Preliminary Questions
1. Is the dot product of two vectors a scalar or a vector?
solution The dot product of two vectors is the sum of products of scalars, hence it is a scalar.
2. What can you say about the angle betweenaandbifaãb<0?
solution Since the cosine of the angle betweenaand bsatisfies cosθ = aaãbb, also cosθ < 0. By definition 0≤θ≤π, but since cosθ <0 thenθis in(π/2, π]. In other words, the angle betweenaandbis obtuse.
3. Which property of dot products allows us to conclude that ifvis orthogonal to bothuandw, thenvis orthogonal to u+w?
solution One property is that two vectors are orthogonal if and only if the dot product of the two vectors is zero. The second property is the Distributive Law. Sincevis orthogonal touandw, we havevãu=0 andvãw=0. Therefore,
vã(u+w)=vãu+vãw=0+0=0 We conclude thatvis orthogonal tou+w.
4. Which is the projection ofvalongv: (a)v or (b)ev? solution The projection ofvalong itself isv, since
v||=vãv vãv
v=v Also, the projection ofvalongevis the same answer,v, because
v||= vãev
evãev
ev= vev=v
5. Letu||be the projection ofualongv. Which of the following is the projectionualong the vector 2vand which is the projection of 2ualongv?
(a) 12u|| (b) u|| (c) 2u||
solution Sinceu||is the projection ofualongv, we have, u||=uãv
vãv
v The projection ofualong the vector 2vis
uã2v 2vã2v
2v=
2uãv 4vãv
2v= 4uãv
4vãv
v=uãv vãv
v=u||
That is,u||is the projection ofualong 2v, so our answer is (b) for the first part. Notice that the projection ofualong vis the projection ofualong the unit vectorev, hence it depends on the direction ofvrather than on the length ofv.
Therefore, the projection ofualongvand along 2vis the same vector.
On the other hand, the projection of 2ualongvis as follows:
2uãv vãv
v=2
uãv vãv
v=2u||
giving us answer (c) for the second part.
6. Which of the following is equal to cosθ, whereθis the angle betweenuandv?
(a) uãv (b) uãev (c) euãev
solution By the Theorems on the Dot Product and the Angle Between Vectors, we have cosθ= uãv
uv= u uã v
v=euãev The correct answer is (c).
Exercises
In Exercises 1–12, compute the dot product.
1. 1,2,1 ã 4,3,5
solution Using the definition of the dot product we obtain
1,2,1 ã 4,3,5 =1ã4+2ã3+1ã5=15 3,−2,2 ã 1,0,1
3. 0,1,0 ã 7,41,−3 solution The dot product is
0,1,0 ã 7,41,−3 =0ã7+1ã41+0ã(−3)=41 1,1,1 ã 6,4,2
5. 3,1 ã 4,−7
solution The dot product of the two vectors is the following scalar:
3,1 ã 4,−7 =3ã4+1ã(−7)=5 1
6,12
ã 3,12 7. kãj
solution By the orthogonality ofjandk, we havekãj=0 kãk
9. (i+j)ã(j+k)
solution By the distributive law and the orthogonality ofi,jandkwe have
(i+j)ã(j+k)=iãj+iãk+jãj+jãk=0+0+1+0=1
May 16, 2011 (3j+2k)ã(i−4k) 11. (i+j+k)ã(3i+2j−5k)
solution We use properties of the dot product to obtain
(i+j+k)ã(3i+2j−5k)=3iãi+2iãj−5iãk+3jãi+2jãj−5jãk+3kãi+2kãj−5kãk
=3i2+2j2−5k2=3ã1+2ã1−5ã1=0
(−k)ã(i−2j+7k)
In Exercises 13–18, determine whether the two vectors are orthogonal and, if not, whether the angle between them is acute or obtuse.
13. 1,1,1, 1,−2,−2
solution We compute the dot product of the two vectors:
1,1,1 ã 1,−2,−2 =1ã1+1ã(−2)+1ã(−2)= −3 Since the dot product is negative, the angle between the vectors is obtuse.
0,2,4, −5,0,0 15. 1,2,1, 7,−3,−1
solution We compute the dot product:
1,2,1 ã 7,−3,−1 =1ã7+2ã(−3)+1ã(−1)=0 The dot product is zero, hence the vectors are orthogonal.
0,2,4, 3,1,0 17. 12
5,−45 , 1
2,−74
solution We find the dot product of the two vectors:
12 5,−4
5
ã 1
2,−7 4
= 12 5 ã1
2+
−4 5
ã
−7 4
= 12 10+28
20 = 13 5 The dot product is positive, hence the angle between the vectors is acute.
