1. **Problem Statement:** A force of magnitude 6 newtons acting North is resolved into two perpendicular components. Find the magnitude of the component in the direction halfway between East and North (Eastern North). 2. **Understanding the direction:** Eastern North is the direction exactly between East and North, forming a 45° angle with North. 3. **Force components:** The original force $F = 6$ newtons acts North (along the y-axis). When resolving, the component in the Eastern North direction is the projection of the force vector on that 45° direction. 4. **Projection calculation:** Since force acts North, the component along the Eastern North direction is: $$ F_{EN} = F \times \cos 45^\circ $$ 5. **Evaluate cosine:** $$ \cos 45^\circ = \frac{\sqrt{2}}{2} $$ 6. **Calculate component:** $$ F_{EN} = 6 \times \frac{\sqrt{2}}{2} = 3\sqrt{2} $$ 7. **Conclusion:** The component of the force in the Eastern North direction is $3\sqrt{2}$ newtons. **Answer:** (b) 3√2