Multiplying a Vector Product by Another Vector

AI Thread Summary
The discussion revolves around calculating the expression (A×B)·C, where vectors A and B lie in the xy-plane and vector C points in the positive z-direction. The calculation of the cross product A×B yields a vector whose magnitude is determined using the sine of the angle between A and B. There is confusion regarding whether to use sine or cosine in the dot product with C, as the dot product typically involves the cosine of the angle between the two vectors. Ultimately, it is clarified that the angle between A×B and C is 0 degrees, leading to the correct use of cos(0) in the final calculation. The discussion highlights the importance of understanding vector directions and the relationships between cross and dot products.
student34
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Homework Statement



The question asks to calculate (AxBC, 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×BC imply a scalar product in which case the last part should be 12.363×6.00×cos90°?
 
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Does the cross product yield a vector or a scalar?
 
student34 said:

Homework Statement



The question asks to calculate (AxBC, 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×BC imply a scalar product in which case the last part should be 12.363×6.00×cos90°?
You mean to say that \vec{A} \times \vec{B} = (5)(4)(sin37°) \widehat{n}, where n-hat is the unit vector orthogonal to both vectors.
 
Mandelbroth said:
You mean to say that \vec{A} \times \vec{B} = (5)(4)(sin37°) \widehat{n}, 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.
 
gneill said:
Does the cross product yield a vector or a scalar?

I don't know. It just asks to calculate (A×BC with the values that I gave above.
 
Mandelbroth said:
You mean to say that \vec{A} \times \vec{B} = (5)(4)(sin37°) \widehat{n}, 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.
 
student34 said:
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?
 
Dick said:
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?
 
student34 said:
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?
 
  • #10
student34 said:
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?
 
  • #11
Dick said:
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.
 
  • #12
haruspex said:
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.
 
  • #13
student34 said:
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)?
 
  • #14
[Deleted]
 
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  • #15
@Puky The question haruspex asked was directed towards the OP to check his understanding.
 
  • #16
My mistake, I thought it was the OP who asked that question.
 
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