Net movement along x-direction S _x = (6 – 4) cos 45° \hat{i}

\(= 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \, \mathrm{km}\)

Net movement along y-direction S _y = (6 + 4) sin 45° \hat{j}

\(= 10 \times \frac{1}{\sqrt{2}} = 5 \sqrt{2} \text{ km}\)

Net movement from starting point

\(\left| \vec{S} \right| = \sqrt{S_x^2 + S_y^2} = \sqrt{(\sqrt{2})^2 + (5\sqrt{2})^2}\) = \(\sqrt{52\,\text{km}}\)

Angle which makes with the east direction

\(\tan \theta = \frac{\text{Y-component}}{\text{X-component}}\) \(= \frac{5 \sqrt{2}}{\sqrt{2}}\)

\ \ \(\theta = \tan^{-1}(5)\)