = √29
|b|=p(1)2+ (5)2+ (−2)2=√1 + 25 + 4 = √30
Step 3: Use the dot product formula to find the angle θbetween vectors a
and b:
cos θ=a·b
|a|·|b|=−15
√29 ·√30
θ= arccos −15
√29 ·√30
θ≈124.8◦
Therefore, the angle between vectors aand bis approximately 124.8◦.
Question 23
Question
Let u = 3
i−2
j+ 4
kand v =−2
i+ 5
j−
k. Find the angle between vectors u
and v.
15
Solution
Step 1: Calculate the dot product of vectors u and v.
u ·v = (3) ·(−2) + (−2) ·(5) + (4) ·(−1)
=−6−10 −4
=−20
Step 2: Find the magnitudes of vectors u and v.
|u|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|v|=p(−2)2+ 52+ (−1)2=√4 + 25 + 1 = √30
Step 3: Calculate the angle between the vectors using the formula cos θ=
u·v
|u||v|.
cos θ=−20
√29 ·√30
Step 4: Solve for the angle θusing the inverse cosine function.
θ= cos−1−20
√29 ·√30
θ≈98.64◦
Therefore, the angle between vectors u and v is approximately 98.64◦.
Question 24
Question
Let u = 2
i−3
j+ 4
kand v =−
i+ 2
j−5
k. Find the angle between vectors u
and v.
Solution
Step 1: Find the dot product of vectors u and v.
u ·v = (2)(−1) + (−3)(2) + (4)(−5) = −2−6−20 = −28
Step 2: Find the magnitudes of vectors u and v.
|u|=p22+ (−3)2+ 42=√4 + 9 + 16 = √29
|v|=p(−1)2+ 22+ (−5)2=√1 + 4 + 25 = √30
Step 3: Use the dot product formula to find the angle between vectors u and
v.
cos(θ) = u ·v
|u|·|v|=−28
√29 ·√30 =−28
√870
16
θ=cos−1−28
√870≈131.86◦
Therefore, the angle between vectors u and v is approximately 131.86◦.
Question 25
Question
Let A=
3
−1
2
,B=
−4
0
1
, and C=
1
5
−3
be three vectors. Determine the
vector projection of Aonto the plane determined by Band C.
Solution
Step 1: Calculate the normal vector of the plane determined by Band C. The
normal vector Nof the plane determined by Band Ccan be found by taking
the cross product of Band C.
N=B×C=
i j k
−4 0 1
1 5 −3
N=
−5
7
20
Step 2: Calculate the projection of Aonto N. The projection of Aonto N
can be found using the inner product formula:
projN(A) = A·N
∥N∥2N
Calculate the dot product A·N:
A·N=
3
−1
2
·
−5
7
20
= 3(−5) + (−1)(7) + 2(20) = −5−7 + 40 = 28
Calculate ∥N∥2:
∥N∥2= (−5)2+ 72+ 202= 25 + 49 + 400 = 474
Calculate projN(A):
projN(A) = 28
474
−5
7
20
=
−140
237
196
237
560
237
17
Therefore, the vector projection of Aonto the plane determined by Band
Cis
−140
237
196
237
560
237
.
Question 26
Question
Let a=
2
−1
3
and b=
4
0
−2
. Find the angle between vectors aand b.
Solution
To find the angle between two vectors, 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 a·b.
a·b= (2)(4) + (−1)(0) + (3)(−2) = 8 −6=2
Step 2: Calculate the magnitudes |a|and |b|.
|a|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|b|=p42+ 02+ (−2)2=√16 + 0 + 4 = √20
Step 3: Substitute the dot product and magnitudes into the formula to
solve for θ.
2 = √14√20 cos θ
cos θ=2
√14√20 =2
√280 =1
√70
Step 4: Find the angle between the vectors.
θ= cos−11
√70≈cos−1(0.1348) ≈81.59◦
Therefore, the angle between vectors aand bis approximately 81.59◦.
Question 27
Question
Let a=
2
−1
3
and b=
4
2
−1
. Find ∥3a−2b∥.
18
Solution
Step 1: Find 3aand 2b.
3a= 3
2
−1
3
=
6
−3
9
2b= 2
4
2
−1
=
8
4
−2
Step 2: Find 3a−2b.
3a−2b=
6
−3
9
−
8
4
−2
=
−2
−7
11
Step 3: Find ∥3a−2b∥.
∥3a−2b∥=
−2
−7
11
=p(−2)2+ (−7)2+ 112
=√4 + 49 + 121
=√174
Therefore, ∥3a−2b∥=√174.
Question 28
Question
Let a= 3i+ 2j−4kand b=−i+ 5j+ 2kbe two vectors. Find the angle
between the two vectors in degrees.
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= (3)(−1) + (2)(5) + (−4)(2) = −3 + 10 −8 = −1
Step 2: Calculate the magnitudes of vectors aand b:
|a|=p32+ 22+ (−4)2=√9 + 4 + 16 = √29
19
|b|=p(−1)2+ 52+ 22=√1 + 25 + 4 = √30
Step 3: Calculate the angle θbetween the two vectors using the formula:
cos(θ) = a·b
|a|·|b|
cos(θ) = −1
√29 ·√30 =−1
√870
Step 4: Find the angle θby taking the arccosine of the calculated value:
θ= arccos −1
√870≈96.25◦
Therefore, the angle between the vectors aand bis approximately 96.25◦.
Question 29
Question
Let u =
2
−1
4
,v =
−3
5
1
, and w =
1
−2
0
. Find the vector x that satisfies the
equation 4u −2v +x = 3 w.
Solution
Step 1: Compute 4u −2v.
4u −2v = 4
2
−1
4
−2
−3
5
1
=
8
−4
16
−
−6
10
2
=
14
−14
14
Step 2: Rewrite the equation 4u −2v +x = 3 w as x = 3 w −(4u −2v).
x = 3
1
−2
0
−
14
−14
14
=
3
−6
0
−
14
−14
14
=
−11
8
−14
Therefore, the vector x that satisfies the equation 4u−2v+x = 3 w is
−11
8
−14
.
Question 30
Question
Let u=
2
−1
3
and v=
−3
4
5
. Find a vector wwhich is orthogonal to both u
and v.
20
Solution
Step 1: To find a vector wwhich is orthogonal to both uand v, we take the
cross product of uand v.
Step 2: The cross product u×vis given by:
u×v=
i j k
2−1 3
−3 4 5
Step 3: Evaluating the determinant gives us:
u×v=
−1(5) −3(4)
−(2(5) −3(−3))
2(4) −(−1)(−3)
=
−17
11
5
Therefore, the vector w=
−17
11
5
is orthogonal to both uand v.
Question 31
Question
Let a= 3i−2j+ 2kand b=−i+ 4j−3k. Determine the angle between vectors
aand b.
Solution
Given vectors aand b, we can find the angle θbetween them using the formula:
cos(θ) = a·b
∥a∥∥b∥
Step 1: Find the dot product of a and b The dot product of two vectors
aand bis given by:
a·b= (3)(−1) + (−2)(4) + (2)(−3)
Calculating the dot product: a·b=−3−8−6 = −17
Step 2: Find the magnitudes of a and b The magnitudes of vectors a
and bare given by:
∥a∥=p32+ (−2)2+ 22=√9 + 4 + 4 = √17
∥b∥=p(−1)2+ 42+ (−3)2=√1 + 16 + 9 = √26
21
Step 3: Calculate the cosine of the angle between a and b Now,
substitute the dot product and magnitudes into the formula for the cosine of
the angle between two vectors:
cos(θ) = −17
√17√26
Step 4: Find the angle Finally, solve for θby taking the arccosine of the
calculated value:
θ= arccos −17
√17√26
Therefore, the angle between vectors aand bis θ≈107.25◦.
Question 32
Question
Let a=
2
−3
1
and b=
−4
2
5
. Find the angle between vectors aand b.
Solution
Step 1: Calculate the dot product of aand b.
a·b= (2)(−4) + (−3)(2) + (1)(5) = −8−6 + 5 = −9
Step 2: Calculate the magnitudes of vectors aand b.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p(−4)2+ 22+ 52=√16 + 4 + 25 = √45
Step 3: Use the dot product to find the angle between aand b.
cos θ=a
·b∥a∥∥b∥=−9
√14√45
θ= cos−1−9
√14√45≈108.18◦
Therefore, the angle between vectors aand bis approximately 108.18◦.
Question 33
Question
Let u=
2
−1
3
and v=
4
2
−1
. Determine the projection of uonto v.
