1. **Problem statement:** Find the length of the perpendicular drawn from point $B(1,-1)$ to the line of action of the force vector $\vec{F} = 4 \hat{i} - 3 \hat{j}$ acting at point $A(2,-1)$. 2. **Understanding the problem:** The force vector $\vec{F}$ defines a line passing through point $A$ in the direction of $\vec{F}$. We want the shortest distance from point $B$ to this line. 3. **Formula for distance from a point to a line in vector form:** $$d = \frac{|(\vec{AB} \times \vec{F})|}{|\vec{F}|}$$ where $\vec{AB} = \vec{B} - \vec{A}$. 4. **Calculate $\vec{AB}$:** $$\vec{AB} = (1 - 2) \hat{i} + (-1 - (-1)) \hat{j} = -1 \hat{i} + 0 \hat{j}$$ 5. **Calculate the cross product magnitude $|\vec{AB} \times \vec{F}|$:** Since vectors are in 2D, treat them as 3D with zero $k$ component: $$\vec{AB} = (-1, 0, 0), \quad \vec{F} = (4, -3, 0)$$ Cross product: $$\vec{AB} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 0 & 0 \\ 4 & -3 & 0 \end{vmatrix} = (0 \cdot 0 - 0 \cdot (-3)) \hat{i} - (-1 \cdot 0 - 0 \cdot 4) \hat{j} + (-1 \cdot (-3) - 0 \cdot 4) \hat{k} = 0 \hat{i} - 0 \hat{j} + 3 \hat{k}$$ Magnitude: $$|\vec{AB} \times \vec{F}| = |3| = 3$$ 6. **Calculate magnitude of $\vec{F}$:** $$|\vec{F}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ 7. **Calculate distance $d$:** $$d = \frac{3}{5}$$ 8. **Answer:** The length of the perpendicular is $\frac{3}{5}$ units. **Final answer:** b) $\frac{3}{5}$