If the constant forces 2 hat{i}-5 hat{j}+6 ha | vedclass

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(C) The resultant force $\overrightarrow{F}$ is the sum of the individual forces: $\overrightarrow{F} = (2 \hat{i}-5 \hat{j}+6 \hat{k}) + (-\hat{i}+2

Vedclass · Accepted Answer

$-15$

Vedclass · Answer

(C) The resultant force $\overrightarrow{F}$ is the sum of the individual forces: $\overrightarrow{F} = (2 \hat{i}-5 \hat{j}+6 \hat{k}) + (-\hat{i}+2 \hat{j}-\hat{k}) = \hat{i}-3 \hat{j}+5 \hat{k}$ The displacement vector $\overrightarrow{d}$ is given by $\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A}$: $\overrightarrow{d} = (6 \hat{i}+\hat{j}-3 \hat{k}) - (4 \hat{i}-3 \hat{j}-2 \hat{k}) = 2 \hat{i}+4 \hat{j}-\hat{k}$ The work done $W$ is the dot product of the force and displacement vectors: $W = \overrightarrow{F} \cdot \overrightarrow{d} = (\hat{i}-3 \hat{j}+5 \hat{k}) \cdot (2 \hat{i}+4 \hat{j}-\hat{k})$ $W = (1)(2) + (-3)(4) + (5)(-1) = 2 - 12 - 5 = -15 	ext{ unit}$

If the constant forces $2 \hat{i}-5 \hat{j}+6 \hat{k}$ and $-\hat{i}+2 \hat{j}-\hat{k}$ act on a particle due to which it is displaced from a point $A(4,-3,-2)$ to a point $B(6,1,-3)$, then the work done by the forces is (in $\text{ unit}$)

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