Answer
(a) $\vec {v_{12}}=12m/s$ and the direction is $53^{\circ}$ north of east.
(b) $\vec {v_{21}}=12m/s$ and the direction is $53^{\circ}$ south of west.
Work Step by Step
(a) First of all, we determine the components of $\vec v_{12}$.
$\vec v_{12}$ represents the velocity of plane 1 relative to plane 2.
$\vec v_{12}=(12m/s)hat y-[(7.5m/s)cos20^{\circ}\hat x+(7.5m/s)sin 20^{\circ}\hat y]$
$\vec v_{12}=(7.0m/s)\hat x+(9.4m/s)\hat y$
Now we can find the direction of $\vec v_{12}$
$\theta=tan^{-1}(\frac{9.4}{7.0})=53^{\circ}$ (north of east).
and the magnitude is given as:
$v_{12}=\sqrt{(v_{12,x})^2+(v_{12,y})^2}$
We plug in the known values to obtain:
$v_{12}=\sqrt{(7.0)^2+(9.4)^2}$
$v_{12}=12m/s$
(b) In this case $\vec {v_{21}}=-\vec {v_{12}}$ which means that the magnitude remains the same while the direction is reversed; that is, $\vec {v_{21}}=12m/s$ and the direction is south of west.
