A particle acted on by constant forces 4-3k- and 3-k- is displaced from the point -2-3k- to the point 5-4-k- The total work done by the forces is

Total work done = [(4î+ĵ 3k̂)+(3î+ĵ k̂)]. [(5î+4ĵ+k̂) (î+2ĵ+3k̂)] =(7î+2ĵ 4k̂).(4î+2ĵ 2k̂)=28+4+8=40 ...

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Standard XI

Mathematics

Applications of Dot Product

A particle ac...

Question

A particle acted on by constant forces $4^i +^j - 3^k$ and $3^i +^j -^k$ is displaced from the point $^i + 2^j + 3^k$ to the point $5^i + 4^j +^k$ . The total work done by the forces is

20 units

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30 units

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40 units

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50 units

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Solution

The correct option is C 40 units
Total work done = $[(4^i +^j - 3^k) + (3^i +^j -^k)] . [(5^i + 4^j +^k) - (^i + 2^j + 3^k)]$
$= (7^i + 2^j - 4^k) . (4^i + 2^j - 2^k) = 28 + 4 + 8 = 40$

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A particle acted on by constant forces 4^i+^j−3^k and 3^i+^j−^k is displaced from the point ^i+2^j+3^k to the point 5^i+4^j+^k. The total work done by the forces is

The correct option is C 40 unitsTotal work done = [(4^i+^j−3^k)+(3^i+^j−^k)].[(5^i+4^j+^k)−(^i+2^j+3^k)] =(7^i+2^j−4^k).(4^i+2^j−2^k)=28+4+8=40

A particle acted on by constant forces $4^i +^j - 3^k$ and $3^i +^j -^k$ is displaced from the point $^i + 2^j + 3^k$ to the point $5^i + 4^j +^k$ . The total work done by the forces is

The correct option is C 40 units
Total work done = $[(4^i +^j - 3^k) + (3^i +^j -^k)] . [(5^i + 4^j +^k) - (^i + 2^j + 3^k)]$
$= (7^i + 2^j - 4^k) . (4^i + 2^j - 2^k) = 28 + 4 + 8 = 40$