Three constant forces −→F1=(2^i−3^j+2^k) N,−→F2=(^i+^j−^k) N and −→F3=(3^i+^j−2^k) N displace a particle from (1,−1,2) m to (−1,−1,3) m and then to (2,3,4) m, displacement being measured in meters. Find the total work done by the forces.

Three constant forces −→F1=(2^i−3^j+2^k) N,−→F2=(^i+^j−^k)

1.	Three constant forces −→F1=(2^i−3^j+2^k) N,−→F2=(^i+^j−^k) N and −→F3=(3^i+^j−2^k) N displace a particle from (1,−1,2) m to (−1,−1,3) m and then to (2,3,4) m, displacement being measured in meters. Find the total work done by the forces.
Answer» Three constant forces $_{1} = (2^i - 3^j + 2^k) N,_{2} = (^i +^j -^k) N$ and $_{3} = (3^i +^j - 2^k) N$ displace a particle from $(1, - 1, 2) m$ to $(- 1, - 1, 3) m$ and then to $(2, 3, 4) m$ , displacement being measured in meters. Find the total work done by the forces.

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Three constant forces −→F1=(2^i−3^j+2^k) N,−→F2=(^i+^j−^k) N and −→F3=(3^i+^j−2^k) N displace a particle from (1,−1,2) m to (−1,−1,3) m and then to (2,3,4) m, displacement being measured in meters. Find the total work done by the forces.

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