22
Solution
To find the projection of uonto v, we use the formula:
projv(u) = u·v
||v||2·v
Step 1: Calculate the dot product of uand v:
u·v=
2
−1
3
·
4
2
−1
= (2 ·4) + (−1·2) + (3 · −1) = 8 −2−3=3
Step 2: Calculate the magnitude of v:
||v|| =p42+ 22+ (−1)2=√16 + 4 + 1 = √21
Step 3: Substitute the values into the projection formula:
projv(u) = 3
21 ·
4
2
−1
=1
7
4
2
−1
=
4
7
2
7
−1
7
Therefore, the projection of uonto vis
4
7
2
7
−1
7
.
Question 34
Question
Let v=
3
−2
5
and w=
−1
4
−3
. Determine the angle between vectors vand w
in radians.
Solution
Step 1: Compute the dot product of vectors vand w.
v·w= (3)(−1) + (−2)(4) + (5)(−3)
v·w=−3−8−15 = −26
Step 2: Compute the magnitudes of vectors vand w.
|v|=p32+ (−2)2+ 52=√9 + 4 + 25 = √38
|w|=p(−1)2+ 42+ (−3)2=√1 + 16 + 9 = √26
23
Step 3: Use the dot product formula to find the angle θbetween vectors v
and w.
v·w=|v||w|cos θ
−26 = √38 ·√26 cos θ
Step 4: Solve for θ.
cos θ=−26
√38 ·√26 =−26
√988
θ= cos−1−26
√988
Therefore, the angle between vectors vand win radians is θ= cos−1−26
√988 .
Question 35
Question
Let v=⟨3,−4,5⟩and w=⟨−1,2,−3⟩be vectors in R3. Find the angle between
vand w.
Solution
To find the angle between two vectors vand w, we can use the formula:
cos θ=v·w
||v|| · ||w||
Where θrepresents the angle between the two vectors, v·wis the dot product
of vand w, and ||v|| and ||w|| are the magnitudes of vand w, respectively.
Step 1: Calculate the dot product. The dot product of vand wis
given by:
v·w= (3)(−1) + (−4)(2) + (5)(−3)
v·w=−3−8−15 = −26
Step 2: Calculate the magnitudes. The magnitude of vis:
||v|| =p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
The magnitude of wis:
||w|| =p(−1)2+ 22+ (−3)2=√1 + 4 + 9 = √14
Step 3: Calculate the angle. Finally, substitute the dot product and
magnitudes into the formula:
cos θ=−26
5√2·√14 =−26
5√28 =−26
5√4·√7=−26
10√7=−13√7
35
Thus, the angle between vand wis θ= cos−1−13√7
35 .
24
Question 2
Question
Let v=
3
−4
5
and w=
2
1
−3
be two vectors in R3. Determine the angle
between the vectors vand w.
Solution
To find the angle between two vectors vand win R3, we can use the dot product
formula:
v·w=|v||w|cos θ
where |v|and |w|are the magnitudes of vectors vand wrespectively, and θis
the angle between the two vectors.
Step 1: Calculate the dot product of vand w:
v·w= (3)(2) + (−4)(1) + (5)(−3) = 6 −4−15 = −13
Step 2: Calculate the magnitudes of vectors vand w:
|v|=p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
|w|=p22+ 12+ (−3)2=√4 + 1 + 9 = √14
Step 3: Substitute the dot product and magnitudes into the formula to find
the cosine of the angle θ:
−13 = (5√2)(√14) cos θ
cos θ=−13
5√28 =−13
10√2=−13√2
20
Step 4: Finally, find the angle θusing the inverse cosine function:
θ= cos−1 −13√2
20 !≈130.32◦
Therefore, the angle between the vectors vand wis approximately 130.32◦.
Question 3
Question
Let u=
2
−1
3
and v=
−4
0
1
. Find ∥u−2v∥.
2
Solution
Step 1: Find 2v.
2v= 2
−4
0
1
=
−8
0
2
Step 2: Find u−2v.
u−2v=
2
−1
3
−
−8
0
2
=
10
−1
1
Step 3: Find the norm of ∥u−2v∥.
∥u−2v∥=p102+ (−1)2+ 12=√100 + 1 + 1 = √102
Therefore, ∥u−2v∥=√102.
Question 4
Question
Let a= 3i−2j−kand b= 2i+ 4j+ 5kbe two vectors. Find the scalar
projection of aonto b.
Solution
Step 1: Find the dot product of aand b.
a·b= (3)(2) + (−2)(4) + (−1)(5) = 6 −8−5 = −7
Step 2: Find the magnitude of b.
|b|=p(2)2+ (4)2+ (5)2=√4 + 16 + 25 = √45
Step 3: Find the scalar projection of aonto busing the formula:
projba=a·b
|b|=−7
√45 =−7√45
45
Therefore, the scalar projection of aonto bis −7√45
45 .
Question 5
Question
Let u =
3
−1
2
and v =
1
4
−2
. Find the angle between vectors u and v to the
nearest degree.
3
Solution
Step 1: The dot product of two vectors u and v is given by:
u ·v =∥u∥∥v∥cos θ
where ∥u∥and ∥v∥are the magnitudes of vectors u and v respectively, and θis
the angle between the two vectors.
Step 2: Calculate the dot product of u and v:
u ·v = (3)(1) + (−1)(4) + (2)(−2) = 3 −4−4 = −5
Step 3: Calculate the magnitudes of vectors u and v:
∥u∥=p32+ (−1)2+ 22=√14
∥v∥=p12+ 42+ (−2)2=√21
Step 4: Substitute the dot product and magnitudes into the formula provided
in Step 1 to solve for the cosine of the angle θ:
−5 = √14 ·√21 ·cos θ
Step 5: Solve for cos θ:
cos θ=−5
√14 ·√21 =−5
√294 =−5√294
294
Step 6: Find the angle between the vectors in degrees using the inverse cosine
function:
θ= arccos −5√294
294 !≈129◦
Therefore, the angle between vectors u and v is approximately 129 degrees.
Question 6
Question
Let a= 2i−3j+kand b=i+ 4j−2k. Determine 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∥
4
where θis the angle between aand b,·denotes the dot product, and ∥ · ∥
denotes the magnitude.
Step 1: Find the dot product a ·b.
a·b= (2i−3j+k)·(i+ 4j−2k)
a·b= 2 ·1+(−3) ·4+1·(−2)
a·b= 2 −12 −2
a·b=−12
Step 2: Find the magnitudes ∥a∥and ∥b∥.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p12+ 42+ (−2)2=√1 + 16 + 4 = √21
Step 3: Find the angle θ.
cos θ=−12
√14 ·√21 =−12
√294
θ= cos−1−12
√294≈104.04◦
Therefore, the angle between vectors aand bis approximately 104.04◦.
Question 7
Question
Let a = 3ˆ
i+ 2ˆ
j+ˆ
kand
b= 2ˆ
i−ˆ
j+ 4ˆ
kbe two vectors. Find the projection of
a onto
b.
Solution
Step 1: The projection of a onto
bcan be calculated using the formula:
proj
b(a) = a ·
b
|
b|2·
b
Step 2: First, we calculate the dot product of a and
b:
a ·
b= (3ˆ
i+ 2ˆ
j+ˆ
k)·(2ˆ
i−ˆ
j+ 4ˆ
k)
= 3(2) + 2(−1) + 1(4) = 6 −2 + 4 = 8
Step 3: Next, we calculate the magnitude of
b:
|
b|=p22+ (−1)2+ 42=√4 + 1 + 16 = √21
5
Step 4: Now, substitute the values into the projection formula:
proj
b(a) = 8
|
b|2·(2ˆ
i−ˆ
j+ 4ˆ
k)
=8
21 ·(2ˆ
i−ˆ
j+ 4ˆ
k) = 16
21ˆ
i−8
21ˆ
j+32
21ˆ
k
Therefore, the projection of a onto
bis 16
21 ˆ
i−8
21 ˆ
j+32
21 ˆ
k.
Question 8
Question
Let u=3
−2and v=−1
4. Find the unit vector in the direction of u+ 2v.
Solution
Step 1: Find u+ 2v.
u+ 2v=3
−2+ 2 −1
4=3
−2+−2
8=1
6
Step 2: Find the magnitude of u+ 2v.
∥u+ 2v∥=p12+ 62=√1 + 36 = √37
Step 3: Find the unit vector in the direction of u+ 2v.
ˆ
u+ 2v=u+ 2v
∥u+ 2v∥=1
√37 1
6="1
√37
6
√37 #
Question 9
Question
Let a = 4ˆ
i−3ˆ
j+ 2ˆ
kand
b= 2ˆ
i+ 5ˆ
j−ˆ
k. Find the angle between vectors a and
b.
