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

[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|)

 

[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)

 

[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?

 

[Dan350]

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?

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

 

[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?

 

[Dan350]

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?

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

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

 

[tiny-tim]

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

solve the equations!

show us how you do it ​

 

[Dan350]

solve the equations!

show us how you do it ​

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,, wouldn't that be vector Bsin(73)?

 

[tiny-tim]

|AxB|sinθ= |A*B|cosθ

no, it's …

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

ok, you have two equations, and you want to solve for one unknown (the angle, θ) …

show us how you do it​

 

[Dan350]

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​

|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?

 

[tiny-tim]

We know cosθ=cos(17)

where does that come from??

sinθ=cosθ
sin^-1(cosθ)=θ

how does that help?

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

 

[Dan350]

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

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

 

[tiny-tim]

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

 

[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 =.

 

[haruspex]

|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?

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.

 

[Dan350]

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.

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

 

[Dan350]

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

that's equal to 45°

 

[tiny-tim]

that's equal to 45°

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?​

 

[Dan350]

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?​

How did I miss that! haha

and for the other part, isn't between 45° and 17° ?? we are know looking for VEctor B

 

[tiny-tim]

… isn't between 45° and 17° ?? we are know looking for VEctor B

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? ​

 

[Dan350]

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? ​

Vector A and B ?

 

[haruspex]

and for the other part, isn't between 45° and 17° ?? we are know looking for VEctor B

You are told:

The Vector A points 17° counterclockwise from the positive x axis.

So 17° is the angle between which two lines?
You have found that 45° is the angle between A and B. So what possibilities are there for the angle B makes to the positive x axis?

You are also told

Vector B lies in the first quadrant of the xy plane.

Which of the possibilities does that select?

 

[Dan350]

You are told:

So 17° is the angle between which two lines?
You have found that 45° is the angle between A and B. So what possibilities are there for the angle B makes to the positive x axis?

You are also told

Which of the possibilities does that select?

So,, if we add 45° to the 17° we get 62° for vector B, and 45° will be between vectors A & B,,

am I right?

 

[haruspex]

So,, if we add 45° to the 17° we get 62° for vector B, and 45° will be between vectors A & B,,

am I right?

Yes!

 

[tiny-tim]

(just got up :zzz:)

yes

but suppose it didn't say …

Vector B lies in the first quadrant of the xy plane.

… do you then know how to show that it must be 62° (and not minus 28°) ?