23. We consider

A with (x, y) components given by (A cos α, A sin α). Similarly,

B

→ (B cos β, B sin β). The

angle (measured from the +x direction) for their vector sum must have a slope given by

tan θ =

A sin α + B sin β

A cos α + B cos β

.

The problem requires that we now consider the orthogonal direction, where tan θ + 90

◦

=

− cot θ. If this

(the negative reciprocal of the above expression) is to equal the slope for their vector difference, then we
must have

−

A cos α + B cos β

A sin α + B sin β

=

A sin α

− B sin β

A cos α

− B cos β

.

Multiplying both sides by A sin α + B sin β and doing the same with A cos α

− B cos β yields

A

2

cos

2

α

− B

2

cos

2

β = A

2

sin

2

α

− B

2

sin

2

β .

Rearranging, using the cos

2

φ + sin

2

φ =1 identity, we obtain

A

2

= B

2

=

⇒ A = B .

In a later section, the scalar (dot) product of vectors is presented and this result can be revisited with
a more compact derivation.