We need to determine the standard angle #theta# for the respective airplane; this will make it easy to use the horizontal distance #d# to compute the #hati# and #hatj# components of the vector for the plane. The #hatk# is easy because it is given as the as the altitude in meters. For any given airplane, #n#, the vector is as follows:
#vecv_n = d_ncos(theta_n)hati+d_nsin(theta_n)hatj+"altitude"_nhatk#
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:
#theta_1 = 180^@+38^@#
#theta_1 = 218^@#
Convert the distance from km to meters:
#d_1 = 19.8" km"#
#d_1 = 19800" m"#
#"altitude"_1 = 900" m"#
#vecv_1 = (19800cos(218^@)" m")hati+(19800sin(218^@)" m")hatj+(900" m")hatk#
#vecv_1= -15602hati -12190hatj + 900hatk# 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:
#theta_2 = 270^@-23^@#
#theta_2 = 247^@#
Convert the distance from km to meters:
#d_2 = 17.7" km"#
#d_2 = 17700" m"#
#"altitude"_2 = 1100" m"#
#vecv_2 = (17700cos(247^@)" m")hati+(17700cos(247^@)" m")hatj+(1100" m")hatk#
#vecv_2 = -6916hati -16293hatj + 1100hatk# in meters
The vector from airplane 1 to airplane 2# is:
#vecv_(1to2) = vecv_2-vecv_1#
#vecv_(1to2) = (-6916- -15602)hati+ (-16293- -12190)hatj+ (1100-900)hatk#
#vecv_(1to2) = 8686hati-4103hatj+ 200hatk# in meters
The distance between the planes is the magnitude of the vector:
#|vecv_(1to2)| = sqrt(8686^2+ (-4103)^2+ 200^2)# in meters
#|vecv_(1to2)| = 9608# meters
