Example 6-4. Given the vectors
A= 2 ˆe 1 + 3 ˆe 2 + 6 ˆe 3 and B =ˆe 1 + 2 ˆe 2 + 2 ˆe 3
Find:
(a) |A|, |B|, A·B, |A+B|
(b) The angle between the vectors A and B
(c) The direction cosines of A and B
(d) A unit vector in the direction C =A−B.
Solution
(a) |A|=
√
(2)^2 + (3)^2 + (6)^2 =
√
49 = 7
|B|=
√
(1)^2 + (2)^2 + (2)^2 =
√
9 = 3
A·B = (2)(1) + (3)(2) + (6)(2) = 20
A+B = 3 ˆe 1 + 5 ˆe 2 + 8 ˆe 3
|A+B|=
√
(3)^2 + (5)^2 + (8)^2 =
√
98
(b) A·B =|A||B|cos θ =⇒ cos θ=
A·B
|A||B|
=^20
7 · 3
=^20
21
θ= arccos (^2021 ) = 0. 3098446 radians = 17. 753 degrees
or one can determine that
tan θ=
√
(21)^2 −(20)^2
20
=
√
41
20
=⇒ θ= 0. 3098446 radians
(c)A unit vector in the direction of the vector A is obtained
by multiplying A by the scalar |A^1 |to obtain
ˆeA=
A
|A|
= cos α 1 ˆe 1 + cos β 1 eˆ 2 + cos γ 1 eˆ 3 =^2
7
ˆe 1 +^3
7
ˆe 2 +^6
7
ˆe 3
which implies the direction cosines are cos α 1 =^2
7
, cos β 1 =^3
7
, cos γ 1 =^6
7
In a similar
fashion one can show ˆeB=
B
|B|
= cos α 2 ˆe 1 + cosβ 2 eˆ 2 + cos γ 2 ˆe 3 =^1
3
ˆe 1 +^2
3
ˆe 2 +^2
3
ˆe 3 which
implies the direction cosines are cos α 2 =
1
3 , cos β^2 =
2
3 , cos γ^2 =
2
3
(d) C=A−B =ˆe 1 +ˆe 2 + 4 ˆe 3 and |C|=|A−B|=
√
(1)^2 + (1)^2 + (4)^2 =
√
18 = 3
√
2 Unit
vector in direction of C is ˆeC=
C
|C|
=eˆ^1 +eˆ^2 + 4 ˆe^3
3
√
2
.Make note of the fact that the
sum of the squares of the direction cosines equals unity.
