Vector components in the 3 dimensions

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The discussion centers on calculating the components of a force vector F = (-20i + 50j + 10k) projected along a pole defined by the position vector r = 3i - 2j + 6k. Participants clarify that instead of finding angles, the dot product formula F·n (where n is the unit vector in the direction of the pole) is more efficient. The unit vector is derived by normalizing the position vector, resulting in u = (3/7)i - (2/7)j + (6/7)k. The final step involves taking the dot product of the force vector with the unit vector to find the component along the pole. This approach simplifies the calculation of the force's projection in three dimensions.
clickcaptain
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Given F = (-20i + 50j = 10k)

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1. The component of the foce projected along the pole AO.
2. The magnitude of the projected component of the F along the pole AO.

I have no idea where to begin, I think I need to find the angles but I'm not sure how in three dimensions.


(please excuse the picture...i know its not the greatest)
 
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Welcome to PF!

clickcaptain said:
I think I need to find the angles but I'm not sure how in three dimensions.)

Hi clickcaptain ! Welcome to PF! :smile:

No, you don't need to find the angles …

you're thinking of the formula F cosθ for the component of force F in a direction at an angle θ.

But you can also use the formula F.n, the direct product of the force and the unit vector in that direction …

and when the question gives you the coordinates, that's a lot easier! :wink:
 
so the unit vector would be ...

position vector of pole = r = 3i - 2j + 6k)
length of r = l = sqrt(3^2 + 2^2 + 6^2) = 7

u = r/l = 3/7i - 2/7 j + 6/7 k

is that right?

then take that times the force vector given?
 
clickcaptain said:
… is that right?

Yup! :biggrin:

(though I would have waited until the next step before dividing by 7 … :wink:)
then take that times the force vector given?

(you mean dot, not times, of course :wink:)

Yes … that gives you the component of force along that direction. :smile:
 
thanks! :-)
 

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