We need to determine the standard angle θ for the respective airplane; this will make it easy to use the horizontal distance d to compute the ˆi and ˆj components of the vector for the plane. The ˆk is easy because it is given as the as the altitude in meters. For any given airplane, n, the vector is as follows:
→vn=dncos(θn)ˆi+dnsin(θn)ˆj+altitudenˆk
For the first airplane convert given heading to a standard angle with the reference point as East.
For airplane 1 we are given 38∘ South of West
West is 180∘ from East and add 38∘ to make it move toward the South:
θ1=180∘+38∘
θ1=218∘
Convert the distance from km to meters:
d1=19.8 km
d1=19800 m
altitude1=900 m
→v1=(19800cos(218∘) m)ˆi+(19800sin(218∘) m)ˆj+(900 m)ˆk
→v1=−15602ˆi−12190ˆj+900ˆk in meters
For the second airplane convert given heading to a standard angle with the reference point as East.
For airplane 2 we are given 23∘ West of South
South is 270∘ from East and subtract 23∘ to make it move toward the West:
θ2=270∘−23∘
θ2=247∘
Convert the distance from km to meters:
d2=17.7 km
d2=17700 m
altitude2=1100 m
→v2=(17700cos(247∘) m)ˆi+(17700cos(247∘) m)ˆj+(1100 m)ˆk
→v2=−6916ˆi−16293ˆj+1100ˆk in meters
The vector from airplane 1 to airplane 2# is:
→v1→2=→v2−→v1
→v1→2=(−6916−−15602)ˆi+(−16293−−12190)ˆj+(1100−900)ˆk
→v1→2=8686ˆi−4103ˆj+200ˆk in meters
The distance between the planes is the magnitude of the vector:
∣∣→v1→2∣∣=√86862+(−4103)2+2002 in meters
∣∣→v1→2∣∣=9608 meters