Solution
Step 1: Calculate the dot product of vectors a and
b
a ·
b= (4)(2) + (−3)(5) + (2)(−1) = 8 −15 −2 = −9
Step 2: Calculate the magnitudes of vectors a and
b
|a|=p42+ (−3)2+ 22=√16 + 9 + 4 = √29
6
|
b|=p22+ 52+ (−1)2=√4 + 25 + 1 = √30
Step 3: Using the dot product formula, find the angle θbetween vectors a
and
b.
cos θ=a ·
b
|a||
b|=−9
√29√30 =−9
√870
Step 4: Solve for θby taking the arccosine of −9
√870 .
θ= arccos −9
√870
Question 10
Question
Let a=3
−2and b=1
5. Determine the magnitude of the vector a−2b.
Solution
Step 1: Find 2b.
Given b=1
5,we have 2b= 2 1
5=2
10.
Step 2: Calculate a−2b.
a−2b=3
−2−2
10=3−2
−2−10=1
−12.
Step 3: Find the magnitude of a−2b. The magnitude of a vector v=x
y
is given by ∥v∥=px2+y2. So, for a−2b=1
−12:
∥a−2b∥=p12+ (−12)2=√1 + 144 = √145.
Therefore, the magnitude of the vector a−2bis √145 .
Question 11
Question
Let v =−3
4and w =1
−2. Find the angle between vectors v and w to the
nearest degree.
7
Solution
Step 1: Calculate the dot product of vectors v and w:
v ·w = (−3)(1) + (4)(−2) = −3−8 = −11
Step 2: Calculate the magnitudes of vectors v and w:
|v|=p(−3)2+ 42=√9 + 16 = √25 = 5
|w|=p12+ (−2)2=√1 + 4 = √5
Step 3: Use the formula cos θ=v·w
|v||w|to find the cosine of the angle between
the vectors:
cos θ=−11
5·√5=−11
5√5
Step 4: Find the angle θby taking the arccosine of −11
5√5:
θ= arccos −11
5√5≈131◦
Therefore, the angle between vectors v and w is approximately 131◦.
Question 12
Question
Let u =
3
−1
4
and v =
−2
5
1
. Determine the projection of v onto u.
Solution
Step 1: Calculate the dot product of u and v.
u ·v = (3 · −2) + (−1·5) + (4 ·1)
u ·v =−6−5+4
u ·v =−7
Step 2: Calculate the magnitude of u.
|u|=p32+ (−1)2+ 42
|u|=√9 + 1 + 16
|u|=√26
8
Step 3: Calculate the projection of v onto u using the formula:
projuv =u ·v
|u|2u
projuv =−7
26
3
−1
4
projuv =
−21
26
7
26
−28
26
Therefore, the projection of v onto u is
−21
26
7
26
−28
26
.
Question 13
Question
A force of 120 N is acting along a direction of 30 degrees east of north. Resolve
this force into two components, one along the north direction and the other
along the east direction.
Solution
Step 1: Draw a diagram to represent the force vector.
Step 2: Resolve the force into two components: Let Fbe the magnitude
of the force and θbe the angle it makes with the north direction. Then, the
force components along the north and east directions can be represented as:
Fn=F·cos(θ) and Fe=F·sin(θ)
Step 3: Calculate the components. Given F= 120 N and θ= 30◦, we have:
Fn= 120 ·cos(30◦)Fe= 120 ·sin(30◦)
Step 4: Solve for the components. Fn= 120 ·cos(30◦) = 120 ·√3
2= 60√3≈
103.92 N Fe= 120 ·sin(30◦) = 120 ·1
2= 60 N
Step 5: Write the final answer. Therefore, the force component along the
north direction is approximately 103.92 N and along the east direction is 60 N.
Question 14
Question
Let a= 3i−2j+kand b= 2i+ 5j−4k. Find the angle between vectors aand
b.
9
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (3i−2j+k)·(2i+ 5j−4k)
= 3(2) + (−2)(5) + 1(−4)
= 6 −10 −4
=−8
Step 2: Calculate the magnitudes of vectors aand b.
|a|=p32+ (−2)2+ 12=√9 + 4 + 1 = √14
|b|=p22+ 52+ (−4)2=√4 + 25 + 16 = √45
Step 3: Use the formula for the angle between two vectors θ= cos−1a·b
|a||b|.
θ= cos−1−8
√14√45
θ= cos−1−8
√630
θ≈cos−1(−0.5131)
θ≈120.47◦
Therefore, the angle between vectors aand bis approximately 120.47◦.
Question 15
Question
Let u =
3
−1
4
and v =
2
5
−2
. Determine a vector w such that w ·u = 2v ·u.
Solution
Step 1: Find w ·u.
w ·u = 2v ·u
w ·u = 2(v ·u)
w ·u = 2(3(2) + (−1)(5) + 4(−2))
w ·u = 2(6 −5−8)
w ·u = 2(−7)
10
w ·u =−14
Step 2: Solve for w. Let w =
x
y
z
. Setting up the dot product equation:
x
y
z
·
3
−1
4
=−14
3x−y+ 4z=−14
3x−y+ 4z=−14
To satisfy the above equation, we have infinitely many solutions. One such
solution is w =
0
−14
0
.
Therefore, w =
0
−14
0
is a vector such that w ·u = 2v ·u.
Question 16
Question
Let u=
3
−2
4
and v=
1
5
2
.
Find u·v(dot product) and ∥u×v∥(magnitude of cross product).
Solution
Step 1: Calculate the dot product u·vusing the formula u·v=u1v1+u2v2+
u3v3.
u·v= (3)(1) + (−2)(5) + (4)(2) = 3 −10 + 8 = 1
Step 2: Calculate the cross product u×vusing the formula
u×v=
u2v3−u3v2
u3v1−u1v3
u1v2−u2v1
u×v=
(−2)(2) −(4)(5)
(4)(1) −(3)(2)
(3)(5) −(−2)(1)
=
−4−20
4−6
15 + 2
=
−24
−2
17
Step 3: Calculate the magnitude of the cross product ∥u×v∥using the
formula ∥u×v∥=p(u2v3−u3v2)2+ (u3v1−u1v3)2+ (u1v2−u2v1)2.
∥u×v∥=p(−24)2+ (−2)2+ 172=√576 + 4 + 289 = √869
11
Question 17
Question
Let a=⟨2,−3,4⟩and b=⟨−1,0,2⟩. Find ∥a+ 2b∥.
Solution
Step 1: Calculate a+ 2bby multiplying bby 2 and adding the result to a.
a+2b=⟨2,−3,4⟩+2⟨−1,0,2⟩=⟨2,−3,4⟩+⟨−2,0,4⟩=⟨2−2,−3+0,4+4⟩=⟨0,−3,8⟩
Step 2: Calculate the magnitude of a+ 2busing the formula for a vector in
3D space ∥⟨x, y, z⟩∥ =px2+y2+z2.
∥a+ 2b∥=p02+ (−3)2+ 82=√0 + 9 + 64 = √73
Therefore, ∥a+ 2b∥=√73.
Question 18
Question
Consider the vectors u =
3
−1
2
,v =
−2
4
−1
, and w =
1
−3
5
. Given that
a = 2u −v and
b=v ·w, determine the value of a ·
b.
Solution
Step 1: Find vector a = 2u −v.
a = 2u −v = 2
3
−1
2
−
−2
4
−1
=
6
−2
4
−
−2
4
−1
=
8
−6
5
Step 2: Find the dot product of
b=v ·w.
b=v ·w =
−2
4
−1
·
1
−3
5
= (−2)(1) + (4)(−3) + (−1)(5) = −2−12 −5 = −19
Step 3: Find the dot product of a ·
b.
a ·
b=
8
−6
5
·−19= (8)(−19) + (−6)(0) + (5)(0) = −152
Therefore, the value of a ·
bis −152 .
12
Question 19
Question
Let a=
2
−3
4
,b=
1
5
−2
, and c=
−3
0
6
. Determine if the vectors a,b, and c
are linearly independent or linearly dependent. If they are linearly dependent,
express one of the vectors as a linear combination of the other two.
Solution
Step 1: To determine if the vectors a,b, and care linearly independent or
dependent, we set up the equation k1a+k2b+k3c=0, where k1,k2, and k3
are scalars.
Step 2: Substituting the given vectors into the equation, we have: k1
2
−3
4
+
k2
1
5
−2
+k3
−3
0
6
=
0
0
0
.
Step 3: This gives us the system of equations: 2k1+k2−3k3= 0, −3k1+
5k2= 0, 4k1−2k2+ 6k3= 0.
