A radar station locates a sinking ship at range 17.3 km and bearing 136 degrees clockwise from north. From the same station a rescue plane is at horizontal range 19.2 km, 153 degrees clockwise from north, with elevation 2.20 km. (a) The vector displacement from plane to ship can be written in the form, a1ˆı + a2ˆ + a3kˆ, where ˆı represents east, ˆ represents north, and kˆ represents up. Find the values of (a) a1, (km) (b) a2, (km) (c) a3. (km) (d) Also, find the distance between the plane and the ship (km).

a1 = 3.28 km

a2 = 4.67 km

a3 = -2.20 km

D = 6.08 km

Explanation:

The position vector from the radar station to the ship will be

S = (17.3 sin 136i + 17.3 cos 136j) km

S = (12i + -12.44j) km

Now, the position vector from the station to the plane will be

S' = (19.2 sin 153i + 19.2 cos 153j + 2.2k) km

S' = (8.72i + -17.11j + 2.2k) km

Finally, in order to fly the ship, the displacement needed is S - S'

Displacement, D = (3.28i + 4.67j - 2.2k) km.

Therefore, the values are

a1 = 3.28 km

a2 = 4.67 km

a3 = -2.20 km

The distance between the plane and the ship is given by

D' = √[(3.28² + 4.67² + (-2.2)²]

D' = √(10.7584 + 21.8089 + 4.4)

D' = √36.9673

D' = 6.08 km

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr