# Is it a scalar product? I'm kind of lost

1. Nov 28, 2013

### Dan350

The Vector A points 17° counterclockwise from the positive x axis. Vector B lues in the first cuadrant of the xy plane. The magnitudes of the cross product and the dot product are the same:
i.e, |AXB|= |A(times)B|
What Angle does B make with the positive x axis?

2. Is ti a scalar product? I'm kind of lost

3. I was thinking of using cos(σ)= (AtimesB)/ (|A||B|)

2. Nov 28, 2013

### tiny-tim

Hi Dan350!

The dot product and the scalar product (of two vectors) are the same thing.

(and are both written A.B not AtimesB)

3. Nov 28, 2013

### CompuChip

The magnitude of the cross product is |AxB| = |A| |B| sin(theta)
The magnitude of the dot product is |A.B| = |A| |B| cos (theta).

What does this tell you when they are equal?

4. Nov 29, 2013

### Dan350

Is A condition given in the problem,
How do I reach this problem?

5. Nov 29, 2013

### 1MileCrash

Look at what CompuChip is saying.

The magnitude of the cross product is |A||B|sin(theta). The scalar yielded by the dot product is |A||B|cos(theta).

You know they are equal. So what does that say about theta?

6. Nov 29, 2013

### Dan350

I'm not quite sure, does that mean that they are 90° apart?

or that I can substitute vector "A" bye sin(17)??

7. Nov 29, 2013

### tiny-tim

solve the equations!!!

show us how you do it

8. Nov 29, 2013

### Dan350

So I have

|AxB|sinθ= |A*B|cosθ

Well since they are giving that Vector A is 17 counter clockwise form the x axis that meas is Acos(17) and vector B is , well I'm stuck there,, wouldnt that be vector Bsin(73)?

9. Nov 29, 2013

### tiny-tim

no, it's …
ok, you have two equations, and you want to solve for one unknown (the angle, θ) …

show us how you do it​

10. Nov 29, 2013

### Dan350

|A| |B| sin(theta)= |A| |B| cos(theta)

sinθ=cosθ
sin^-1(cosθ)=θ

We know cosθ=cos(17)
so θ= 73
Am I right?

11. Nov 29, 2013

### tiny-tim

where does that come from??
how does that help?

sinθ=cosθ … if you can't think of a way of solving it, draw a diagram or a graph

12. Nov 29, 2013

### Dan350

I thought that since is counterclockwise form the x axis,, is going to be cos(17)
and I solved for theta in "sin^-1(cosθ)=θ"

I drew it, i only have my Vector A 17 raising counterclockwise form the x axis,, the vector B is inthe same cuadrant

Any clue?

Thanks

13. Nov 29, 2013

### tiny-tim

no, just draw sinθ = cosθ, and then solve it

14. Nov 29, 2013

### Panphobia

At what angle does sinθ = cosθ, you know that at cos(0) = 1 and sin(0) = 0, and cos(90) = 0, sin(90) = 1, so since they are continuous functions, you can estimate at what θ they will be =.

15. Nov 29, 2013

### haruspex

You're confusing two angles.
The question asks you to find the angle B makes with the positive x axis. The angle θ in |A.B| = |A| |B| cos(θ) and |AxB| = |A| |B| sin(θ) is the angle between A and B. So, first find θ (which will not involve the 17 degrees), then use θ and the 17 degrees to find the angle B makes with the positive x axis.

16. Nov 29, 2013

### Dan350

So for sinθ=cosθ the angle will be 45°
Now how do I find angle B using the 17°??

17. Nov 29, 2013

### Dan350

that's equal to 45°

18. Nov 29, 2013

### tiny-tim

correct

(btw, the quickest way of solving that would be sinθ = cosθ, so tanθ = 1, so θ = 45°)

ok, so the angle θ is 45° …

between what and what is θ the angle?​

19. Nov 29, 2013

### Dan350

How did I miss that! haha

and for the other part, isnt between 45° and 17° ?? we are know looking for VEctor B

20. Nov 29, 2013

### tiny-tim

i think i know what you mean, but that doesn't actually make sense, does it?

an angle is between two lines

what are the lines?