Dot Product stuff ;-(

  • #1
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Homework Statement


A constant force of 1i - 5j -8k moves (1,-4,2) (-3,2,-1), what is the work done on the particle?


Homework Equations


Avector*Bvector=ABsinθ
?? I think?

The Attempt at a Solution


I really am quite lost... but I found the coordinates for the position vector...

(-3-1)i + (2--4j) + (-1-2)k and the magnitude of the position vector is 7.81
so the unit vector is -.512i + .768j -.384k... am I on the right track here?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


A constant force of 1i - 5j -8k moves (1,-4,2) (-3,2,-1), what is the work done on the particle?


Homework Equations


Avector*Bvector=ABsinθ
?? I think?

The Attempt at a Solution


I really am quite lost... but I found the coordinates for the position vector...

(-3-1)i + (2--4j) + (-1-2)k and the magnitude of the position vector is 7.81
so the unit vector is -.512i + .768j -.384k... am I on the right track here?
For a constant force F, the work done is just the dot product of the force with the displacement. Use that.
 
  • #3
784
11
so the force is 1i - 5j -8k and the displacement is -4i +6j -3k? so dot product would be -4 -30 - 24 = -58?
 
  • #4
784
11
How do I find the angel between these vectors?
 
  • #5
Dick
Science Advisor
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so the force is 1i - 5j -8k and the displacement is -4i +6j -3k? so dot product would be -4 -30 - 24 = -58?
Yes, if the force isn't constant you need to work harder and integrate, but if it's constant, it's that easy.
 
  • #6
Dick
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How do I find the angel between these vectors?
Why do you think you need the angle?
 
  • #7
HallsofIvy
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If you are required, separately, to find the angle between vectors, the dot product can be defined as
[tex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/tex]

where [itex]\theta[/itex] is the angle between the vectors.
 

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