Plane to Ship Displacement (using component method)

AI Thread Summary
The discussion focuses on calculating the vector displacement from a plane to a ship using the component method. The initial calculations for the ship's and plane's coordinates were corrected, emphasizing the use of sine and cosine functions for determining the x (east) and y (north) components. After adjustments, the displacement vector was determined to be (-3.20i, -8.41j, 2.06k) km, indicating the direction and magnitude needed to reach the ship from the plane. A final clarification highlighted that the displacement should be expressed as (3.2i, 8.41j, -2.06k) km, reflecting the correct movement direction. The conversation underscores the importance of accurately applying trigonometric functions in vector calculations.
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Homework Statement


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.

Homework Equations


The Attempt at a Solution


Shipx=15.4cos123 = -8.39i km
Shipy=15.4sin123 = 12.92j kmPlanex=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

Any help you could give me would be greatly appreciated!
 
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Shipx=15.4cos123 = -8.39i km
Shipy=15.4sin123 = 12.92j km
You seem to have these reversed; the sine gives the i (east) value.
 
Delphi51 said:
You seem to have these reversed; the sine gives the i (east) value.

Really? So Sin is for the X direction, and cos is for the Y? That's not what my textbook says?
 
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I hate to make a general statement about this with angles greater than 90 degrees.
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.
 
Delphi51 said:
I hate to make a general statement about this with angles greater than 90 degrees.
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.
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??
 
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Therefore Displacement PtoS = (P-S)
=(-3.20i , -8.41j , 2.06k) km
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).
 
Delphi51 said:
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).

Makes sense. Thanks again for the help! :)
 
Most welcome.
 
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