A particle moves from (1, 0, 3) to the point | vedclass

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(C) The initial position vector of the particle is $\vec{r}_1 = \hat{i} + 3 \hat{k}$.The final position vector of the particle is $\vec{r}_2 = -3 \hat

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$6$

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(C) The initial position vector of the particle is $\vec{r}_1 = \hat{i} + 3 \hat{k}$.The final position vector of the particle is $\vec{r}_2 = -3 \hat{i} + 4 \hat{j} + 5 \hat{k}$.The force acting on the particle is $\vec{F} = \hat{i} + 5 \hat{k}$.The displacement vector $\vec{s}$ is given by $\vec{r}_2 - \vec{r}_1$:$\vec{s} = (-3 \hat{i} + 4 \hat{j} + 5 \hat{k}) - (\hat{i} + 3 \hat{k})$$\vec{s} = -4 \hat{i} + 4 \hat{j} + 2 \hat{k}$.Work done $W$ is the dot product of force and displacement:$W = \vec{F} \cdot \vec{s} = (\hat{i} + 5 \hat{k}) \cdot (-4 \hat{i} + 4 \hat{j} + 2 \hat{k})$.Calculating the dot product:$W = (1)(-4) + (0)(4) + (5)(2)$$W = -4 + 0 + 10 = 6 \ J$.

$A$ particle moves from $(1, 0, 3)$ to the point $(-3, 4, 5)$,when a force $F = \hat{i} + 5 \hat{k}$ acts on it. The amount of work done in joules is:

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