We need to find the vector \( \overline{OP} \), which represents the displacement from \( O \) to \( P \). \[\] The direction of north-east corresponds to an angle of \( 45^\circ \) with the positive \( x \)-axis. \[\] The unit vector in the north-east direction is: \[ \hat{v}_{NE} = \frac{1}{\sqrt{2}} (\hat{\imath} + \hat{\jmath}). \] Thus, after walking 3 units in the north-east direction, the displacement vector is: \[ \vec{OP}_1 = 3 \times \frac{1}{\sqrt{2}} (\hat{\imath} + \hat{\jmath}) = \frac{3}{\sqrt{2}} (\hat{\imath} + \hat{\jmath}). \] The direction of north-west corresponds to an angle of \( 135^\circ \) with the positive \( x \)-axis. \[\] The unit vector in the north-west direction is: \[ \hat{v}_{NW} = \frac{1}{\sqrt{2}} (-\hat{\imath} + \hat{\jmath}). \] Thus, after walking 4 units in the north-west direction, the displacement vector is: \[ \vec{OP}_2 = 4 \times \frac{1}{\sqrt{2}} (-\hat{\imath} + \hat{\jmath}) = \frac{4}{\sqrt{2}} (-\hat{\imath} + \hat{\jmath}). \] The total displacement vector \( \overline{OP} \) is the sum of the two displacement vectors: \[ \overline{OP} = \vec{OP}_1 + \vec{OP}_2 = \frac{3}{\sqrt{2}} (\hat{\imath} + \hat{\jmath}) + \frac{4}{\sqrt{2}} (-\hat{\imath} + \hat{\jmath}). \] Now, let's combine the terms: \[ \overline{OP} = \frac{1}{\sqrt{2}} \left[ 3 (\hat{\imath} + \hat{\jmath}) + 4 (-\hat{\imath} + \hat{\jmath}) \right]. \] Simplifying: \[ \overline{OP} = \frac{1}{\sqrt{2}} \left[ (3 - 4) \hat{\imath} + (3 + 4) \hat{\jmath} \right] \] \[ \overline{OP} = \frac{1}{\sqrt{2}} (-\hat{\imath} + 7 \hat{\jmath}). \] Thus, the displacement vector \( \overline{OP} \) is: \[ {\frac{1}{\sqrt{2}} (-\hat{\imath} + 7 \hat{\jmath})}. \]