# Answer to Question #862 in Mechanics | Relativity for duha

Question #862

A radar station locates a sinking ship at range 15.6 km and bearing 136° clockwise from north. From the same station, a rescue plane is at horizontal range 19.6 km, 160° clockwise from north, with elevation 2.10 km.

(a) Write the displacement vector from plane to ship, letting represent i east, j north, and k up.

(b) How far apart are the plane and ship?

(a) Write the displacement vector from plane to ship, letting represent i east, j north, and k up.

(b) How far apart are the plane and ship?

Expert's answer

i - east, j - north, k - up:

projections for the ship:

S

S

S

Thus the vector coordinates would be (-11.2, 10.8, 0)

for the plane:

P

P

the coordinates in vector form are (-18.1, 6.6, 2.1)

The displacement vector from plane to ship equals to

S - P =(-11.2, 10.8, 0) - (-18.1, 6.6, 2.1) = (6.9, 4.2, -2.1)

The distance equals to the modulus of this vector:

sqrt(6.9

projections for the ship:

S

_{i}= - 15.6 * cos(180°-136°) = - 11.2;S

_{j}= 15.6 * sin(136°) = 10.8.S

_{k}= 0;Thus the vector coordinates would be (-11.2, 10.8, 0)

for the plane:

P

_{i}= - 19.6 * cos(180° - 160°) = - 18.1;P_{j}= 19.6 * sin (160°) = 6.6;P

_{k}= 2.1the coordinates in vector form are (-18.1, 6.6, 2.1)

The displacement vector from plane to ship equals to

S - P =(-11.2, 10.8, 0) - (-18.1, 6.6, 2.1) = (6.9, 4.2, -2.1)

The distance equals to the modulus of this vector:

sqrt(6.9

^{2}+4.2^{2}+(-2.1)^{2}) = sqrt(69.66) = 8.43 km
## Comments

Assignment Expert19.02.14, 16:4519.6*cos(20)=-7.99

Gary11.02.14, 04:31-19.6 * cos 20 does not equal to -18.1

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