Two forces vec{F_1} = 2hat{i} - 5hat{j} + 6ha | vedclass

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(D) The resultant force $\vec{F}$ is given by $\vec{F} = \vec{F_1} + \vec{F_2} = (2\hat{i} - 5\hat{j} + 6\hat{k}) + (-\hat{i} + 2\hat{j} - \hat{k}) =

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$-15$

Vedclass · Answer

(D) The resultant force $\vec{F}$ is given by $\vec{F} = \vec{F_1} + \vec{F_2} = (2\hat{i} - 5\hat{j} + 6\hat{k}) + (-\hat{i} + 2\hat{j} - \hat{k}) = \hat{i} - 3\hat{j} + 5\hat{k}$.The displacement vector $\vec{d}$ is given by $\vec{d} = \vec{PQ} = \vec{OQ} - \vec{OP} = (6\hat{i} + \hat{j} + 3\hat{k}) - (4\hat{i} - 3\hat{j} + 2\hat{k}) = 2\hat{i} + 4\hat{j} + \hat{k}$.The work done $W$ is the dot product of the resultant force and the displacement vector: $W = \vec{F} \cdot \vec{d} = (\hat{i} - 3\hat{j} + 5\hat{k}) \cdot (2\hat{i} + 4\hat{j} + \hat{k})$.Calculating the dot product: $W = (1)(2) + (-3)(4) + (5)(1) = 2 - 12 + 5 = -5$ units.

Two forces $\vec{F_1} = 2\hat{i} - 5\hat{j} + 6\hat{k}$ and $\vec{F_2} = -\hat{i} + 2\hat{j} - \hat{k}$ act on a particle. The particle is displaced from point $P(4\hat{i} - 3\hat{j} + 2\hat{k})$ to point $Q(6\hat{i} + \hat{j} + 3\hat{k})$. The work done by the forces is ............. units.

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