# Vector question - North-east direction

1. Aug 4, 2016

### gracy

1. The problem statement, all variables and given/known data
A particle undergoes three successive displacements given by $S_1$=√2 m North- East, $S_2$=2m due south and $S_3$=4m 30 degree north of west , then find the magnitude of net displacement?

2. Relevant equations
$S$= $S_1$ + $S_2$ + $S_3$

3. The attempt at a solution
I don't understand how to represent these displacement in the form of vectors.
I do have solution of this problem. According to the given solution
$S_1$ = (√2 cos 45)i + (√2sin 45) j
I can't comprehend how the simple North-east turned into this equation? And I think the question not at all mentions angle 45 . Is there any convention or thumb rule that I am missing while dealing with directions?

2. Aug 4, 2016

### A.T.

It mentions "North-East".

3. Aug 4, 2016

### gracy

Last edited: Aug 4, 2016
4. Aug 4, 2016

### gracy

Is this correct representation of the displacements in question?

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5. Aug 4, 2016

Yes.

6. Aug 4, 2016

### gracy

In solution $S_3$ is given to be (-4cos30)i +(4sin30)j
I did not understand why is there negative sign in (-4 cos30)i

7. Aug 4, 2016

### cnh1995

Sign convention. Second quadrant has -ve x and +ve y. That way, S2 should be -2j.

8. Aug 4, 2016

### Ray Vickson

Look at your diagram in post #4. Is the x-component of $S_3$ positive, negative, or zero?

9. Aug 4, 2016

### David Lewis

It may be helpful to draw the vectors to scale, and represent coordinate axes with a thin line, and the vectors with a thick line. (That way you can tell them apart.) Also, if you arrange the vectors head to tail, you will be able to get an approximate answer quickly.

10. Aug 5, 2016

### gracy

Exactly, but in solution it's given to be +2j.

11. Aug 5, 2016

### cnh1995

It should be -2j. There might be a typo in the solution provided by the book.

12. Aug 5, 2016

### Nidum

13. Aug 15, 2016

### gracy

I encountered one more problem
Vector A points vertically upward and B points towards north.Vector product AXB is ?
a)along west
b)zero
c)vertically downward
d)along east
I've been drawing north as an upward direction .

Going with that approach angle between A and B would be zero

14. Aug 15, 2016

### cnh1995

The wording implies north is not vertically upwards. Imagine you're standing on a ground. What does 'vertically upward' mean to you in this case?

15. Aug 15, 2016

### gracy

Towards the sky?

16. Aug 15, 2016

### cnh1995

Yep..

17. Aug 15, 2016

### gracy

So here shall I consider north in the usual way and vertically upward as coming out from page?

18. Aug 15, 2016

### cnh1995

Yes.

19. Aug 15, 2016

### gracy

That would be perpendicular to the plane containing A and B . It's either west or right. Using right hand screw rule I am getting along west as my answer .

20. Aug 15, 2016

### cnh1995

Are you sure?