Vector question - North-east direction

[gracy]

Homework Statement

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?

Homework Equations

$S$= $S_1$ + $S_2$ + $S_3$

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?

 

[A.T.]

And I think the question not at all mentions angle 45

It mentions "North-East".

 

[gracy]

It mentions "North-East".

http://www.physicsclassroom.com/class/vectors/Lesson-1/Vectors-and-Direction
This site is helpful.

 

[gracy]

Is this correct representation of the displacements in question?

 

[cnh1995]

View attachment 104283
Is this correct representation of the displacements in question?

Yes.

 

[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

 

[cnh1995]

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

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

 

[Ray Vickson]

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

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

 

[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.

 

[gracy]

That way, S2 should be -2j.

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

 

[cnh1995]

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

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

 

[Nidum]

 

[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

 

[cnh1995]

Vector A points vertically upward and B points towards north.Vector product AXB is ?

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?

 

[gracy]

Imagine you're standing on a ground. What does 'vertically upward

Towards the sky?

 

[cnh1995]

Towards the sky?

Yep..

 

[gracy]

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

 

[cnh1995]

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

Yes.

 

[gracy]

.Vector product AXB is ?

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 .

 

[cnh1995]

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 .

Are you sure?
Edit: I (mis)read the problem as A being towards north. Your answer is right.

 

[gracy]

Here goes one more problem
Vector A is pointing eastwards and vector B northwards then direction of (A - B) ?
1)North - east
2)vertically upward
3)vertically downward
4)None of the above
I tried to draw it

I think none of the above is the right option.

 

[cnh1995]

I don't think none of the above is the right option.

It is. Check you drawing. Have you drawn A-B correctly?

 

[gracy]

I think none of the above is the right option.

I have edited my post.

 

[gracy]

Have you drawn A-B correctly?

I think yes. What's wrong ?

 

[cnh1995]

I think yes. What's wrong ?

B is towards north. So what would be the direction of -B?

 

[gracy]

Oh! what I have shown as A - B is actually - (A + B)

 

[cnh1995]

Oh! what I have shown as A - B is actually - (A + B)

Yes.

 

[gracy]

I still think none of the above is right option.

 

[cnh1995]

View attachment 104749
I still think none of the above is right option.

Yes, it is.

 

[gracy]

Direction of (A + B) is given to be northeast northeast means 45 clockwise from north . But the question does not mention magnitude of A and B is equal.. So why will their resultant be at 45 degrees?

 

[cnh1995]

So why will their resultant be at 45 degrees?

Because

Direction of (A + B) is given to be northeast

 

[gracy]

Direction of (A + B) is given to be northeast

No, I meant in the solution direction of (A + B) is given to be northeast.

 

[gracy]

Vector A is pointing eastwards and vector B northwards then direction of (A × B) × (A × B)
1)North - east
2)vertically upward
3)vertically downward
4)None of the above
I tried to draw it[/QUOTE]
I can't differentiate signs of usual multiplication and vector product here.

 

[haruspex]

No, I meant in the solution direction of (A + B) is given to be northeast.

Yes, I see your difficulty. Two ways out: interpret the question as asking which could be right; interpret the answer "north - east" as meaning in that general direction, i.e. in the NE quadrant. Neither very satisfactory I agree.

 

[haruspex]

(A × B) × (A × B)

What is the angle between the two vector arguments to the central x?

 

[gracy]

What is the angle between the two vector arguments to the central x?

Zero. Sin 0=0

 

[haruspex]

Zero. Sin 0=0

So what is (a x b) x (a x b)?

 

[gracy]

So what is (a x b) x (a x b)?

zero

 

[haruspex]

zero

Right.