1. **State the problem:** We are given two vectors $\mathbf{a} = 2\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 3\mathbf{i} + 4\mathbf{k}$. We need to find the direction cosines of each vector. 2. **Formula and explanation:** The direction cosines of a vector are the cosines of the angles that the vector makes with the coordinate axes. For a vector $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, the direction cosines are given by: $$\cos \alpha = \frac{x}{|\mathbf{v}|}, \quad \cos \beta = \frac{y}{|\mathbf{v}|}, \quad \cos \gamma = \frac{z}{|\mathbf{v}|}$$ where $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$ is the magnitude of the vector. 3. **Calculate magnitude of $\mathbf{a}$:** $$|\mathbf{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$ 4. **Calculate direction cosines of $\mathbf{a}$:** $$\cos \alpha_a = \frac{2}{3}, \quad \cos \beta_a = \frac{2}{3}, \quad \cos \gamma_a = \frac{-1}{3}$$ 5. **Calculate magnitude of $\mathbf{b}$:** $$|\mathbf{b}| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5$$ 6. **Calculate direction cosines of $\mathbf{b}$:** $$\cos \alpha_b = \frac{3}{5}, \quad \cos \beta_b = \frac{0}{5} = 0, \quad \cos \gamma_b = \frac{4}{5}$$ **Final answer:** - Direction cosines of $\mathbf{a}$ are $\left(\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}\right)$. - Direction cosines of $\mathbf{b}$ are $\left(\frac{3}{5}, 0, \frac{4}{5}\right)$.