Explanation

(c) Work done by the gravitational field is zero, when displacement is perpendicular to gravitational field. Here, gravitational field, $\overrightarrow{\mathbf{I}} = 4\widehat{\mathbf{i}} + \widehat{\mathbf{j}}.$ if $\theta_{1}$ is the angle which makes with positive $x$-axis, then
$\tan\theta_{1} = \frac{1}{4}$ or $\theta_{1} = \tan^{- 1}{\left( \frac{1}{4} \right) = 14{^\circ}6'}$
If $\theta_{2}$ is the angle which the line $y + 4x = 6$ makes with positive $x$-axis, then $\theta_{2} = \tan^{- 1}{( - 4)} = 75{^\circ}56'$ so $\theta_{1} + \theta_{2} = 90{^\circ}$
$ie,$ the line $y + 4x = 6$ is perpendicular to $I$