What is the displacement vector FROM the first plane TO the second plane, letting hatiˆi represent east, hatjˆj north, and hatkˆk up?

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1 Answer
Feb 5, 2018

(a) vecv_(1to2) = 8686hati-4103hatj+ 200hatkv12=8686ˆi4103ˆj+200ˆk in meters

(b) |vecv_(1to2)| = 9608v12=9608 meters

Explanation:

We need to determine the standard angle thetaθ for the respective airplane; this will make it easy to use the horizontal distance dd to compute the hatiˆi and hatjˆj components of the vector for the plane. The hatkˆk is easy because it is given as the as the altitude in meters. For any given airplane, nn, the vector is as follows:

vecv_n = d_ncos(theta_n)hati+d_nsin(theta_n)hatj+"altitude"_nhatkvn=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^@38 South of West

West is 180^@180 from East and add 38^@38 to make it move toward the South:

theta_1 = 180^@+38^@θ1=180+38

theta_1 = 218^@θ1=218

Convert the distance from km to meters:

d_1 = 19.8" km"d1=19.8 km

d_1 = 19800" m"d1=19800 m

"altitude"_1 = 900" m"altitude1=900 m

vecv_1 = (19800cos(218^@)" m")hati+(19800sin(218^@)" m")hatj+(900" m")hatkv1=(19800cos(218) m)ˆi+(19800sin(218) m)ˆj+(900 m)ˆk

vecv_1= -15602hati -12190hatj + 900hatkv1=15602ˆi12190ˆj+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^@23 West of South

South is 270^@270 from East and subtract 23^@23 to make it move toward the West:

theta_2 = 270^@-23^@θ2=27023

theta_2 = 247^@θ2=247

Convert the distance from km to meters:

d_2 = 17.7" km"d2=17.7 km

d_2 = 17700" m"d2=17700 m

"altitude"_2 = 1100" m"altitude2=1100 m

vecv_2 = (17700cos(247^@)" m")hati+(17700cos(247^@)" m")hatj+(1100" m")hatkv2=(17700cos(247) m)ˆi+(17700cos(247) m)ˆj+(1100 m)ˆk

vecv_2 = -6916hati -16293hatj + 1100hatkv2=6916ˆi16293ˆj+1100ˆk in meters

The vector from airplane 1 to airplane 2# is:

vecv_(1to2) = vecv_2-vecv_1v12=v2v1

vecv_(1to2) = (-6916- -15602)hati+ (-16293- -12190)hatj+ (1100-900)hatkv12=(691615602)ˆi+(1629312190)ˆj+(1100900)ˆk

vecv_(1to2) = 8686hati-4103hatj+ 200hatkv12=8686ˆi4103ˆj+200ˆk in meters

The distance between the planes is the magnitude of the vector:

|vecv_(1to2)| = sqrt(8686^2+ (-4103)^2+ 200^2)v12=86862+(4103)2+2002 in meters

|vecv_(1to2)| = 9608v12=9608 meters