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.