Step 4: To determine if the vectors are linearly independent, we solve this
system of equations. One possible way to proceed is by row reducing the aug-
mented matrix [A — 0]:
2 1 −3|0
−3 5 0 |0
4−2 6 |0
Step 5: After performing row operations, the matrix is row reduced to:
102|0
011|0
000|0
Step 6: The system has an infinite number of solutions, indicating that the
vectors a,b, and care linearly dependent.
Step 7: To express one of the vectors as a linear combination of the other
two, we can express cas a linear combination of aand b. From the row-reduced
matrix, we have k1=−2k3and k2=−k3. Thus, we can write c= 2a+b.
Question 20
Question
Let a = 3ˆ
i−2ˆ
j+ 4ˆ
kand
b=ˆ
i+ 3ˆ
j−2ˆ
k. Find the angle between a and
bin
degrees.
13
Solution
Step 1: Calculate the dot product of a and
b.
a ·
b= (3)(1) + (−2)(3) + (4)(−2) = 3 −6−8 = −11
Step 2: Calculate the magnitudes of a and
b.
|a|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|
b|=p12+ 32+ (−2)2=√1 + 9 + 4 = √14
Step 3: Calculate the angle between a and
busing the formula cos(θ) = a·
b
|a||
b|.
cos(θ) = −11
√29√14
Step 4: Find the angle between a and
b.
θ= cos−1−11
√29√14≈128.6◦
Therefore, the angle between a and
bis approximately 128.6◦.
Question 21
Question
Given two vectors u = 2ˆ
i−3ˆ
j+ 4ˆ
kand v =−ˆ
i+ 2ˆ
j−5ˆ
k, find the angle between
the two vectors.
Solution
To find the angle between two vectors, we can use the dot product formula:
u ·v =|u||v|cos(θ), where θis the angle between the vectors.
Step 1: Calculate the dot product
u ·v = (2)(−1) + (−3)(2) + (4)(−5)
=−2−6−20
=−28
Step 2: Calculate the magnitudes of the vectors
|u|=p22+ (−3)2+ 42
=√4 + 9 + 16
=√29
|v|=p(−1)2+ 22+ (−5)2
=√1 + 4 + 25
=√30
14
Step 3: Find the angle between the vectors
cos(θ) = u ·v
|u||v|
cos(θ) = −28
√29 ·√30
θ= cos−1−28
√29 ·√30
Therefore, the angle between the vectors u and v is θ≈109.46◦.
Question 22
Question
Let a= 3i−2j+ 4kand b=i+ 5j−2k. Determine the angle between vectors
aand b.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (3)(1) + (−2)(5) + (4)(−2) = 3 −10 −8 = −15
Step 2: Find the magnitudes of vectors aand b.
|a|=p(3)2+ (−2)2+ (4)2=√9 + 4 + 16 = √29
|b|=p(1)2+ (5)2+ (−2)2=√1 + 25 + 4 = √30
Step 3: Use the dot product formula to find the angle θbetween vectors a
and b:
cos θ=a·b
|a|·|b|=−15
√29 ·√30
θ= arccos −15
√29 ·√30
θ≈124.8◦
Therefore, the angle between vectors aand bis approximately 124.8◦.
Question 23
Question
Let u = 3
i−2
j+ 4
kand v =−2
i+ 5
j−
k. Find the angle between vectors u
and v.
15
Solution
Step 1: Calculate the dot product of vectors u and v.
u ·v = (3) ·(−2) + (−2) ·(5) + (4) ·(−1)
=−6−10 −4
=−20
Step 2: Find the magnitudes of vectors u and v.
|u|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|v|=p(−2)2+ 52+ (−1)2=√4 + 25 + 1 = √30
Step 3: Calculate the angle between the vectors using the formula cos θ=
u·v
|u||v|.
cos θ=−20
√29 ·√30
Step 4: Solve for the angle θusing the inverse cosine function.
θ= cos−1−20
√29 ·√30
θ≈98.64◦
Therefore, the angle between vectors u and v is approximately 98.64◦.
Question 24
Question
Let u = 2
i−3
j+ 4
kand v =−
i+ 2
j−5
k. Find the angle between vectors u
and v.
Solution
Step 1: Find the dot product of vectors u and v.
u ·v = (2)(−1) + (−3)(2) + (4)(−5) = −2−6−20 = −28
Step 2: Find the magnitudes of vectors u and v.
|u|=p22+ (−3)2+ 42=√4 + 9 + 16 = √29
|v|=p(−1)2+ 22+ (−5)2=√1 + 4 + 25 = √30
Step 3: Use the dot product formula to find the angle between vectors u and
v.
cos(θ) = u ·v
|u|·|v|=−28
√29 ·√30 =−28
√870
16
θ=cos−1−28
√870≈131.86◦
Therefore, the angle between vectors u and v is approximately 131.86◦.
Question 25
Question
Let A=
3
−1
2
,B=
−4
0
1
, and C=
1
5
−3
be three vectors. Determine the
vector projection of Aonto the plane determined by Band C.
Solution
Step 1: Calculate the normal vector of the plane determined by Band C. The
normal vector Nof the plane determined by Band Ccan be found by taking
the cross product of Band C.
N=B×C=
i j k
−4 0 1
1 5 −3
N=
−5
7
20
Step 2: Calculate the projection of Aonto N. The projection of Aonto N
can be found using the inner product formula:
projN(A) = A·N
∥N∥2N
Calculate the dot product A·N:
A·N=
3
−1
2
·
−5
7
20
= 3(−5) + (−1)(7) + 2(20) = −5−7 + 40 = 28
Calculate ∥N∥2:
∥N∥2= (−5)2+ 72+ 202= 25 + 49 + 400 = 474
Calculate projN(A):
projN(A) = 28
474
−5
7
20
=
−140
237
196
237
560
237
17
Therefore, the vector projection of Aonto the plane determined by Band
Cis
−140
237
196
237
560
237
.
Question 26
Question
Let a=
2
−1
3
and b=
4
0
−2
. Find the angle between vectors aand b.
Solution
To find the angle between two vectors, 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 a·b.
a·b= (2)(4) + (−1)(0) + (3)(−2) = 8 −6=2
Step 2: Calculate the magnitudes |a|and |b|.
|a|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|b|=p42+ 02+ (−2)2=√16 + 0 + 4 = √20
Step 3: Substitute the dot product and magnitudes into the formula to
solve for θ.
2 = √14√20 cos θ
cos θ=2
√14√20 =2
√280 =1
√70
Step 4: Find the angle between the vectors.
θ= cos−11
√70≈cos−1(0.1348) ≈81.59◦
Therefore, the angle between vectors aand bis approximately 81.59◦.
Question 27
Question
Let a=
2
−1
3
and b=
4
2
−1
. Find ∥3a−2b∥.
18
Solution
Step 1: Find 3aand 2b.
3a= 3
2
−1
3
=
6
−3
9
2b= 2
4
2
−1
=
8
4
−2
Step 2: Find 3a−2b.
3a−2b=
6
−3
9
−
8
4
−2
=
−2
−7
11
Step 3: Find ∥3a−2b∥.
∥3a−2b∥=
−2
−7
11
=p(−2)2+ (−7)2+ 112
=√4 + 49 + 121
=√174
Therefore, ∥3a−2b∥=√174.
Question 28
Question
Let a= 3i+ 2j−4kand b=−i+ 5j+ 2kbe two vectors. Find the angle
between the two vectors in degrees.
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= (3)(−1) + (2)(5) + (−4)(2) = −3 + 10 −8 = −1
Step 2: Calculate the magnitudes of vectors aand b:
|a|=p32+ 22+ (−4)2=√9 + 4 + 16 = √29
19
|b|=p(−1)2+ 52+ 22=√1 + 25 + 4 = √30
Step 3: Calculate the angle θbetween the two vectors using the formula:
cos(θ) = a·b
|a|·|b|
cos(θ) = −1
√29 ·√30 =−1
√870
Step 4: Find the angle θby taking the arccosine of the calculated value:
θ= arccos −1
√870≈96.25◦
Therefore, the angle between the vectors aand bis approximately 96.25◦.
Question 29
Question
Let u =
2
−1
4
,v =
−3
5
1
, and w =
1
−2
0
. Find the vector x that satisfies the
equation 4u −2v +x = 3 w.
Solution
Step 1: Compute 4u −2v.
