A force of magnitude 30, N acting along hat{i | vedclass

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(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector in the direction of $\hat{i} + \hat{j} + \hat{k}$.Unit vector $

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$30\sqrt{3}\, J$

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(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector in the direction of $\hat{i} + \hat{j} + \hat{k}$.Unit vector $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.So,$\vec{F} = 30 	imes \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = 10\sqrt{3}(\hat{i} + \hat{j} + \hat{k})\, N$.The displacement vector $\vec{S}$ is the difference between the final and initial position vectors:$\vec{S} = (3-2)\hat{i} + (5-4)\hat{j} + (2-1)\hat{k} = \hat{i} + \hat{j} + \hat{k}\, m$.The work done $W$ is the dot product of $\vec{F}$ and $\vec{S}$:$W = \vec{F} \cdot \vec{S} = [10\sqrt{3}(\hat{i} + \hat{j} + \hat{k})] \cdot [\hat{i} + \hat{j} + \hat{k}]$$W = 10\sqrt{3} (1 	imes 1 + 1 	imes 1 + 1 	imes 1) = 10\sqrt{3} 	imes 3 = 30\sqrt{3}\, J$.

$A$ force of magnitude $30\, N$ acting along $\hat{i} + \hat{j} + \hat{k}$ displaces a particle from point $(2, 4, 1)$ to $(3, 5, 2)$. The work done during this displacement is:

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