12,6, 2,−4
In Exercises 19–22, find the cosine of the angle between the vectors.
19. 0,3,1, 4,0,0
solution Since0,3,1 ã 4,0,0 =0ã4+3ã0+1ã0=0, the vectors are orthogonal, that is, the angle between them isθ=90◦and cosθ=0.
1,1,1, 2,−1,2 21. i+j, j+2k
solution We use the formula for the cosine of the angle between two vectors. Letv=i+jandw = j+2k. We compute the following values:
v = i+j =
12+12=√ 2 w = j+2k =
12+22=√ 5
vãw=(i+j)ã(j+2k)=iãj+2iãk+jãj+2jãk= j2=1 Hence,
cosθ= vãw
vw= 1
√2√
5 = 1
√10.
3i+k, i+j+k
In Exercises 23–28, find the angle between the vectors. Use a calculator if necessary.
23.
2,√ 2
, 1+√
2,1−√ 2 solution We writev=
2,√ 2
andw= 2,√
2
. To use the formula for the cosine of the angleθbetween two vectors we need to compute the following values:
v =√
4+2=√ 6 w =
(1+√
2)2+(1−√ 2)2=√
6 vãw=2+2√
2+√
2−2=3√ 2
Hence,
cosθ= vãw
vw = 3√
√ 2 6√
6=
√2 2 and so,
θ=cos−1
√2 2 =π/4 5,√
3
, √
3,2 25. 1,1,1, 1,0,1
solution We denotev= 1,1,1andw= 1,0,1. To use the formula for the cosine of the angleθbetween two vectors we need to compute the following values:
v =
12+12+12=√ 3 w =
12+02+12=√ 2 vãw=1+0+1=2 Hence,
cosθ= vãw
vw = 2
√3√ 2=
√6 3 and so,
θ=cos−1
√6 3 ≈0.615 3,1,1, 2,−4,2
27. 0,1,1, 1,−1,0
solution We denotev= 0,1,1andw= 1,−1,0. To use the formula for the cosine of the angleθbetween two vectors we need to compute the following values:
v =
02+12+12=√ 2 w =
12+(−1)2+02=√ 2 vãw=0+(−1)+0= −1 Hence,
cosθ= vãw
vw= −1
√2√ 2 = −1
2 and so,
θ=cos−1−1 2= 2π
3 1,1,−1, 1,−2,−1
29. Find all values ofbfor which the vectors are orthogonal.
(a) b,3,2, 1, b,1 (b) 4,−2,7, b2, b,0 solution
(a) The vectors are orthogonal if and only if the scalar product is zero. That is, b,3,2 ã 1, b,1 =0
bã1+3ãb+2ã1=0
4b+2=0 ⇒ b= −1 2 (b) We set the scalar product of the two vectors equal to zero and solve forb. This gives
4,−2,7 ã b2, b,0 =0 4b2−2b+7ã0=0
2b(2b−1)=0 ⇒ b=0 or b= 1 2
May 16, 2011
Find a vector that is orthogonal to−1,2,2.
31. Find two vectors that are not multiples of each other and are both orthogonal to2,0,−3.
solution We denote bya, b, c, a vector orthogonal to2,0,−3. Hence, a, b, c ã 2,0,−3 =0
2a+0−3c=0
2a−3c=0 ⇒ a= 3 2c Thus, the vectors orthogonal to2,0,−3are of the form
3 2c, b, c
.
We may find two such vectors by settingc=0,b=1 andc=2,b=2. We obtain v1= 0,1,0, v2= 3,2,2. Find a vector that is orthogonal tov= 1,2,1but not tow= 1,0,−1.
33. Findvãewherev =3,eis a unit vector, and the angle betweeneandvis2π3.
solution Sincevãe= vecos 2π/3, andv =3 ande =1, we havevãe=3ã1ã(−1/2)= −3/2.
Assume thatvlies in theyz-plane. Which of the following dot products is equal to zero for all choices ofv?
(a) vã 0,2,1 (b) vãk
(c) vã −3,0,0 (d) vãj
In Exercises 35–38, simplify the expression.