4u −2v = 4
2
−1
4
−2
−3
5
1
=
8
−4
16
−
−6
10
2
=
14
−14
14
Step 2: Rewrite the equation 4u −2v +x = 3 w as x = 3 w −(4u −2v).
x = 3
1
−2
0
−
14
−14
14
=
3
−6
0
−
14
−14
14
=
−11
8
−14
Therefore, the vector x that satisfies the equation 4u−2v+x = 3 w is
−11
8
−14
.
Question 30
Question
Let u=
2
−1
3
and v=
−3
4
5
. Find a vector wwhich is orthogonal to both u
and v.
20
Solution
Step 1: To find a vector wwhich is orthogonal to both uand v, we take the
cross product of uand v.
Step 2: The cross product u×vis given by:
u×v=
i j k
2−1 3
−3 4 5
Step 3: Evaluating the determinant gives us:
u×v=
−1(5) −3(4)
−(2(5) −3(−3))
2(4) −(−1)(−3)
=
−17
11
5
Therefore, the vector w=
−17
11
5
is orthogonal to both uand v.
Question 31
Question
Let a= 3i−2j+ 2kand b=−i+ 4j−3k. Determine the angle between vectors
aand b.
Solution
Given vectors aand b, we can find the angle θbetween them using the formula:
cos(θ) = a·b
∥a∥∥b∥
Step 1: Find the dot product of a and b The dot product of two vectors
aand bis given by:
a·b= (3)(−1) + (−2)(4) + (2)(−3)
Calculating the dot product: a·b=−3−8−6 = −17
Step 2: Find the magnitudes of a and b The magnitudes of vectors a
and bare given by:
∥a∥=p32+ (−2)2+ 22=√9 + 4 + 4 = √17
∥b∥=p(−1)2+ 42+ (−3)2=√1 + 16 + 9 = √26
21
Step 3: Calculate the cosine of the angle between a and b Now,
substitute the dot product and magnitudes into the formula for the cosine of
the angle between two vectors:
cos(θ) = −17
√17√26
Step 4: Find the angle Finally, solve for θby taking the arccosine of the
calculated value:
θ= arccos −17
√17√26
Therefore, the angle between vectors aand bis θ≈107.25◦.
Question 32
Question
Let a=
2
−3
1
and b=
−4
2
5
. Find the angle between vectors aand b.
Solution
Step 1: Calculate the dot product of aand b.
a·b= (2)(−4) + (−3)(2) + (1)(5) = −8−6 + 5 = −9
Step 2: Calculate the magnitudes of vectors aand b.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p(−4)2+ 22+ 52=√16 + 4 + 25 = √45
Step 3: Use the dot product to find the angle between aand b.
cos θ=a
·b∥a∥∥b∥=−9
√14√45
θ= cos−1−9
√14√45≈108.18◦
Therefore, the angle between vectors aand bis approximately 108.18◦.
Question 33
Question
Let u=
2
−1
3
and v=
4
2
−1
. Determine the projection of uonto v.
22
Solution
To find the projection of uonto v, we use the formula:
projv(u) = u·v
||v||2·v
Step 1: Calculate the dot product of uand v:
u·v=
2
−1
3
·
4
2
−1
= (2 ·4) + (−1·2) + (3 · −1) = 8 −2−3=3
Step 2: Calculate the magnitude of v:
||v|| =p42+ 22+ (−1)2=√16 + 4 + 1 = √21
Step 3: Substitute the values into the projection formula:
projv(u) = 3
21 ·
4
2
−1
=1
7
4
2
−1
=
4
7
2
7
−1
7
Therefore, the projection of uonto vis
4
7
2
7
−1
7
.
Question 34
Question
Let v=
3
−2
5
and w=
−1
4
−3
. Determine the angle between vectors vand w
in radians.
Solution
Step 1: Compute the dot product of vectors vand w.
v·w= (3)(−1) + (−2)(4) + (5)(−3)
v·w=−3−8−15 = −26
Step 2: Compute the magnitudes of vectors vand w.
|v|=p32+ (−2)2+ 52=√9 + 4 + 25 = √38
|w|=p(−1)2+ 42+ (−3)2=√1 + 16 + 9 = √26
23
Step 3: Use the dot product formula to find the angle θbetween vectors v
and w.
v·w=|v||w|cos θ
−26 = √38 ·√26 cos θ
Step 4: Solve for θ.
cos θ=−26
√38 ·√26 =−26
√988
θ= cos−1−26
√988
Therefore, the angle between vectors vand win radians is θ= cos−1−26
√988 .
Question 35
Question
Let v=⟨3,−4,5⟩and w=⟨−1,2,−3⟩be vectors in R3. Find the angle between
vand w.
Solution
To find the angle between two vectors vand w, we can use the formula:
cos θ=v·w
||v|| · ||w||
Where θrepresents the angle between the two vectors, v·wis the dot product
of vand w, and ||v|| and ||w|| are the magnitudes of vand w, respectively.
Step 1: Calculate the dot product. The dot product of vand wis
given by:
v·w= (3)(−1) + (−4)(2) + (5)(−3)
v·w=−3−8−15 = −26
Step 2: Calculate the magnitudes. The magnitude of vis:
||v|| =p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
The magnitude of wis:
||w|| =p(−1)2+ 22+ (−3)2=√1 + 4 + 9 = √14
Step 3: Calculate the angle. Finally, substitute the dot product and
magnitudes into the formula:
cos θ=−26
5√2·√14 =−26
5√28 =−26
5√4·√7=−26
10√7=−13√7
35
Thus, the angle between vand wis θ= cos−1−13√7
35 .
24
Question 2
Question
Let v=
3
−4
5
and w=
2
1
−3
be two vectors in R3. Determine the angle
between the vectors vand w.
Solution
To find the angle between two vectors vand win R3, we can use the dot product
formula:
v·w=|v||w|cos θ
where |v|and |w|are the magnitudes of vectors vand wrespectively, and θis
the angle between the two vectors.
Step 1: Calculate the dot product of vand w:
v·w= (3)(2) + (−4)(1) + (5)(−3) = 6 −4−15 = −13
Step 2: Calculate the magnitudes of vectors vand w:
|v|=p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
|w|=p22+ 12+ (−3)2=√4 + 1 + 9 = √14
Step 3: Substitute the dot product and magnitudes into the formula to find
the cosine of the angle θ:
−13 = (5√2)(√14) cos θ
cos θ=−13
5√28 =−13
10√2=−13√2
20
Step 4: Finally, find the angle θusing the inverse cosine function:
θ= cos−1 −13√2
20 !≈130.32◦
Therefore, the angle between the vectors vand wis approximately 130.32◦.
Question 3
Question
Let u=
2
−1
3
and v=
−4
0
1
. Find ∥u−2v∥.
2
Solution
Step 1: Find 2v.
2v= 2
−4
0
1
=
−8
0
2
Step 2: Find u−2v.
u−2v=
2
−1
3
−
−8
0
2
=
10
−1
1
Step 3: Find the norm of ∥u−2v∥.
∥u−2v∥=p102+ (−1)2+ 12=√100 + 1 + 1 = √102
Therefore, ∥u−2v∥=√102.
Question 4
Question
Let a= 3i−2j−kand b= 2i+ 4j+ 5kbe two vectors. Find the scalar
projection of aonto b.
Solution
Step 1: Find the dot product of aand b.
a·b= (3)(2) + (−2)(4) + (−1)(5) = 6 −8−5 = −7
Step 2: Find the magnitude of b.
|b|=p(2)2+ (4)2+ (5)2=√4 + 16 + 25 = √45
Step 3: Find the scalar projection of aonto busing the formula:
projba=a·b
|b|=−7
√45 =−7√45
45
Therefore, the scalar projection of aonto bis −7√45
45 .
Question 5
Question
Let u =
3
−1
2
and v =
1
4
−2
. Find the angle between vectors u and v to the
nearest degree.
3
Solution
Step 1: The dot product of two vectors u and v is given by:
u ·v =∥u∥∥v∥cos θ
where ∥u∥and ∥v∥are the magnitudes of vectors u and v respectively, and θis
the angle between the two vectors.
Step 2: Calculate the dot product of u and v:
u ·v = (3)(1) + (−1)(4) + (2)(−2) = 3 −4−4 = −5
Step 3: Calculate the magnitudes of vectors u and v:
∥u∥=p32+ (−1)2+ 22=√14
∥v∥=p12+ 42+ (−2)2=√21
Step 4: Substitute the dot product and magnitudes into the formula provided
in Step 1 to solve for the cosine of the angle θ:
−5 = √14 ·√21 ·cos θ
Step 5: Solve for cos θ:
cos θ=−5
√14 ·√21 =−5
√294 =−5√294
294
Step 6: Find the angle between the vectors in degrees using the inverse cosine
function:
θ= arccos −5√294
294 !≈129◦
Therefore, the angle between vectors u and v is approximately 129 degrees.
