# Homework Help: Finding the distance using vectors

1. Sep 20, 2008

### physics19

Oasis B is 25 km due east of oasis A. Starting from oasis A, a camel walks 24 km in a direction 19° south of east and then walks 5.5 km due north. How far is the camel then from oasis B?

For this problem I tried to set 3 different vectors. I named the vector A,B,C. Two of the vectors had only a y component while another had both x and y components. After I found out what all the components were I used the data to try to find the x and y components of vector r. After this I took the magnitude of the vector. I thought that I was doing the correct procedure but I am not coming up with the correct response. Help would be greatly appreciated.

2. Sep 20, 2008

### cepheid

Staff Emeritus
I would do this question in two steps.

1. Figure out how far east the camel went. Figure out how far south the camel went (including the northward part). This involves resolving his displacement vector (whose magnitude and direction you are given) into eastward and southward components.

2. Based on these numbers you know:

how much farther east he has to go to get to oasis B

how much farther north he has to go to get to oasis B

So in other words, you know the components of the vector that describes his displacement from point B. Now you can just use pythagoras to figure out the magnitude of that vector (which is the distance to oasis B). Make sure you draw a picture, it really helps.

3. Sep 20, 2008

### LowlyPion

Isn't it really an A, B, C, D ? Where:

$$\vec{A} + \vec{B} + \vec{C} = \vec{D}$$

You are given A and B and D. Hence you need to solve for C which is the remaining vector to get you to your destination D.

$$\vec{C} = \vec{D} + (-\vec{A}) + (-\vec{B})$$

4. Sep 20, 2008

### physics19

Ok am I doing this right???

Vector A: -25j
Vector B: 22.69i -7.8136j
Vector C: 5.5j

Vector D= 22.69i -27.3136j

sqrt((22.69^2 )+ (-27.3136^2))= Distance

5. Sep 20, 2008

### cepheid

Staff Emeritus
your directions aren't consistent. The way you have written vector C suggests that north is in the positive y direction. However, the way that you have written vector A suggests that east is in the negative y direction. Both of these statements can't be true.

6. Sep 20, 2008

### physics19

Ok thank you.. makes sense.. Did I do everything else right though??

7. Sep 20, 2008

### physics19

well i fixed A and C to being the same positive values and I still do not get the right answer.

A: 25j
B: 22.6924i -7.813636j
C: 5.5 j

Distance= sqrt ((22.6924^2)+(22.6863^2)

8. Sep 20, 2008

### cepheid

Staff Emeritus
Umm, no, you didn't fix it. A is the direction EAST. C is NORTH. You have EAST and NORTH BOTH going in the positive j direction. Can you see that EAST and NORTH are not the same directions?

9. Sep 20, 2008

### cepheid

Staff Emeritus
Let's CHOOSE a convention. Let's decide that the x axis runs east-west and that east is the positive direction, and the y axis runs north-south and that north is the positive direction. Then we have the following vectors given (oasis A is at the origin):

vector A: a vector running from oases A to B: +25i (note that is an *i*)

vector B: the camel's first walk: 24cos(19)i - 24sin(19)j

vector C: the camel's second walk: +5.5j

ADD together the camel's two walks to get his current position:

B + C = 24cos(19)i +[5.5 - 24sin(19)]j

Call this new position vector D. What is the distance between D (the camel's position after walking) and oasis B (whose location is given by position vector A)?

10. Sep 20, 2008

### LowlyPion

This was not what I meant.

The destination Oasis is Due East. That is where you want to get to. That is given simply by:

D = 25i

The first vector A is the one at the angle so A = (22.7)i + (-7.8)j

The second due north walk is B = 5.5j

You are asked to find C = D +(-A) +(-B)

25 i + 0j = D
-22.7i + 7.8j = -A
0 i - 5.5j = -B

Hence C = 2.3i + 2.3j

11. Sep 20, 2008

### physics19

ok thank you all for the help. i understand what i was doing wrong.