Two velocities `vecv_1 and vecv_2` are 3m/s towards north and 4m/s towards east , respectively. Find `vecv_1-vecv_2`.

Explanation

Let us choose , x -axis along east and y-axis along north.
`therefore vec(v_1)=3hatjm//s and vec(v_2)=4hatim//s`
`therefore vecv_1-vecv_2=-4hati+3hatj` (between west and north )
`|vecv_1-vecv_2|=sqrt((-4)^2+3^2)=5m//s`
If `vecv_1-vecv_2` makes an angle `theta` with east, then
`tan theta =3/(-4)=tan 143.1^@=tan(180^@-36.9^@)`
`therefore vecv_1-vecv_2` is inclined at an angle of `143.1 ^@` north of east, i.e.,`36.9^@` north of west.