Question 6
Question
Let a= 2i−3j+kand b=i+ 4j−2k. Determine 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∥
4
where θis the angle between aand b,·denotes the dot product, and ∥ · ∥
denotes the magnitude.
Step 1: Find the dot product a ·b.
a·b= (2i−3j+k)·(i+ 4j−2k)
a·b= 2 ·1+(−3) ·4+1·(−2)
a·b= 2 −12 −2
a·b=−12
Step 2: Find the magnitudes ∥a∥and ∥b∥.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p12+ 42+ (−2)2=√1 + 16 + 4 = √21
Step 3: Find the angle θ.
cos θ=−12
√14 ·√21 =−12
√294
θ= cos−1−12
√294≈104.04◦
Therefore, the angle between vectors aand bis approximately 104.04◦.
Question 7
Question
Let a = 3ˆ
i+ 2ˆ
j+ˆ
kand
b= 2ˆ
i−ˆ
j+ 4ˆ
kbe two vectors. Find the projection of
a onto
b.
Solution
Step 1: The projection of a onto
bcan be calculated using the formula:
proj
b(a) = a ·
b
|
b|2·
b
Step 2: First, we calculate the dot product of a and
b:
a ·
b= (3ˆ
i+ 2ˆ
j+ˆ
k)·(2ˆ
i−ˆ
j+ 4ˆ
k)
= 3(2) + 2(−1) + 1(4) = 6 −2 + 4 = 8
Step 3: Next, we calculate the magnitude of
b:
|
b|=p22+ (−1)2+ 42=√4 + 1 + 16 = √21
5
Step 4: Now, substitute the values into the projection formula:
proj
b(a) = 8
|
b|2·(2ˆ
i−ˆ
j+ 4ˆ
k)
=8
21 ·(2ˆ
i−ˆ
j+ 4ˆ
k) = 16
21ˆ
i−8
21ˆ
j+32
21ˆ
k
Therefore, the projection of a onto
bis 16
21 ˆ
i−8
21 ˆ
j+32
21 ˆ
k.
Question 8
Question
Let u=3
−2and v=−1
4. Find the unit vector in the direction of u+ 2v.
Solution
Step 1: Find u+ 2v.
u+ 2v=3
−2+ 2 −1
4=3
−2+−2
8=1
6
Step 2: Find the magnitude of u+ 2v.
∥u+ 2v∥=p12+ 62=√1 + 36 = √37
Step 3: Find the unit vector in the direction of u+ 2v.
ˆ
u+ 2v=u+ 2v
∥u+ 2v∥=1
√37 1
6="1
√37
6
√37 #
Question 9
Question
Let a = 4ˆ
i−3ˆ
j+ 2ˆ
kand
b= 2ˆ
i+ 5ˆ
j−ˆ
k. Find the angle between vectors a and
b.
Solution
Step 1: Calculate the dot product of vectors a and
b
a ·
b= (4)(2) + (−3)(5) + (2)(−1) = 8 −15 −2 = −9
Step 2: Calculate the magnitudes of vectors a and
b
|a|=p42+ (−3)2+ 22=√16 + 9 + 4 = √29
6
|
b|=p22+ 52+ (−1)2=√4 + 25 + 1 = √30
Step 3: Using the dot product formula, find the angle θbetween vectors a
and
b.
cos θ=a ·
b
|a||
b|=−9
√29√30 =−9
√870
Step 4: Solve for θby taking the arccosine of −9
√870 .
θ= arccos −9
√870
Question 10
Question
Let a=3
−2and b=1
5. Determine the magnitude of the vector a−2b.
Solution
Step 1: Find 2b.
Given b=1
5,we have 2b= 2 1
5=2
10.
Step 2: Calculate a−2b.
a−2b=3
−2−2
10=3−2
−2−10=1
−12.
Step 3: Find the magnitude of a−2b. The magnitude of a vector v=x
y
is given by ∥v∥=px2+y2. So, for a−2b=1
−12:
∥a−2b∥=p12+ (−12)2=√1 + 144 = √145.
Therefore, the magnitude of the vector a−2bis √145 .
Question 11
Question
Let v =−3
4and w =1
−2. Find the angle between vectors v and w to the
nearest degree.
7
Solution
Step 1: Calculate the dot product of vectors v and w:
v ·w = (−3)(1) + (4)(−2) = −3−8 = −11
Step 2: Calculate the magnitudes of vectors v and w:
|v|=p(−3)2+ 42=√9 + 16 = √25 = 5
|w|=p12+ (−2)2=√1 + 4 = √5
Step 3: Use the formula cos θ=v·w
|v||w|to find the cosine of the angle between
the vectors:
cos θ=−11
5·√5=−11
5√5
Step 4: Find the angle θby taking the arccosine of −11
5√5:
θ= arccos −11
5√5≈131◦
Therefore, the angle between vectors v and w is approximately 131◦.
Question 12
Question
Let u =
3
−1
4
and v =
−2
5
1
. Determine the projection of v onto u.
Solution
Step 1: Calculate the dot product of u and v.
u ·v = (3 · −2) + (−1·5) + (4 ·1)
u ·v =−6−5+4
u ·v =−7
Step 2: Calculate the magnitude of u.
|u|=p32+ (−1)2+ 42
|u|=√9 + 1 + 16
|u|=√26
8
Step 3: Calculate the projection of v onto u using the formula:
projuv =u ·v
|u|2u
projuv =−7
26
3
−1
4
projuv =
−21
26
7
26
−28
26
Therefore, the projection of v onto u is
−21
26
7
26
−28
26
.
Question 13
Question
A force of 120 N is acting along a direction of 30 degrees east of north. Resolve
this force into two components, one along the north direction and the other
along the east direction.
Solution
Step 1: Draw a diagram to represent the force vector.
Step 2: Resolve the force into two components: Let Fbe the magnitude
of the force and θbe the angle it makes with the north direction. Then, the
force components along the north and east directions can be represented as:
Fn=F·cos(θ) and Fe=F·sin(θ)
Step 3: Calculate the components. Given F= 120 N and θ= 30◦, we have:
Fn= 120 ·cos(30◦)Fe= 120 ·sin(30◦)
Step 4: Solve for the components. Fn= 120 ·cos(30◦) = 120 ·√3
2= 60√3≈
103.92 N Fe= 120 ·sin(30◦) = 120 ·1
2= 60 N
Step 5: Write the final answer. Therefore, the force component along the
north direction is approximately 103.92 N and along the east direction is 60 N.
Question 14
Question
Let a= 3i−2j+kand b= 2i+ 5j−4k. Find the angle between vectors aand
b.
9
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (3i−2j+k)·(2i+ 5j−4k)
= 3(2) + (−2)(5) + 1(−4)
= 6 −10 −4
=−8
Step 2: Calculate the magnitudes of vectors aand b.
|a|=p32+ (−2)2+ 12=√9 + 4 + 1 = √14
|b|=p22+ 52+ (−4)2=√4 + 25 + 16 = √45
Step 3: Use the formula for the angle between two vectors θ= cos−1a·b
|a||b|.
θ= cos−1−8
√14√45
θ= cos−1−8
√630
θ≈cos−1(−0.5131)
θ≈120.47◦
Therefore, the angle between vectors aand bis approximately 120.47◦.
Question 15
Question
Let u =
3
−1
4
and v =
2
5
−2
. Determine a vector w such that w ·u = 2v ·u.
Solution
Step 1: Find w ·u.
w ·u = 2v ·u
w ·u = 2(v ·u)
w ·u = 2(3(2) + (−1)(5) + 4(−2))
w ·u = 2(6 −5−8)
w ·u = 2(−7)
10
w ·u =−14
Step 2: Solve for w. Let w =
x
y
z
. Setting up the dot product equation:
x
y
z
·
3
−1
4
=−14
3x−y+ 4z=−14
3x−y+ 4z=−14
To satisfy the above equation, we have infinitely many solutions. One such
solution is w =
0
−14
0
.
Therefore, w =
0
−14
0
is a vector such that w ·u = 2v ·u.
Question 16
Question
Let u=
3
−2
4
and v=
1
5
2
.
Find u·v(dot product) and ∥u×v∥(magnitude of cross product).
