Solution: Let $ \vecOA = \mathbfa $, $ \vecOB = \mathbfb $. Then $ \overrightarrowOC = m\mathbfa + n\mathbfb $. For $ \overrightarrowOC \perp (\mathbfa - \mathbfb) $, their dot product is zero: $ (m\mathbfa + n\mathbfb) \cdot (\mathbfa - \mathbfb) = 0 $. Expand: $ m\|\mathbfa\|^2 - m\mathbfa \cdot \mathbfb + n\mathbfb \cdot \mathbfa - n\|\mathbfb\|^2 = 0 $. Substitute $ \|\mathbfa\| = 2 $, $ \|\mathbfb\| = 3 $, $ \mathbfa \cdot \mathbfb = 2 \cdot 3 \cdot \cos 60^\circ = 3 $: $ 4m - 3m + 3n - 9n = 0 \Rightarrow m - 6n = 0 $. Choose $ n = 1 $, then $ m = 6 $. Normalize if needed, but the relation is $ m = 6n $. For simplicity, take $ n = 1 $, $ m = 6 $. - Veritas Home Health