anonymous
  • anonymous
Vectors A→ and B→ have scalar product -3.00 and their vector product has magnitude 6.00. What is the angle between these two vectors?
Physics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\bar A·\bar B=\left| \bar A \right|\left| \bar B \right|\cos(\theta)=-3\]\[\left| \bar A \times \bar B \right|=\left| \bar A \right|\left| \bar B \right|\left| \sin(\theta) \right|=6\] Then you have that \[\frac{ \left| \sin(\theta) \right| }{ \cos(\theta) }=-2\] As cos(theta)<0, angle must be either in the second or third quadrant, then: \[\theta=\pi-Atan(2)\]or \[\theta =\pi+Atan(2)\]
theEric
  • theEric
\(\theta\) is just the angle between them, so there is no coordinate system involved. So I think you just go from \(\tan\theta=-2\) to \(\theta=\left|\arctan (-2)\right|\). Do you agree @CarlosGP ? Or am I mistaken?
anonymous
  • anonymous
this is wrong because \[\theta=\left| Atan(-2) \right|=Atan(2)\] and Atan(2) corresponds to an angle that might be in the first or third quadrant, and this is not the case becuase for the first quadrant cos(theta)>0

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anonymous
  • anonymous
On the other hand, the two results i have given are posititve angles so considering the magnitude is unnecessary
theEric
  • theEric
Are we defining the quadrants as \(\vec A\) is in the direction of the positive \(x\)-axis? Does it work like that?
anonymous
  • anonymous
No, quadrants are defined counterclockwise 1,2,3,and 4
anonymous
  • anonymous
And they refer to angles. Imagine a clock and rotate counterclockwise: 1st is between 3 and 12, second is between 12 and 9, third between 9 and 6 and fourth between 6 and 3
theEric
  • theEric
How can \(\left|\left|\vec A\right|\right|~~\left|\left|\vec B\right|\right|~~\cos(\theta)\) be negative since, \(\left|\left|\vec A\right|\right|\) and \(\left|\left|\vec B\right|\right|\) are positive? Then \(\cos\theta\) must be negative. So, if \(\theta\) is measured from the positive \(x\)-axis, then \(\theta\) would be in the second or third quadrant, as \(\cos\theta<0\implies \frac{\pi}{2}\lt \theta\lt\frac{3\pi}{2}\). |dw:1378151978048:dw| So I see that much! And still we don't use \(\arctan(-2)\)? This trig confused me before, and I'm distracted at the moment. I might ask something in the math section later. Thanks for replying earlier! |dw:1378152915086:dw|
goformit100
  • goformit100
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