1. Problema: Determinar un ángulo entre los vectores indicados. Para encontrar el ángulo $\theta$ entre dos vectores $\mathbf{a}$ y $\mathbf{b}$, usamos la fórmula: $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$$ Donde $\mathbf{a} \cdot \mathbf{b}$ es el producto punto y $\|\mathbf{a}\|$, $\|\mathbf{b}\|$ son las magnitudes de los vectores. **a.** $\mathbf{a} = 3\mathbf{i} - \mathbf{k}$, $\mathbf{b} = 2\mathbf{i} + 2\mathbf{k}$ 1. Producto punto: $$\mathbf{a} \cdot \mathbf{b} = (3)(2) + (0)(0) + (-1)(2) = 6 - 2 = 4$$ 2. Magnitudes: $$\|\mathbf{a}\| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$$ $$\|\mathbf{b}\| = \sqrt{2^2 + 0^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8}$$ 3. Cálculo de $\cos \theta$: $$\cos \theta = \frac{4}{\sqrt{10} \times \sqrt{8}} = \frac{4}{\sqrt{80}} = \frac{4}{4\sqrt{5}} = \frac{\cancel{4}}{\cancel{4}\sqrt{5}} = \frac{1}{\sqrt{5}}$$ 4. Ángulo: $$\theta = \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$$ **b.** $\mathbf{a} = 2\mathbf{i} + \mathbf{j}$, $\mathbf{b} = -3\mathbf{i} - 4\mathbf{j}$ 1. Producto punto: $$\mathbf{a} \cdot \mathbf{b} = (2)(-3) + (1)(-4) = -6 - 4 = -10$$ 2. Magnitudes: $$\|\mathbf{a}\| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$$ $$\|\mathbf{b}\| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ 3. Cálculo de $\cos \theta$: $$\cos \theta = \frac{-10}{\sqrt{5} \times 5} = \frac{-10}{5\sqrt{5}} = \frac{\cancel{-10}}{\cancel{5}\sqrt{5}} = \frac{-2}{\sqrt{5}}$$ 4. Ángulo: $$\theta = \cos^{-1} \left( \frac{-2}{\sqrt{5}} \right)$$ **c.** $\mathbf{a} = (2,4,1)$, $\mathbf{b} = (-1,-1,4)$ 1. Producto punto: $$\mathbf{a} \cdot \mathbf{b} = 2(-1) + 4(-1) + 1(4) = -2 - 4 + 4 = -2$$ 2. Magnitudes: $$\|\mathbf{a}\| = \sqrt{2^2 + 4^2 + 1^2} = \sqrt{4 + 16 + 1} = \sqrt{21}$$ $$\|\mathbf{b}\| = \sqrt{(-1)^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$$ 3. Cálculo de $\cos \theta$: $$\cos \theta = \frac{-2}{\sqrt{21} \times \sqrt{18}} = \frac{-2}{\sqrt{378}} = \frac{-2}{3\sqrt{42}}$$ 4. Ángulo: $$\theta = \cos^{-1} \left( \frac{-2}{3\sqrt{42}} \right)$$