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.