Forces acting on a particle have magnitudes $5, 3, 1 k g . w t$ and act in the directions of the vectors $6 i + 2 j + 3 k, 3 i - 2 j + 6 k$ and $2 i - 3 j - 6 k$ respectively.These remain constant while the particle is displaced from A (4, -2, -6) to B (7,-2,-2). Find the work done.

21

units.

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33

units.

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47

units.

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52

units.

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Solution

The correct option is C $33$ units.
Unit vectors in the direction of the forces are $\frac{6 i + 2 j + 3 k}{\sqrt{(36 + 4 + 9)}}, \frac{3 i - 2 j + 6 k}{7}, \frac{2 i - 3 j - 6 k}{7}$
Hence if $F_{1}, F_{2}, F_{3},$ be the forces of magnitudes 5, 3, 1, then they are
$\frac{5}{7} (6 i + 2 j + 3 k), \frac{3}{7} (3 i - 2 j + 6 k)$ and $\frac{1}{7} (2 i - 3 j - 6 k) .$
Again if $R$ be their resultant then
$R = F_{1} + F_{2} + F_{3} .$ $∴ R = \frac{1}{7} [(30 + 9 + 2) i + (10 - 6 - 3) j + (15 + 18 - 6) k]$ $= \frac{1}{7} (41 i + j + 27 k) .$ $= \frac{1}{7} (41 i + j + 27 k) .$
Again $\to A B = P . V$ of $B - P . V$ A $= (4 i - 2 j - 6 k) - (7 i - 2 j - 2 k) = - 3 i + 0 j - 4 k$
The work done by various forces is equal to work done by their resultant.
$∴$ Work done $= R . \to A B$ $= \frac{1}{7} [41 i + j + 27 k] . [- 3 i - 4 k]$ $= \frac{1}{7} [- 123 + 0 - 108] = \frac{- 231}{7} = - 33$
So work done is $| - 33 | = 33$ units.

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Similar questions

5, 3

1

6 i + 2 j + 3 k, 3 i - 2 j + 6 k

2 i - 3 j - 6 k,

A (2, - 1, - 3)

B (5, - 1, 1)

P = 2 i - 5 j + 6 k

Q = - i + 2 j - k

4 i - 3 j - 2 k

6 i + j - 3 k .

A unit vector perpendicular to the plane determined by the vectors $4^i + 3^j -^k a n d 2^i - 6^j - 3^k$

2 \hat{i} - 3 \hat{j} + 6 \hat{k}

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