13.3 The Dot Product
1.Definition (Dot product) Suppose u =< a1 , b1 , c1 > and v =< a2 , b2 , c2 > are vectors in R3 . The dot product of u
and v is the scalar defined as follows
u · v =< a1 , b1 , c1 > · < a2 , b2 , c2 >= a1 a2 + b1 b2 + c1 c2
(e.g) Let u =< −1, 1, 0 > and v =< 3, 3, 1 >. Find
(a) u · v
(b) v · u
(c) (u + v) · (u − v)
(d) (3u + v) · (−u + v)
2.Properties Suppose u, v, w are vectors and α, β are scalars. Then the following properties hold.
(a) u · v = v · u
(b) u · (v + w) = u · v + u · w
(c) (αu) · v = u · (αv) = α(u · v)
(Hence, we simply write αu · v without worrying about the order of operations.)
(d) u · (αv + βw) = αu · v + βu · w
(e) u · u = kuk2
(Proof) Suppose u =< a1 , b1 , c1 >, v =< a2 , b2 , c2 >, and w =< a3 , b3 , c3 >.
(a) u · v = a1 a2 + b1 b2 + c1 c2 = a2 a1 + b2 b1 + c2 c1 = v · u
(b) u · (v + w) = a1 (a2 + a3 ) + b1 (b2 + b3 ) + c1 (c2 + c3 ) = (a1 a2 + b1 b2 + c1 c2 ) + (a1 a3 + b1 b3 + c1 c3 ) = u · v + u · w
(c) (αu)·v =< αa1 , αb1 , αc1 > · < a2 , b2 , c2 >= (αa1 )a2 +(αb1 )b2 +(αc1 )c2 =

 a1 (αa2 ) + b1 (αb2 ) + c1 (αc2 ) = u · (αv)
 α(a a + b b + c c ) = α(u · v)
1 2
1 2
1 2
(d) u · u = a21 + b21 + c21 =
q
a21 + b21 + c21
2
= kuk2
3.Example Let u =< −1, 1, 0 > and v =< 3, 3, 1 >. Use the properties of dot products to evaluate
(a) (u + v) · (u − v)
(b) (3u + v) · (−u + v)
(Answers)
(a) u · u − u · v + v · u − v · v = u · u − v · v = 2 − 19 = −11
(b) −3u · u + 3u · v − v · u + v · v = −3u · u + 2u · v + v · v = −6 + 0 + 19 = 13.
1
4.Theorem Let u and v be vectors in R2 or R3 and θ be the smaller angle between them. Then the following formula
provides us with another way of evaluating u · v :
u · v = kukkvk cos θ
(Proof) By the law of cosines,
ku − vk2
=
kuk2 + kvk2 − 2kukkvk cos θ
(u − v) · (u − v)
=
kuk2 + kvk2 − 2kukkvk cos θ
u · u − 2u · v + v · v
=
kuk2 + kvk2 − 2kukkvk cos θ
−2u · v
=
−2kukkvk cos θ
u·v
=
kukkvk cos θ
5.Example Find the angle between the vectors :
(a) a =< 6, 0, 2 >, b =< 5, 3, −2 >
(b) u = i + j + 2k, v = 2j − 3k
(Answers)
(a) a · b = 30 − 4 =
√
40
(b) u · v = 0 + 2 − 6 =
√
√
;
38 cos θ
1+1+4
√
cos θ = √
0 + 4 + 9 cos θ
26
√
40
;
38
;
θ = cos−1
13
√
2 95
−4
cos θ = √ √
6 13
;
θ = cos−1
4
−√
78
6.Definition (Orthogonal or perpendicular vectors) We say two nonzero vectors u and v are orthogonal if
u·v =0
7.Definition (Direction angles, direction cosines)
Let v be a nonzero vector in R3 . The angles α, β, γ in the interval [0, π]
between v and the positive x, y, and z axes, respectively, are called the
direction angles of v.
When α, β, and γ are direction angles of a vector v, we call cos α, cos β and
cos γ the direction cosines of v.
8.Properties Let v =< a, b, c > be a nonzero vector in R3 and α, β, γ be direction angles of v.
(a) We can achieve the angles α, β, γ by finding the angles between v and i, j, k respectively. That is,

v·i









v·j









v·k
=
kvk kik cos α
=
kvk kjk cos β
=
kvk kkk cos γ
⇒


cos α








cos β








 cos γ
(b) As a consequence of (a), we achieve the identity
2
=
√
=
√
=
√
a
a2 + b2 + c2
a2
b
+ b2 + c2
a2
c
+ b2 + c2
cos2 α + cos2 β + cos2 γ = 1
(c) The unit vector in the direction of v is
v
1
< a, b, c >=< cos α, cos β, cos γ >
=√
2
kvk
a + b2 + c2
9.Example Find the direction angles of the vector and use them to find the unit vector in the direction of the vector.
(a) p =< 1, 1, 1 >
(b) q =< 0, π, 3 >
−−→
(c) AB when A(−1, 2, 0) and B(0, 0, 3)
(Answers)
1
1
cos α = cos β = cos γ = √
; α = β = γ = cos−1 √
3
3

π
−1
cos α = 0
; α = cos (0) =



2






π
π
q
1
cos β = √
; β = cos−1 √
(b)
= √
< 0, π, 3 > ;
π2 + 9
π2 + 9

kqk
π2 + 9






3
3


; γ = cos−1 √
cos γ = √
π2 + 9
π2 + 9

1
−1

√
 α = cos


14





−
−
→

2
1
AB
−
−
→
β
= cos−1 − √
(c) AB =< 1, −2, 3 > ;
< 1, −2, 3 > ;
−
−
→ = √

14
14

kABk





3


 γ
= cos−1 √
14
(a)
p
1
= √ < 1, 1, 1 >
kpk
3
;
10.Definition (Projections)
Let u, v be vectors in R2 or R3 with the same initial point P and distinct
terminal points R, Q respectively. Let S be the perpendicular foot from R to
the line through P Q. Then the vector P S is called the projection of v onto u
and it is denoted by
proju v
11.Theorem Let u and v be vectors in R2 or R3 . then
(a) proju v =
u·v
u
u·u
(b) v can be written as the sum of a vector parallel to u and a vector perpendicular to u in the following
way
v = proju v + (v − proju v),
3
where
proju v \\ u
v − proju v ⊥ u
(Proof)
u
is the unit vector in the direction of u. Since the length of the projection proju v is kvk cos θ where θ
kuk
is the smaller angle between u and v,
(a) Notice
proju v
=
(kvk cos θ)
kukkvk cos θ
u·v
u
=
u=
u
kuk
kuk2
u·u
the formula in (a) is proved.
(b) It suffices to show that v − proju v is perpendicular to u. It is seen because
(v − proju v) · u
=
v · u − proju v · u = v · u −
u·v
u·u=v·u−v·u=0
u·u
12.Definition (Component of v along u) We define the component of v along u, denoted by compu v, as
compu v =
u·v
kuk
Notice if u and v make an acute angle compu v = kproju vk, and if u and v make an obtuse angle
compu v = −kproju vk.
13.Example Find the projection of v onto u. Then decompose v into the sum of a vector parallel to u and a perpendicular to u.
(a) u =< −3, 1, 2 >, v =< 1, −1, 2 >
(b) u = i + k, v = i − j
(Answers)
0
v·u
u= u=0
u·u
6
v·u
1
(b) proju v =
u=
< 1, 0, 1 >
u·u
2
(a) proju v =
4