Solution
Step 1: Calculate the dot product u·vusing the formula u·v=u1v1+u2v2+
u3v3.
u·v= (3)(1) + (−2)(5) + (4)(2) = 3 −10 + 8 = 1
Step 2: Calculate the cross product u×vusing the formula
u×v=
u2v3−u3v2
u3v1−u1v3
u1v2−u2v1
u×v=
(−2)(2) −(4)(5)
(4)(1) −(3)(2)
(3)(5) −(−2)(1)
=
−4−20
4−6
15 + 2
=
−24
−2
17
Step 3: Calculate the magnitude of the cross product ∥u×v∥using the
formula ∥u×v∥=p(u2v3−u3v2)2+ (u3v1−u1v3)2+ (u1v2−u2v1)2.
∥u×v∥=p(−24)2+ (−2)2+ 172=√576 + 4 + 289 = √869
11
Question 17
Question
Let a=⟨2,−3,4⟩and b=⟨−1,0,2⟩. Find ∥a+ 2b∥.
Solution
Step 1: Calculate a+ 2bby multiplying bby 2 and adding the result to a.
a+2b=⟨2,−3,4⟩+2⟨−1,0,2⟩=⟨2,−3,4⟩+⟨−2,0,4⟩=⟨2−2,−3+0,4+4⟩=⟨0,−3,8⟩
Step 2: Calculate the magnitude of a+ 2busing the formula for a vector in
3D space ∥⟨x, y, z⟩∥ =px2+y2+z2.
∥a+ 2b∥=p02+ (−3)2+ 82=√0 + 9 + 64 = √73
Therefore, ∥a+ 2b∥=√73.
Question 18
Question
Consider the vectors u =
3
−1
2
,v =
−2
4
−1
, and w =
1
−3
5
. Given that
a = 2u −v and
b=v ·w, determine the value of a ·
b.
Solution
Step 1: Find vector a = 2u −v.
a = 2u −v = 2
3
−1
2
−
−2
4
−1
=
6
−2
4
−
−2
4
−1
=
8
−6
5
Step 2: Find the dot product of
b=v ·w.
b=v ·w =
−2
4
−1
·
1
−3
5
= (−2)(1) + (4)(−3) + (−1)(5) = −2−12 −5 = −19
Step 3: Find the dot product of a ·
b.
a ·
b=
8
−6
5
·−19= (8)(−19) + (−6)(0) + (5)(0) = −152
Therefore, the value of a ·
bis −152 .
12
Question 19
Question
Let a=
2
−3
4
,b=
1
5
−2
, and c=
−3
0
6
. Determine if the vectors a,b, and c
are linearly independent or linearly dependent. If they are linearly dependent,
express one of the vectors as a linear combination of the other two.
Solution
Step 1: To determine if the vectors a,b, and care linearly independent or
dependent, we set up the equation k1a+k2b+k3c=0, where k1,k2, and k3
are scalars.
Step 2: Substituting the given vectors into the equation, we have: k1
2
−3
4
+
k2
1
5
−2
+k3
−3
0
6
=
0
0
0
.
Step 3: This gives us the system of equations: 2k1+k2−3k3= 0, −3k1+
5k2= 0, 4k1−2k2+ 6k3= 0.
Step 4: To determine if the vectors are linearly independent, we solve this
system of equations. One possible way to proceed is by row reducing the aug-
mented matrix [A — 0]:
2 1 −3|0
−3 5 0 |0
4−2 6 |0
Step 5: After performing row operations, the matrix is row reduced to:
102|0
011|0
000|0
Step 6: The system has an infinite number of solutions, indicating that the
vectors a,b, and care linearly dependent.
Step 7: To express one of the vectors as a linear combination of the other
two, we can express cas a linear combination of aand b. From the row-reduced
matrix, we have k1=−2k3and k2=−k3. Thus, we can write c= 2a+b.
Question 20
Question
Let a = 3ˆ
i−2ˆ
j+ 4ˆ
kand
b=ˆ
i+ 3ˆ
j−2ˆ
k. Find the angle between a and
bin
degrees.
13
Solution
Step 1: Calculate the dot product of a and
b.
a ·
b= (3)(1) + (−2)(3) + (4)(−2) = 3 −6−8 = −11
Step 2: Calculate the magnitudes of a and
b.
|a|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|
b|=p12+ 32+ (−2)2=√1 + 9 + 4 = √14
Step 3: Calculate the angle between a and
busing the formula cos(θ) = a·
b
|a||
b|.
cos(θ) = −11
√29√14
Step 4: Find the angle between a and
b.
θ= cos−1−11
√29√14≈128.6◦
Therefore, the angle between a and
bis approximately 128.6◦.
Question 21
Question
Given two vectors u = 2ˆ
i−3ˆ
j+ 4ˆ
kand v =−ˆ
i+ 2ˆ
j−5ˆ
k, find the angle between
the two vectors.
Solution
To find the angle between two vectors, we can use the dot product formula:
u ·v =|u||v|cos(θ), where θis the angle between the vectors.
Step 1: Calculate the dot product
u ·v = (2)(−1) + (−3)(2) + (4)(−5)
=−2−6−20
=−28
Step 2: Calculate the magnitudes of the vectors
|u|=p22+ (−3)2+ 42
=√4 + 9 + 16
=√29
|v|=p(−1)2+ 22+ (−5)2
=√1 + 4 + 25
=√30
14
Step 3: Find the angle between the vectors
cos(θ) = u ·v
|u||v|
cos(θ) = −28
√29 ·√30
θ= cos−1−28
√29 ·√30
Therefore, the angle between the vectors u and v is θ≈109.46◦.
Question 22
Question
Let a= 3i−2j+ 4kand b=i+ 5j−2k. Determine the angle between vectors
aand b.
Solution
Step 1: Find the dot product of vectors aand b.
a·b= (3)(1) + (−2)(5) + (4)(−2) = 3 −10 −8 = −15
Step 2: Find the magnitudes of vectors aand b.
|a|=p(3)2+ (−2)2+ (4)2=√9 + 4 + 16 = √29
|b|=p(1)2+ (5)2+ (−2)2=√1 + 25 + 4 = √30
Step 3: Use the dot product formula to find the angle θbetween vectors a
and b:
cos θ=a·b
|a|·|b|=−15
√29 ·√30
θ= arccos −15
√29 ·√30
θ≈124.8◦
Therefore, the angle between vectors aand bis approximately 124.8◦.
Question 23
Question
Let u = 3
i−2
j+ 4
kand v =−2
i+ 5
j−
k. Find the angle between vectors u
and v.
15
Solution
Step 1: Calculate the dot product of vectors u and v.
u ·v = (3) ·(−2) + (−2) ·(5) + (4) ·(−1)
=−6−10 −4
=−20
Step 2: Find the magnitudes of vectors u and v.
|u|=p32+ (−2)2+ 42=√9 + 4 + 16 = √29
|v|=p(−2)2+ 52+ (−1)2=√4 + 25 + 1 = √30
Step 3: Calculate the angle between the vectors using the formula cos θ=
u·v
|u||v|.
cos θ=−20
√29 ·√30
Step 4: Solve for the angle θusing the inverse cosine function.
θ= cos−1−20
√29 ·√30
θ≈98.64◦
Therefore, the angle between vectors u and v is approximately 98.64◦.
Question 24
Question
Let u = 2
i−3
j+ 4
kand v =−
i+ 2
j−5
k. Find the angle between vectors u
and v.
Solution
Step 1: Find the dot product of vectors u and v.
u ·v = (2)(−1) + (−3)(2) + (4)(−5) = −2−6−20 = −28
Step 2: Find the magnitudes of vectors u and v.
|u|=p22+ (−3)2+ 42=√4 + 9 + 16 = √29
|v|=p(−1)2+ 22+ (−5)2=√1 + 4 + 25 = √30
Step 3: Use the dot product formula to find the angle between vectors u and
v.
cos(θ) = u ·v
|u|·|v|=−28
√29 ·√30 =−28
√870
16
θ=cos−1−28
√870≈131.86◦
Therefore, the angle between vectors u and v is approximately 131.86◦.
Question 25
Question
Let A=
3
−1
2
,B=
−4
0
1
, and C=
1
5
−3
be three vectors. Determine the
vector projection of Aonto the plane determined by Band C.
Solution
Step 1: Calculate the normal vector of the plane determined by Band C. The
normal vector Nof the plane determined by Band Ccan be found by taking
the cross product of Band C.
N=B×C=
i j k
−4 0 1
1 5 −3
N=
−5
7
20
Step 2: Calculate the projection of Aonto N. The projection of Aonto N
can be found using the inner product formula:
projN(A) = A·N
∥N∥2N
Calculate the dot product A·N:
A·N=
3
−1
2
·
−5
7
20
= 3(−5) + (−1)(7) + 2(20) = −5−7 + 40 = 28
Calculate ∥N∥2:
∥N∥2= (−5)2+ 72+ 202= 25 + 49 + 400 = 474
Calculate projN(A):
projN(A) = 28
474
−5
7
20
=
−140
237
196
237
560
237
17
Therefore, the vector projection of Aonto the plane determined by Band
Cis
−140
237
196
237
560
237
.
