# Vector question HELP

1. Sep 24, 2006

### parwana

An airplane flies 200 km due west from city A to city B and then 305 km in the direction of 34.0° north of west from city B to city C.

(a) In straight-line distance, how far is city C from city A?

(b) Relative to city A, in what direction is city C?

for a cant u do a^2 + b^2= c^2???
I tried doing that and it doesnt work, help

or is it 305- 200??

for b I dont know what to do?

2. Sep 24, 2006

For A, you need to find a vector that points from city A to city C, lets call it $\vec V_{AC}$. If you find the magnitude of this vectors ($| \vec V_{AC} |$) that gives you the length of the vector (ie the straight line distance from A to C.

What would subtracting 305-200 give you?

Instead you need to subtract the components individually.

For B,
I'm not really sure what they are asking. You'll have to wait for someone else to help you. I could guess, but I don't want to confuse the situation.

3. Sep 24, 2006

### parwana

how do I find magnitude?

4. Sep 24, 2006

Lets say you have a vector $\vec v = \hat i(1) + \hat j(2)$
I don't know what notation you use to represent components.

The modulus (magnitude) of the vector is $|\vec v| = \sqrt{1^2 + 2^2}$

Or in general,
$$\vec A = (x,y)$$
$$| \vec A | = \sqrt{x^2 + y^2}$$

5. Sep 25, 2006

### natarae

Did you try drawing the diagram out?

C is 34.0 degrees north of west, so that means C is on a bearing of (270 + 34) from B.

Now you've got 305km and 200km and the angle in between in, which "entitles" you to use the cosine rule.

So for part A,

CA^2 = AC^2 + AB^2 - 2(AB)(AC)cos(ABC)

and for part B, they want you to find the angle CAB, and write it in the form eg. 34.0 degrees north of west from xxx.

using the sine rule and the result of part A, i.e CA,

(sin ABC)/CA = (sin CAB)/CB
angle CAB = sin inverse ((BCsinABC)/CA)

therefore, C is XXX degrees north of west from A.

Last edited: Sep 25, 2006
6. Sep 25, 2006

### parwana

whats AC? in the equation natarae? and whats the value of ABC when u do cos ABC. Sorry I am really bad with physics

7. Sep 25, 2006

### RoyalewithCh33s3

I believe AC is the line between points A and C. Thats one leg of the triangle ABC.

8. Sep 25, 2006

### parwana

yeah but I am not given AC?

9. Sep 25, 2006

### RoyalewithCh33s3

AC was what you figured out in part A.

And to figure out angle ABC you could draw a straight line with another line coming off of it at a 34 degree angle. So what do you get for angle ABC? What you get for that you can plug into the equations that natarae gave you because then you'll have CA and angle ABC.

Is that what you were asking?

10. Sep 25, 2006

### parwana

But I didnt figure out part A yet. Arent I trying to find AC in part A itself?

11. Sep 25, 2006

### RoyalewithCh33s3

Well I want to know if natarae might have made a mistake in his equation to figure out AC

they wrote:

But I think the AC^2 and the AC are meant to be BC^2 and BC respectively, which would make it possible to find AC.

12. Sep 25, 2006

### parwana

I see, thats what I was getting confused on

now I understand, but still whats angle ABC. Wouldnt it just be 34?

13. Sep 25, 2006

### RoyalewithCh33s3

Well since the line is 34 degrees north of west, I believe you would have to look at the line like this ___34_\________ so you would have to figure out the supplementary angle, and that would be angle ABC

14. Sep 25, 2006

### parwana

Oh I see I drew it wrong, in the other direction, so it would be 146

15. Sep 25, 2006

### parwana

thank u so so much royalewith. I understand

One last question, why do we use cosine instead of sine for the angle?

16. Sep 25, 2006

### RoyalewithCh33s3

Because we are using an adjacent leg and the hypoenuse, so according to my favorite math acronym...or whatever its called:

SOHCAHTOA

We are using A and H so we would use cos.

17. Sep 25, 2006

### parwana

thank u so much

18. Sep 25, 2006

### RoyalewithCh33s3

[spam]

No problem, glad to help...now if only I could figure out my own physics :rofl:

[/spam]