# Thread: Simple Vector Question

1. I solved this question but I am unable to find the direction? Please help, tell me the method on how to solve.

A woman walks eastward for 5 km and then north-ward for 10 km. How far is she from her starting point? If she had walked directly to her destination, in what direction would she have headed?

I solved the question and I got R = 11.18km

but I am unable to get the direction. Someone explain it.

My guess is,

R Sin x = 10
11 Sin x = 10
Sin x = 10/11

Is that correct?

2.

3. You,re not asking a simple question maybe you ask for a simple answer.

The question is not simple because easth west lines run parallel and south easth lines only do that at the ecquator.

Also who are you asking for : for someone who doesn,t travel or the travelling woman ?

4. Originally Posted by isuru
I solved this question but I am unable to find the direction? Please help, tell me the method on how to solve.

A woman walks eastward for 5 km and then north-ward for 10 km. How far is she from her starting point? If she had walked directly to her destination, in what direction would she have headed?

I solved the question and I got R = 11.18km

but I am unable to get the direction. Someone explain it.

My guess is,

R Sin x = 10
11 Sin x = 10
Sin x = 10/11

Is that correct?
Welcom to the forum. Don't pay any attention to Ghrasp. Fools like him inhabit all public forums.

The angle you are looking for is acrtangent(10/5) which is 63.4 degrees.
Note that the woman walked two legs of a right triangle (5 and 10). You calculated the hypotenuse correctly. The tangent of the angle between her starting and ending points is (10/5). So take the acrtangent of that value to get the angle.

5. The arcsin of 10/11.18 will get you to the same answer, so I think you were on the right track.

6. Thank you everyone! Thanks for helping me to figure it out!

7. Originally Posted by mikelizzi
Originally Posted by isuru
I solved this question but I am unable to find the direction? Please help, tell me the method on how to solve.

A woman walks eastward for 5 km and then north-ward for 10 km. How far is she from her starting point? If she had walked directly to her destination, in what direction would she have headed?

I solved the question and I got R = 11.18km

but I am unable to get the direction. Someone explain it.

My guess is,

R Sin x = 10
11 Sin x = 10
Sin x = 10/11

Is that correct?
Welcom to the forum. Don't pay any attention to Ghrasp. Fools like him inhabit all public forums.

The angle you are looking for is acrtangent(10/5) which is 63.4 degrees.
Note that the woman walked two legs of a right triangle (5 and 10). You calculated the hypotenuse correctly. The tangent of the angle between her starting and ending points is (10/5). So take the acrtangent of that value to get the angle.

I apologize for the insulting remark. That was immature.

8. Offtopic : Being a fool but not simple or simple and not being a fool is a matter of preference also and I prefer being a fool to some sometimes.

In the meanwhile the simple answers have not yet defined to which line these answers refer to : north south or easth west and it's not given by the question either.

Is Isuro (or the women) suppoced to throw dice on this ?

Simple answer (for when the lines all run parallel); The answer for direction is allready given by the question as a ratio.

The angle to one of the two componing directions can be calculated as inv-tan for tan=0,5 or inv-tan for tan=2 dependant to which line you want the angle to relate to. Inv tan 0,5 would relate to east direction Inv tan 2 to north direction.
But keep it in the back of you,re head the direction for the north-south line is not constant along the trajektory. It changes for the travelling person compared to for someone who doesn,t travel and sees this direction as constant.

Because of this maps are always deformed which is not something I found out.

9. In school I learned these three terms:

SOH
CAH
TOA

SOH stands for

CAH stands for

TOA stands for

Works for all right-angled triangles. Makes it easier to remember. :wink:

10. That,s what I learned also. But arc-tan is the easiest here I think. Tangens can,t be used if there is not an angle. You need inv-tan then.

To remind which line the angle revers to : inv-tan 5/10, 5 : 10 or "five to ten" the angle relates to the ten.

11. That,s what I learned also. But arc-tan is the easiest here I think. Tangens can,t be used if there is not an angle. You need arc tan then. Arc tan 5/10 or 5 : 10 is one direction 5 relating to the other being 10.

Arc-tan 5/10 or "five to ten" the angle also relates to the ten ; north-direction here.

