The concept of relative velocity can be easily extended to include motion in a plane or in three dimensions. Suppose that two objects A and B are moving with velocities $\mathbf{v}_{\mathbf{A}}$ and $\mathbf{v}_{\mathbf{B}}$ (each with respect to some common frame of reference, say ground.). Then, the velocity of object A relative to that of B is:
$
\mathbf{v}_{\mathrm{AB}}=\mathbf{v}_{\mathrm{A}}-\mathbf{v}_{\mathrm{B}}
$
and similarly, the velocity of object B relative to that of $\mathrm{A}$ is :
$
\mathbf{v}_{B A}=\mathbf{v}_{B}-\mathbf{v}_{\mathrm{A}}
$
Therefore, $\mathbf{v}_{\mathrm{AB}}=-\mathbf{v}_{\mathrm{BA}}$
and, $\left|\mathbf{v}_{\mathrm{AB}}\right|=\left|\mathbf{v}_{\mathrm{BA}}\right|$

Example: Rain is falling vertically with a speed of $35 \mathrm{~m} \mathrm{~s}^{-1}$. A woman rides a bicycle with a speed of $12 \mathrm{~m} \mathrm{~s}^{-1}$ in east to west direction. What is the direction in which she should hold her umbrella?

Solution: In Figure 4.9 below $\mathbf{v}_{\mathrm{r}}$ represents the velocity of rain and $\mathbf{v}_{b}$, the velocity of the bicycle, the woman is riding. Both these velocities are with respect to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by her is the velocity of rain relative to the velocity of the bicycle she is riding.

Figure-4.9

That is $\mathbf{v}_{\mathrm{rb}}=\mathbf{v}_{\mathrm{r}}-\mathbf{v}_{\mathrm{b}}$ This relative velocity vector as shown in Fig. $4.9$ makes an angle $\theta$ with the vertical. It is given by
$
\tan \theta=\frac{v_{b}}{v_{r}}=\frac{12}{35}=0.343
$
Or, $\quad \theta \cong 19^{\circ}$
Therefore, the woman should hold her umbrella at an angle of about $19^{\circ}$ with the vertical towards the west.

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