# Homework Help: Vector displacement between ship and plane

1. Oct 16, 2009

### waldvocm

1. A radar station locates a ship at range 17.3km and bearing 136 degrees clockwise from north. From the same station a rescue plane is at horizontal range 21.5km, 153 degrees clockwise from north, with elevation 2.20km. The vector displacement from plane to ship can be written in the form, a1i+a2j+a3k, where i is east, j is north, k is up. Find the values of a1, a2 a3 and the distance between the plane and ship.

2.Ax=Acos(-136)=-12
Ay=Asin(-136)=-12
Bx=Bcos(-153)=-19
By=Bsin(-153)=-9
Cx= Do I find the "up" components the same way? I know the elevation is 2.20km, is the angle 90 degrees?

3. The attempt at a solution

Last edited: Oct 16, 2009
2. Oct 16, 2009

### Andrew Mason

Write the co-ordinates for each displacement vector. For the ship it is (x1, y1, 0) and for the plane it is (x2, y2, 2.2) where x1 = 17.3cos(-46) and y1 = 17.3 sin(-46). x2 =? and y2 =?

Then it is just a matter of subtracting the two vectors to find the difference. Use pythagoras to calculate the length of the resulting difference vector.

AM

3. Oct 16, 2009

### waldvocm

A radar station locates a ship at range 17.3km and bearing 136 degrees clockwise from north. From the same station a plane is at horizontal range 21.5km, 153 degrees clockwise from north, with elevation 2.20km. The vector displacement from plane to ship can be written in the form, a1i+a2j+a3k, where i is east, j is north, k is up. Find the values of a1, a2, a3 and the distance between the plane and the ship.

Ax=Acos(-135)=-12
Ay=Asin(-135)=-12
Bx=Bcos(-153)=-19
By=Bsin(-153)=-9

How so I find the components of the third vector? I know the magnitude is 2.20km up. Is the angle 90 degrees? Cx=Ccos(90)=0 Cy=Csin(90)=2.2

What is my next step? Do is add all of the x components for the a1i value, the y for a2j and what for k?

a1i=-12+-19+0=-31
a2j=-12+-9+2.2=-19
a3k=???????????????

OPPPS! I thought my original question didn't show up

4. Oct 16, 2009

### waldvocm

ship=(12,-12,0)
plane=(9,-19,2.2)

a1i=12-9=3
a2j=-12--19=7
a3k=0-2.2=-2.2

The distance would be.............a^2+b^2=c^2 17.3^2+23.7^2=c^2 = The distance between the plane and ship = 29.34km

Is this correct?

5. Oct 16, 2009

### waldvocm

ok, I am rethinking my distance.

I know that one side must be 2.20 since that is the side that is creating the right triangle. I am confused as far as the other sides to use in a^2+b^2=c^2.

6. Oct 16, 2009

### waldvocm

ok to find the magnitude of the displacement do I take

R=3i+7j-2.2k

R=sqrt3^2+7^2+-2.2^2=

Magnitude of displacement is 2.79

2.20^2+2.79^2=c^2 The distance =3.55km

Is this correct?

7. Oct 16, 2009

### Andrew Mason

First you have to determine the correct (x,y,z) co-ordinates of the ship and the plane.

136 deg clockwise from north is an angle of -46 degrees using the direction of the x axis as 0 degrees. So the x co-ordinate is 17.3 cos(-46) = +12 km. The y co-ordinate is 17.3 sin(-46) = -12.4 km. The z co-ordinate is 0. So the ship vector is (12, -12.4, 0)

For the plane it is a little trickier. The z component is easy: 2.2 km. To resolve the x and y components you use the horizontal range (ie. this is the projection of the vector from the origin to the plane onto the x,y plane). So the x co-ordinate is 21.5 cos(-63) = +9.8 km. The y co-ordinate is 21.5 sin(-63) = -19.2 km. The z co-ordinate is 2.2. So the plane's vector is (9.8, -19.2, 2.2)

To determine displacement of the ship vector from the plane vector just subtract the plane's co-ordinates from the ships (ie. this difference vector when added to the plane's vector results in the ship's vector):

Ship vector - plane vector = plane-to-ship vector

(12, -12.4, 0) - (9.8, -19.2, 2.2) = (?, ?, ?)

The length of that vector is now very easy to calculate. I get 7.48 km.

AM