Solution: Let $ \vec{OA} = \mathbf{a} $, $ \vec{OB} = \mathbf{b} $. Then $ \overrightarrow{OC} = m\mathbf{a} + n\mathbf{b} $. For $ \overrightarrow{OC} \perp (\mathbf{a} - \mathbf{b}) $, their dot product is zero: $ (m\mathbf{a} + n\mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = 0 $. Expand: $ m\|\mathbf{a}\|^2 - m\mathbf{a} \cdot \mathbf{b} + n\mathbf{b} \cdot \mathbf{a} - n\|\mathbf{b}\|^2 = 0 $. Substitute $ \|\mathbf{a}\| = 2 $, $ \|\mathbf{b}\| = 3 $, $ \mathbf{a} \cdot \mathbf{b} = 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 $.

Mon, 04 May 2026 01:48:00 +0000

Content under review