1. We are given two vectors \( \mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \) and \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \), and we need to find \( \cos\theta \) and \( \sin\theta \) where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \). 2. First, compute the dot product \( \mathbf{a} \cdot \mathbf{b} \): $$ \mathbf{a} \cdot \mathbf{b} = (2)(1) + (-3)(2) + (4)(2) = 2 - 6 + 8 = 4 $$ 3. Compute the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \): $$ |\mathbf{a}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} $$ $$ |\mathbf{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 $$ 4. Use the formula for cosine of the angle between two vectors: $$ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{4}{\sqrt{29} \times 3} = \frac{4}{3\sqrt{29}} $$ 5. Using the identity \( \sin^2\theta + \cos^2\theta = 1 \), compute \( \sin\theta \): $$ \sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \left( \frac{4}{3\sqrt{29}} \right)^2} = \sqrt{1 - \frac{16}{9 \times 29}} = \sqrt{1 - \frac{16}{261}} = \sqrt{\frac{261 - 16}{261}} = \sqrt{\frac{245}{261}} $$ 6. Simplify \( \sin\theta \) as much as possible: $$ \sin\theta = \frac{\sqrt{245}}{\sqrt{261}} = \frac{7\sqrt{5}}{\sqrt{261}} $$ Therefore, the final answers are: $$ \cos\theta = \frac{4}{3\sqrt{29}} $$ $$ \sin\theta = \frac{7\sqrt{5}}{\sqrt{261}} $$