A particle is at rest at the origin. A force $\vec{A} = (a \hat{i} + b \hat{j} + 6 \hat{k}) N$ is applied on the particle and it is displaced from the origin to the point (1, 2, 3) metre. For what values of ‘a’ and ‘b’, the particle will gain maximum kinetic energy?

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$a = 2, b = 4$

$a = 3, b = 2$

$a = 4, b = 3$

$a = 1, b = 2$

answer is A.

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Detailed Solution

displacement, $\vec{d} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) m$

$W = \vec{F} . \vec{d} = F d \cos θ$

By work-energy theorem kinetic energy gained will be maximum, work done on the particle is maximum i.e., when $θ = 0^{o}$ .

i.e. when $\vec{F} and \vec{d}$ are parallel.

In that case, $\frac{a}{1} = \frac{b}{2} = \frac{c}{3}$

$\Rightarrow a = 2 and b = 4$

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