Multiplying a Vector Product by Another Vector

[student34]

Homework Statement

The question asks to calculate (AxB)·C, where A's magnitude is 5.00, B's magnitude is 4.00, and they are both in the xy-plane. B is 37° counter clockwise from A. C has a magnitude of 6.00 and is in the +z-direction.

Homework Equations

(A×B) = ABsinθ = D; D·C = DCcosσ = R

The Attempt at a Solution

(A×B) = 5×4×sin37° = 12.363 = D. D·C = 12.363×6.00×sin90° = 72.2 = R which is the correct answer. But doesn't the dot in (A×B)·C imply a scalar product in which case the last part should be 12.363×6.00×cos90°?

 

[gneill]

Does the cross product yield a vector or a scalar?

 

[Mandelbroth]

Homework Statement

The question asks to calculate (AxB)·C, where A's magnitude is 5.00, B's magnitude is 4.00, and they are both in the xy-plane. B is 37° counter clockwise from A. C has a magnitude of 6.00 and is in the +z-direction.

Homework Equations

(A×B) = ABsinθ = D; D·C = DCcosσ = R

The Attempt at a Solution

(A×B) = 5×4×sin37° = 12.363 = D. D·C = 12.363×6.00×sin90° = 72.2 = R which is the correct answer. But doesn't the dot in (A×B)·C imply a scalar product in which case the last part should be 12.363×6.00×cos90°?

You mean to say that [itex]\vec{A} \times \vec{B} = (5)(4)(sin37°) \widehat{n}[/itex], where n-hat is the unit vector orthogonal to both vectors.

 

[haruspex]

You mean to say that [itex]\vec{A} \times \vec{B} = (5)(4)(sin37°) \widehat{n}[/itex], where n-hat is the unit vector orthogonal to both vectors.

... and don't forget to use the right-hand rule to get the direction of the product in relation to the z axis sign.

 

[student34]

Does the cross product yield a vector or a scalar?

I don't know. It just asks to calculate (A×B)·C with the values that I gave above.

 

[student34]

You mean to say that [itex]\vec{A} \times \vec{B} = (5)(4)(sin37°) \widehat{n}[/itex], where n-hat is the unit vector orthogonal to both vectors.

We haven't seen that n yet, so I don't think we are suppose to use it in this question.

 

[Dick]

We haven't seen that n yet, so I don't think we are suppose to use it in this question.

Mandelbroth just means n to be the direction that AxB points. You have to know that to find the angle between AxB and C. How is the direction AxB points related to the directions A and B point?

 

[student34]

Mandelbroth just means n to be the direction that AxB points. You have to know that to find the angle between AxB and C. How is the direction AxB points related to the directions A and B point?

It would be in the positive direction, but isn't A×B perpindicular to C either way? And why does the answer seem to use sin90° instead of cos90°; doesn't the function (·) mean a scalar product?

 

[Dick]

It would be in the positive direction, but isn't A×B perpindicular to C either way? And why does the answer seem to use sin90° instead of cos90°; doesn't the function (·) mean a scalar product?

What do you mean by positive direction? Which positive? C points in the positive z direction. Which positive direction does AxB point? I don't think they are perpendicular. If the answer says sin(90), it really shouldn't. That's misleading. The answer should contain a cos. cos of what?

 

[haruspex]

It would be in the positive direction, but isn't A×B perpindicular to C either way?

By definition, AxB is perpendicular to both A and B. What does that tell you about its direction in relation to C?

 

[student34]

What do you mean by positive direction? Which positive? C points in the positive z direction. Which positive direction does AxB point? I don't think they are perpendicular. If the answer says sin(90), it really shouldn't. That's misleading. The answer should contain a cos. cos of what?

Ah, I got it now. Oh ya, and the answer did not have sin90°, I just fudged that in there to try to make sense of it all, but I realize now that it is cos0° - thanks.

 

[student34]

By definition, AxB is perpendicular to both A and B. What does that tell you about its direction in relation to C?

Thanks, I go it now.

 

[haruspex]

Ah, I got it now. Oh ya, and the answer did not have sin90°, I just fudged that in there to try to make sense of it all, but I realize now that it is cos0° - thanks.

OK, but just to check... why cos(0) and not cos(180 degrees)?

 

[Puky]

[Deleted]

 

[CAF123]

@Puky The question haruspex asked was directed towards the OP to check his understanding.

 

[Puky]

My mistake, I thought it was the OP who asked that question.