A radar station locates a sinking ship at range 173 mathrmkm...

A radar station locates a sinking ship at range 17.3 $\mathrm{km}$ and bearing $136^{\circ}$ clockwise from north. From the same station, a rescue plane is at horizontal range $19.6 \mathrm{km},$ $153^{\circ}$ clockwise from north, with elevation 2.20 $\mathrm{km}$ . (a) Write the position vector for the ship relative to the plane, letting i represent east, \hat{j} north, and $\hat{\mathbf{k}}$ up. (b) How far apart are the plane and ship?

Transcript

00:01 In this problem, a ship is sinking 17 .3 kilometers away from a radio tower, 136 degrees off of north, clockwise, off of north.

00:13 And there's a rescue plane that's 19 .6 kilometers away from the radio tower, 153 degrees clockwise from north.

00:21 And we want to determine this vector between the location of the plane and the location of the ship.

00:29 It's also important to note that the plane is flying 2 .2 kilometers above sea level, and the 19 .6 kilometers is just towards its shadow, the sea level projection of the airplane.

00:45 So to do this, what we're going to do is we're going to break it down into components.

00:50 We're going to find this y value of the ship and the x value of the ship, and we're going to compare that to the x value of the plane.

01:02 And this y value of the plane.

01:09 So i'm going to draw some triangles.

01:12 So this is a radio tower.

01:14 Then draw it just to the ship.

01:20 There we go.

01:21 So we know this here is this angle, which is 136 minus the angle between north and east, which is 90 degrees.

01:31 So 136 degrees minus 90 degrees is 46 degrees.

01:37 So this is a 46 degree angle and this is 17 .3 kilometers long.

01:42 We want to find this x1 and y1 will say.

01:50 To do this it's fairly simple trigonometry.

01:54 We can use um so kottoa, which is a acronym i like quite a bit.

02:00 So for the adjacent component the x it's um we're using product because it's a a unit and we have the hypotenuse.

02:09 So it's, it's, the cosine of 46 degrees is equal to adjacent over our partners x1 over 17 .3.

02:18 Or we can say that x1 here is equal to 17 .3 times cosine of 46 degrees.

02:27 And for y, we'll use sign.

02:32 Sign of 46 degrees is equal to y1 over 17 .3.

02:39 Or we can say y1 is equal to 17 .3 sine of 46 degrees.

02:46 And we can solve these using calculator, 17 .3 times cosine of 46 is equal to about 12 .02 kilometers.

02:58 So this here is 12 .02 kilometers and 17 .36 or 17 .3 times sine of 46 degrees is equal to 12 .44 kilometers.

03:15 Again, this is 12 .44 kilometers.

03:20 That is the x and y of our boat and the boat said sea level.

03:28 So we can we know that the z if we want to keep track of it is equal to zero.

03:34 Now for the plane we'll do the same thing.

03:41 This is actually a bit bigger this time, which is 153 minus 90.

03:47 So that is 63 degrees...

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