12. I'm thinking it's a simple question with a simple answer: North North East.

13. Originally Posted by mikelizzi
The angle you are looking for is acrtangent(10/5) which is 63.4 degrees.
aka arctan(2)

or as Farsight said NNE

Your recommendation to ignore Ghrasp was right on target. Only someone like that could make a mountain out of this mole hill, and then enshroud the mountain in a bank of fog.

14. You can,t projekt such a question to a rounded surface and be exact and the question asks for exact answers. asking the question this way leads to the fogginess. If I showed that you may call me foggy.

The woman will travel a curved trajektory if she walks a constant angle this way. Waste of energy.
It,s only a short distance but principle counts.

@ Isuru : I assumed this was a question from a book or for school and this type of didactic approach just irritates me. So don,t take it personal it was not meant to be.

15. Curved trajectories??? Please explain.

16. The angle between two lines south-north and east west (randomly chosen) is always perpendicular. But the north south lines meet at the north- and southpole while the east west lines stay parallel. If the woman orientates in a math way that fits a flat euclidean space this always causes problems. If you make two maps this way and one she takes on a travel the other stays behind these maps sorta deform to each other.

This problem and making a map the problems are different but come from the same problem of projektion.

This is why sometimes planes have to fly fly north first to get to a destiny more south then where they depart from that distances it becomes more significant. Thay choose the shortest trajektory and that can be only one.

During the travel the angle to the east - west direktion changes during the flight. So any trajektory where it stays a constant angle has to be longer..

It can simply be tested with a globe or a sphere where you define a north and southpole. Draw what you think is a straight trajektory between two dots on it . This is not difficult to do, ask ten people to do this and they all do it right.

If you think all latitude lines will have constant angle to this line try and draw a series of such lines on different latitude with same angle to the trajektory. Comparing these lines with the latitude lines you see the angle to the latitude lines is not the same. Same as with a plane. The constant angle to latitude lines can,t be the shortest trajektory if a trajektory where the angle changes is allready found to be the shortest. Not being the shortest it curves more then the shortest line.

Therefor travelling by kompas you always need to adapt the angle to travel a straight line.

Normally this is explained with the idea that the magnetic field has different direktion for different location....it,s the other way round..... Going by a constant angle to a kompas is what makes the trajektory curve as can be shown topological with the globe or sphere and letting a pen or pencil travel the globe..

17. Planes sometimes travel north first to go to the south for the shortest trajektory. So obviously the angle to the latitudal lines and/or longitudal lines (lokal they are constantly perpendicular) can,t be constant angle and same time fly the shortest trajektory with the least fuell needed. For the shortest distance the angle has to change during a flight except for flights south ór north and west ór east.

Only one trajektory can be the shortest not two or more. A line that is not the shortest obviously has more curvation then necessary.

Take a simple sphere somewhat close to an absolute sphere (which is impossible to make) with an up and downside (there is always an up a top and bottom, up and down to it as to anything). Draw the lines for latitude all parallel to each other (all horizontal) and the lines for longitude coming together at top and bottom. Drawing, the pen travels the sphere.

Then let a pen travel "north-north-east" pulling a straight line over the sphere... Somehwere it will become a tangent for a latitude line meaning the angle has become zero to this stripe. Apparently the angle changed it decreased to become a tangent and then increases again but with reversed angle to the latitude line.

Any other line between two points is curved to this shortest trajektory the shortest trajektory is also curved but less and that,s what we normally refer to as walking a straight line.

18. Originally Posted by Ghrasp
Only one trajektory can be the shortest not two or more. A line that is not the shortest obviously has more curvation then necessary.
wrong

There are infinitely many geodesic paths joining the poles of a sphere. Two points do not determine a line in all geometries.

19. There are infinitely many geodesic paths joining the poles of a sphere.
If you mean she can constantly orientate on the local longitudal line you are right...she can (but won,t help). The local longitudal line (north -south) is still at perpendicular angle with the local latitudal line (easth west). A factor to adapt the kompas also changes both directions of the kompas to the magnet. In fact changes the angle.

You need such a factor (used by more advanced kompasses) with a perfect sphere just as well.

With constant angle alpha to longitudal direction she automatically also travels (90 - alpha) at constant angle to to the local lines for latitude (horizontal round for the sphere) Eventually - if the trajektory is extended, r will always spiral to one of the poles.

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