25. Without loss of generality, we assume a points along the +x axis, and that b is at θ measured counter-
clockwise from a. We wish to verify that
r
2
= a
2
+ b
2
+ 2ab cos θ
where a =
|a| = a
x
(we’ll call it a for simplicity) and b =
|b| =
b
2
x
+ b
2
y
. Since
r = a + b then
r =
|r| =
(a + b
x
)
2
+ b
2
y
. Thus, the above expression becomes
(a + b
x
)
2
+ b
2
y
2
=
a
2
+
b
2
x
+ b
2
y
2
+ 2ab cos θ
a
2
+ 2ab
x
+ b
2
x
+ b
2
y
=
a
2
+ b
2
x
+ b
2
y
+ 2ab cos θ
which makes a valid equality since (the last term) 2ab cos θ is indeed the same as 2ab
x
(on the left-hand
side). In a later section, the scalar (dot) product of vectors is presented and this result can be revisited
with a somewhat different-looking derivation.