Forces 3i + 2j + 5k and 2i + j - 3k are actin | vedclass

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(C) The resultant force $\vec{F}$ is the sum of the individual forces:$\vec{F} = (3i + 2j + 5k) + (2i + j - 3k) = (3+2)i + (2+1)j + (5-3)k = 5i + 3j +

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$24$

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(C) The resultant force $\vec{F}$ is the sum of the individual forces:$\vec{F} = (3i + 2j + 5k) + (2i + j - 3k) = (3+2)i + (2+1)j + (5-3)k = 5i + 3j + 2k$.The displacement vector $\vec{d}$ is the difference between the final position vector and the initial position vector:$\vec{d} = (4i - 3j + 7k) - (2i - j - 3k) = (4-2)i + (-3 - (-1))j + (7 - (-3))k = 2i - 2j + 10k$.Work done $W$ is the dot product of the force and displacement vectors:$W = \vec{F} \cdot \vec{d} = (5i + 3j + 2k) \cdot (2i - 2j + 10k)$$W = (5 	imes 2) + (3 	imes -2) + (2 	imes 10)$$W = 10 - 6 + 20 = 24 \, 	ext{units}$.

Forces $3i + 2j + 5k$ and $2i + j - 3k$ are acting on a particle and displace it from the point $2i - j - 3k$ to the point $4i - 3j + 7k$. The work done by the forces is ............... $unit$.

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