# Plane to Ship Displacement (using component method)

1. Sep 27, 2009

### joeseppe

1. The problem statement, all variables and given/known data
A coastguard station locates a ship at range 15.4 km and bearing 123° clockwise from north.

From the same station a plane is at horizontal range 19.4 km, 150° clockwise from north, with elevation 2.06 km.

What is the vector displacement from plane to ship, let i represent east, j north, and k up.

2. Relevant equations

3. The attempt at a solution
Shipx=15.4cos123 = -8.39i km
Shipy=15.4sin123 = 12.92j km

Planex=19.4cos150 = -16.8i km
Planey=19.4sin150 = 9.7j km
Planez=2.06 k km

Therefore Displacement PtoS = (P-S)
=(-8.41i , -3.22j , 2.06k) km

Last edited: Sep 27, 2009
2. Sep 27, 2009

### Delphi51

You seem to have these reversed; the sine gives the i (east) value.

3. Sep 27, 2009

### joeseppe

Really? So Sin is for the X direction, and cos is for the Y? That's not what my textbook says?

Last edited: Sep 27, 2009
4. Sep 27, 2009

### Delphi51

But if you draw the diagram and note that the ship is 57 degrees away from the south line, then you would naturally say that sin(57) = x/15.4 so x = 15.4*sin(57) = 12.9 to the east.

5. Sep 27, 2009

### joeseppe

Yeah I did draw a diagram, I just don't have a scanner to upload it.
I never thought to look at it as two triangles though.

So what I actually should have is:

Shipx=15.4sin57 = 12.9i
Shipy=15.4cos57 = -8.39j

Planex=19.4sin30 = 9.7i km
Planey=19.4cos30 = -16.8j km
Planez=2.06k km

Therefore Displacement PtoS = (P-S)
=(-3.20i , -8.41j , 2.06k) km

Right??

Last edited: Sep 27, 2009
6. Sep 27, 2009

### Delphi51

Hmm, from the plane you would have to go east, north and down to get to the ship.
Therefore I think it should be (3.2i, 8.41j, -2.06k).

7. Sep 27, 2009

### joeseppe

Makes sense. Thanks again for the help! :)

8. Sep 27, 2009

### Delphi51

Most welcome.