Question 26
Question
Let a=
2
−1
3
and b=
4
0
−2
. Find the angle between vectors aand b.
Solution
To find the angle between two vectors, 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 a·b.
a·b= (2)(4) + (−1)(0) + (3)(−2) = 8 −6=2
Step 2: Calculate the magnitudes |a|and |b|.
|a|=p22+ (−1)2+ 32=√4 + 1 + 9 = √14
|b|=p42+ 02+ (−2)2=√16 + 0 + 4 = √20
Step 3: Substitute the dot product and magnitudes into the formula to
solve for θ.
2 = √14√20 cos θ
cos θ=2
√14√20 =2
√280 =1
√70
Step 4: Find the angle between the vectors.
θ= cos−11
√70≈cos−1(0.1348) ≈81.59◦
Therefore, the angle between vectors aand bis approximately 81.59◦.
Question 27
Question
Let a=
2
−1
3
and b=
4
2
−1
. Find ∥3a−2b∥.
18
Solution
Step 1: Find 3aand 2b.
3a= 3
2
−1
3
=
6
−3
9
2b= 2
4
2
−1
=
8
4
−2
Step 2: Find 3a−2b.
3a−2b=
6
−3
9
−
8
4
−2
=
−2
−7
11
Step 3: Find ∥3a−2b∥.
∥3a−2b∥=
−2
−7
11
=p(−2)2+ (−7)2+ 112
=√4 + 49 + 121
=√174
Therefore, ∥3a−2b∥=√174.
Question 28
Question
Let a= 3i+ 2j−4kand b=−i+ 5j+ 2kbe two vectors. Find the angle
between the two vectors in degrees.
Solution
Step 1: Calculate the dot product of vectors aand b:
a·b= (3)(−1) + (2)(5) + (−4)(2) = −3 + 10 −8 = −1
Step 2: Calculate the magnitudes of vectors aand b:
|a|=p32+ 22+ (−4)2=√9 + 4 + 16 = √29
19
|b|=p(−1)2+ 52+ 22=√1 + 25 + 4 = √30
Step 3: Calculate the angle θbetween the two vectors using the formula:
cos(θ) = a·b
|a|·|b|
cos(θ) = −1
√29 ·√30 =−1
√870
Step 4: Find the angle θby taking the arccosine of the calculated value:
θ= arccos −1
√870≈96.25◦
Therefore, the angle between the vectors aand bis approximately 96.25◦.
Question 29
Question
Let u =
2
−1
4
,v =
−3
5
1
, and w =
1
−2
0
. Find the vector x that satisfies the
equation 4u −2v +x = 3 w.
Solution
Step 1: Compute 4u −2v.
4u −2v = 4
2
−1
4
−2
−3
5
1
=
8
−4
16
−
−6
10
2
=
14
−14
14
Step 2: Rewrite the equation 4u −2v +x = 3 w as x = 3 w −(4u −2v).
x = 3
1
−2
0
−
14
−14
14
=
3
−6
0
−
14
−14
14
=
−11
8
−14
Therefore, the vector x that satisfies the equation 4u−2v+x = 3 w is
−11
8
−14
.
Question 30
Question
Let u=
2
−1
3
and v=
−3
4
5
. Find a vector wwhich is orthogonal to both u
and v.
20
Solution
Step 1: To find a vector wwhich is orthogonal to both uand v, we take the
cross product of uand v.
Step 2: The cross product u×vis given by:
u×v=
i j k
2−1 3
−3 4 5
Step 3: Evaluating the determinant gives us:
u×v=
−1(5) −3(4)
−(2(5) −3(−3))
2(4) −(−1)(−3)
=
−17
11
5
Therefore, the vector w=
−17
11
5
is orthogonal to both uand v.
Question 31
Question
Let a= 3i−2j+ 2kand b=−i+ 4j−3k. Determine the angle between vectors
aand b.
Solution
Given vectors aand b, we can find the angle θbetween them using the formula:
cos(θ) = a·b
∥a∥∥b∥
Step 1: Find the dot product of a and b The dot product of two vectors
aand bis given by:
a·b= (3)(−1) + (−2)(4) + (2)(−3)
Calculating the dot product: a·b=−3−8−6 = −17
Step 2: Find the magnitudes of a and b The magnitudes of vectors a
and bare given by:
∥a∥=p32+ (−2)2+ 22=√9 + 4 + 4 = √17
∥b∥=p(−1)2+ 42+ (−3)2=√1 + 16 + 9 = √26
21
Step 3: Calculate the cosine of the angle between a and b Now,
substitute the dot product and magnitudes into the formula for the cosine of
the angle between two vectors:
cos(θ) = −17
√17√26
Step 4: Find the angle Finally, solve for θby taking the arccosine of the
calculated value:
θ= arccos −17
√17√26
Therefore, the angle between vectors aand bis θ≈107.25◦.
Question 32
Question
Let a=
2
−3
1
and b=
−4
2
5
. Find the angle between vectors aand b.
Solution
Step 1: Calculate the dot product of aand b.
a·b= (2)(−4) + (−3)(2) + (1)(5) = −8−6 + 5 = −9
Step 2: Calculate the magnitudes of vectors aand b.
∥a∥=p22+ (−3)2+ 12=√4 + 9 + 1 = √14
∥b∥=p(−4)2+ 22+ 52=√16 + 4 + 25 = √45
Step 3: Use the dot product to find the angle between aand b.
cos θ=a
·b∥a∥∥b∥=−9
√14√45
θ= cos−1−9
√14√45≈108.18◦
Therefore, the angle between vectors aand bis approximately 108.18◦.
Question 33
Question
Let u=
2
−1
3
and v=
4
2
−1
. Determine the projection of uonto v.
22
Solution
To find the projection of uonto v, we use the formula:
projv(u) = u·v
||v||2·v
Step 1: Calculate the dot product of uand v:
u·v=
2
−1
3
·
4
2
−1
= (2 ·4) + (−1·2) + (3 · −1) = 8 −2−3=3
Step 2: Calculate the magnitude of v:
||v|| =p42+ 22+ (−1)2=√16 + 4 + 1 = √21
Step 3: Substitute the values into the projection formula:
projv(u) = 3
21 ·
4
2
−1
=1
7
4
2
−1
=
4
7
2
7
−1
7
Therefore, the projection of uonto vis
4
7
2
7
−1
7
.
Question 34
Question
Let v=
3
−2
5
and w=
−1
4
−3
. Determine the angle between vectors vand w
in radians.
Solution
Step 1: Compute the dot product of vectors vand w.
v·w= (3)(−1) + (−2)(4) + (5)(−3)
v·w=−3−8−15 = −26
Step 2: Compute the magnitudes of vectors vand w.
|v|=p32+ (−2)2+ 52=√9 + 4 + 25 = √38
|w|=p(−1)2+ 42+ (−3)2=√1 + 16 + 9 = √26
23
Step 3: Use the dot product formula to find the angle θbetween vectors v
and w.
v·w=|v||w|cos θ
−26 = √38 ·√26 cos θ
Step 4: Solve for θ.
cos θ=−26
√38 ·√26 =−26
√988
θ= cos−1−26
√988
Therefore, the angle between vectors vand win radians is θ= cos−1−26
√988 .
Question 35
Question
Let v=⟨3,−4,5⟩and w=⟨−1,2,−3⟩be vectors in R3. Find the angle between
vand w.
Solution
To find the angle between two vectors vand w, we can use the formula:
cos θ=v·w
||v|| · ||w||
Where θrepresents the angle between the two vectors, v·wis the dot product
of vand w, and ||v|| and ||w|| are the magnitudes of vand w, respectively.
Step 1: Calculate the dot product. The dot product of vand wis
given by:
v·w= (3)(−1) + (−4)(2) + (5)(−3)
v·w=−3−8−15 = −26
Step 2: Calculate the magnitudes. The magnitude of vis:
||v|| =p32+ (−4)2+ 52=√9 + 16 + 25 = √50 = 5√2
The magnitude of wis:
||w|| =p(−1)2+ 22+ (−3)2=√1 + 4 + 9 = √14
Step 3: Calculate the angle. Finally, substitute the dot product and
magnitudes into the formula:
cos θ=−26
5√2·√14 =−26
5√28 =−26
5√4·√7=−26
10√7=−13√7
35
Thus, the angle between vand wis θ= cos−1−13√7
35 .
24
