1. **State the problem:** Given vectors $ \vec{AB} = -3\mathbf{i} + 6\mathbf{j} $ and $ \vec{AC} = 10\mathbf{i} - 2\mathbf{j} $, find: (a) The size of angle $ \angle BAC $ in degrees to 1 decimal place. (b) The exact area of triangle $ \triangle ABC $. 2. **Find the size of $ \angle BAC $:** The angle between two vectors $ \vec{u} $ and $ \vec{v} $ is given by: $$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$$ Calculate the dot product: $$\vec{AB} \cdot \vec{AC} = (-3)(10) + (6)(-2) = -30 - 12 = -42$$ Calculate the magnitudes: $$|\vec{AB}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$$ $$|\vec{AC}| = \sqrt{10^2 + (-2)^2} = \sqrt{100 + 4} = \sqrt{104} = 2\sqrt{26}$$ Calculate $ \cos \theta $: $$\cos \theta = \frac{-42}{3\sqrt{5} \times 2\sqrt{26}} = \frac{-42}{6 \sqrt{130}} = \frac{-7}{\sqrt{130}}$$ Find $ \theta $ in degrees: $$\theta = \cos^{-1} \left( \frac{-7}{\sqrt{130}} \right) \approx 129.5^\circ$$ 3. **Find the exact area of $ \triangle ABC $:** The area of the triangle formed by vectors $ \vec{AB} $ and $ \vec{AC} $ is: $$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$$ In 2D, the magnitude of the cross product is: $$|\vec{AB} \times \vec{AC}| = |x_1 y_2 - y_1 x_2|$$ Calculate: $$|\vec{AB} \times \vec{AC}| = |-3 \times (-2) - 6 \times 10| = |6 - 60| = | -54 | = 54$$ Therefore, $$\text{Area} = \frac{1}{2} \times 54 = 27$$ **Final answers:** (a) $ \angle BAC \approx 129.5^\circ $ (b) Area of $ \triangle ABC = 27 $