35. (v−w)ãv+vãw
solution By properties of the dot product we obtain
(v−w)ãv+vãw=vãv−wãv+vãw= v2−vãw+vãw= v2 (v+w)ã(v+w)−2vãw
37. (v+w)ãv−(v+w)ãw
solution We use properties of the dot product to write
(v+w)ãv−(v+w)ãw=vãv+wãv−vãw−wãw
= v2+wãv−wãv− w2= v2− w2
(v+w)ãv−(v−w)ãw
In Exercises 39–42, use the properties of the dot product to evaluate the expression, assuming thatuãv=2,u =1, andv =3.
39. uã(4v)
solution Using properties of the dot product we get
uã(4v)=4(uãv)=4ã2=8.
(u+v)ãv 41. 2uã(3u−v)
solution By properties of the dot product we obtain
2uã(3u−v)=(2u)ã(3u)−(2u)ãv=6(uãu)−2(uãv)
=6u2−2(uãv)=6ã12−2ã2=2 (u+v)ã(u−v)
43. Find the angle betweenvandwifvãw= −v w.
solution Using the formula for dot product, and the given equationvãw= −v w, we get:
v wcosθ= −v w, which implies cosθ= −1, and so the angle between the two vectors isθ=π.
Find the angle betweenvandwifvãw= 12v w.
45. Assume thatv =3,w =5 and that the angle betweenvandwisθ= π3.
(a) Use the relationv+w2=(v+w)ã(v+w)to show thatv+w2=32+52+2vãw.
(b) Findv+w.
solution For part (a), we use the distributive property to get:
v+w2=(v+w)ã(v+w)
=vãv+vãw+wãv+wãw
= v2+2vãw+ w2
=32+52+2vãw
For part (b), we use the definition of dot product on the previous equation to get:
v+w2=32+52+2vãw
=34+2ã3ã5ãcosπ/3
=34+15=49 Thus,v+w =√
49=7.
Assume thatv =2,w =3, and the angle betweenvandwis 120◦. Determine:
(a) vãw (b) 2v+w (c) 2v−3w
47. Show that ifeandfare unit vectors such thate+f = 32, thene−f =√27.Hint:Show thateãf = 18. solution We use the relation of the dot product with length and properties of the dot product to write
9/4= e+f2=(e+f)ã(e+f)=eãe+eãf+fãe+fãf
= e2+2eãf+ f2=12+2eãf+12=2+2eãf We now findeãf:
9/4=2+2eãf ⇒ eãf =1/8 Hence, using the same method as above, we have:
e−f2=(e−f)ã(e−f)=eãe−eãf−fãe+fãf
= e2−2eãf+ f2=12−2eãf+12=2−2eãf =2−2/8=7/4.
Taking square roots, we get:
e−f =
√7 2
Find2e−3fassuming thateandf are unit vectors such thate+f =√ 49. Find the angleθin the triangle in Figure 12. 3/2.
x y
(0, 10)
(10, 8)
(3, 2) FIGURE 12 solution We denote byuandvthe vectors in the figure.
x y
(0, 10)
(10, 8)
(3, 2) v
u
Hence,
cosθ= vãu
vu (1)
We find the vectorsvandu, and then compute their length and the dot productvãu. This gives v= 0−10,10−8 = −10,2
u= 3−10,2−8 = −7,−6 v =
(−10)2+22=√ 104 u =
(−7)2+(−6)2=√ 85
vãu= −10,2 ã −7,−6 =(−10)ã(−7)+2ã(−6)=58
May 16, 2011
Substituting these values in (1) yields
cosθ= 58
√104√
85 ≈0.617 Hence the angle of the triangle is 51.91◦.
Find all three angles in the triangle in Figure 13.
In Exercises 51–58, find the projection ofualongv.
51. u= 2,5, v= 1,1
solution We first compute the following dot products:
uãv= 2,5 ã 1,1 =7 vãv= v2=12+12=2 The projection ofualongvis the following vector:
u||=uãv vãv
v= 7 2v=
7 2,7
2
u= 2,−3, v= 1,2 53. u= −1,2,0, v= 2,0,1
solution The projection ofualongvis the following vector:
u||=uãv vãv
v We compute the values in this expression:
uãv= −1,2,0 ã 2,0,1 = −1ã2+2ã0+0ã1= −2 vãv= v2=22+02+12=5
Hence,
u||= −2
52,0,1 =
−4 5,0,−2
5
.
u= 1,1,1, v= 1,1,0 55. u=5i+7j−4k, v=k
solution The projection ofualongvis the following vector:
u||=uãv vãv
v We compute the dot products:
uãv=(5i+7j−4k)ãk= −4kãk= −4 vãv= v2= k2=1
Hence,
u||= −4
1 k= −4k u=i+29k, v=j
57. u= a, b, c, v=i