Quick force question

devious_

Points A and B have position vectors (3i + j)m and (5i + 4j + 2k)m respectively. A particle moves from rest at the point A to the point B under the action of a constant force F newtons only. Given that the work done by the force in moving the bead from A to B is 34 Nm, find F.

I've found that $|\textbf{F}| = \sqrt{68}$, which is correct, but I'm supposed to find it in vector form: $\textbf{F} = 4\textbf{i} + 6\textbf{j} + 4\textbf{k}$. How do I do that?

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whozum

Draw a vector from point A to point B, this is vector F. Now break it down into its directional components

devious_

Isn't that the position vector of B relative to A?

whozum

Yes, thats where the force acted on the particle.

devious_

Oh! I see what you're getting at:
$$\textbf{F} = \sqrt{\frac{68}{17}} (2\textbf{i} + 3\textbf{j} + 2\textbf{k})$$

And that gives the required answer. Thanks!

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