1. **Problem statement:** Find the vector perpendicular (orthogonal) to the vector $\vec{v} = (3, -4)$ in $\mathbb{R}^2$. 2. **Formula and concept:** In $\mathbb{R}^2$, a vector perpendicular to $(x, y)$ is $(-y, x)$ or $(y, -x)$ because their dot product is zero: $$ (x, y) \cdot (-y, x) = x(-y) + yx = -xy + xy = 0 $$ 3. **Calculate the perpendicular vector:** For $\vec{v} = (3, -4)$, the perpendicular vectors are: $$ (-(-4), 3) = (4, 3) \quad \text{or} \quad (-4, -3) $$ 4. **Normalize the perpendicular vector:** To get the unit vector (length 1), calculate the magnitude: $$ \| (4, 3) \| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $$ 5. **Unit perpendicular vector:** $$ \left( \frac{4}{5}, \frac{3}{5} \right) \quad \text{or} \quad \left( -\frac{4}{5}, -\frac{3}{5} \right) $$ 6. **Match with options:** The option closest to the unit perpendicular vector is (C) $\left( \frac{4}{5}, -\frac{3}{5} \right)$, which corresponds to the vector perpendicular to $(3, -4)$. **Final answer:** (C) $\left( \frac{4}{5}, -\frac{3}{5} \right)$