1. **Problem statement:** Find the unit vector $\vec{C}$ that makes an angle of 60° with $2\hat{i} + 2\hat{j} - \hat{k}$ and an angle of 45° with $\hat{i} - \hat{k}$. Then compute $\vec{C} + \left(-\frac{1}{2}\hat{i} + \frac{1}{3\sqrt{2}}\hat{j} - \frac{\sqrt{2}}{3}\hat{k}\right)$. 2. **Recall the formula for angle between vectors:** If $\vec{a}$ and $\vec{b}$ are vectors, the angle $\theta$ between them satisfies $$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ Since $\vec{C}$ is a unit vector, $|\vec{C}|=1$. 3. **Set up equations using given angles:** Let $\vec{C} = x\hat{i} + y\hat{j} + z\hat{k}$ with $x^2 + y^2 + z^2 = 1$. - Angle 60° with $\vec{A} = 2\hat{i} + 2\hat{j} - \hat{k}$: $$\cos 60^\circ = \frac{\vec{C} \cdot \vec{A}}{|\vec{A}|} = \frac{2x + 2y - z}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{2x + 2y - z}{3} = \frac{1}{2}$$ Multiply both sides by 3: $$2x + 2y - z = \frac{3}{2}$$ - Angle 45° with $\vec{B} = \hat{i} - \hat{k}$: $$\cos 45^\circ = \frac{\vec{C} \cdot \vec{B}}{|\vec{B}|} = \frac{x - z}{\sqrt{1^2 + (-1)^2}} = \frac{x - z}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ Multiply both sides by $\sqrt{2}$: $$x - z = 1$$ 4. **Write the system of equations:** $$\begin{cases} 2x + 2y - z = \frac{3}{2} \\ x - z = 1 \\ x^2 + y^2 + z^2 = 1 \end{cases}$$ 5. **Express $z$ from second equation:** $$z = x - 1$$ 6. **Substitute $z$ into first equation:** $$2x + 2y - (x - 1) = \frac{3}{2} \implies 2x + 2y - x + 1 = \frac{3}{2} \implies x + 2y = \frac{3}{2} - 1 = \frac{1}{2}$$ So, $$2y = \frac{1}{2} - x \implies y = \frac{1}{4} - \frac{x}{2}$$ 7. **Substitute $y$ and $z$ into unit vector equation:** $$x^2 + \left(\frac{1}{4} - \frac{x}{2}\right)^2 + (x - 1)^2 = 1$$ Expand terms: $$x^2 + \left(\frac{1}{16} - \frac{x}{4} + \frac{x^2}{4}\right) + (x^2 - 2x + 1) = 1$$ Combine like terms: $$x^2 + \frac{1}{16} - \frac{x}{4} + \frac{x^2}{4} + x^2 - 2x + 1 = 1$$ $$\left(x^2 + \frac{x^2}{4} + x^2\right) + \left(-\frac{x}{4} - 2x\right) + \left(\frac{1}{16} + 1\right) = 1$$ $$\frac{9x^2}{4} - \frac{9x}{4} + \frac{17}{16} = 1$$ 8. **Multiply entire equation by 16 to clear denominators:** $$16 \times \frac{9x^2}{4} - 16 \times \frac{9x}{4} + 16 \times \frac{17}{16} = 16 \times 1$$ $$36x^2 - 36x + 17 = 16$$ 9. **Simplify:** $$36x^2 - 36x + 17 - 16 = 0 \implies 36x^2 - 36x + 1 = 0$$ 10. **Solve quadratic equation:** $$x = \frac{36 \pm \sqrt{(-36)^2 - 4 \times 36 \times 1}}{2 \times 36} = \frac{36 \pm \sqrt{1296 - 144}}{72} = \frac{36 \pm \sqrt{1152}}{72}$$ $$\sqrt{1152} = \sqrt{256 \times 4.5} = 16 \sqrt{4.5} = 16 \times \frac{3\sqrt{2}}{2} = 24\sqrt{2}$$ So, $$x = \frac{36 \pm 24\sqrt{2}}{72} = \frac{1}{2} \pm \frac{1}{3} \sqrt{2}$$ 11. **Find corresponding $y$ and $z$ for each $x$:** - For $x = \frac{1}{2} + \frac{1}{3} \sqrt{2}$: $$y = \frac{1}{4} - \frac{x}{2} = \frac{1}{4} - \frac{1}{2} \left(\frac{1}{2} + \frac{1}{3} \sqrt{2}\right) = \frac{1}{4} - \frac{1}{4} - \frac{\sqrt{2}}{6} = -\frac{\sqrt{2}}{6}$$ $$z = x - 1 = \left(\frac{1}{2} + \frac{1}{3} \sqrt{2}\right) - 1 = -\frac{1}{2} + \frac{1}{3} \sqrt{2}$$ - For $x = \frac{1}{2} - \frac{1}{3} \sqrt{2}$: $$y = \frac{1}{4} - \frac{x}{2} = \frac{1}{4} - \frac{1}{2} \left(\frac{1}{2} - \frac{1}{3} \sqrt{2}\right) = \frac{1}{4} - \frac{1}{4} + \frac{\sqrt{2}}{6} = \frac{\sqrt{2}}{6}$$ $$z = x - 1 = \left(\frac{1}{2} - \frac{1}{3} \sqrt{2}\right) - 1 = -\frac{1}{2} - \frac{1}{3} \sqrt{2}$$ 12. **Write the two possible unit vectors $\vec{C}$:** $$\vec{C}_1 = \left(\frac{1}{2} + \frac{1}{3} \sqrt{2}\right) \hat{i} - \frac{\sqrt{2}}{6} \hat{j} + \left(-\frac{1}{2} + \frac{1}{3} \sqrt{2}\right) \hat{k}$$ $$\vec{C}_2 = \left(\frac{1}{2} - \frac{1}{3} \sqrt{2}\right) \hat{i} + \frac{\sqrt{2}}{6} \hat{j} + \left(-\frac{1}{2} - \frac{1}{3} \sqrt{2}\right) \hat{k}$$ 13. **Add the given vector $\vec{D} = -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k}$ to $\vec{C}$:** - For $\vec{C}_1 + \vec{D}$: $$\left(\frac{1}{2} + \frac{1}{3} \sqrt{2} - \frac{1}{2}\right) \hat{i} + \left(-\frac{\sqrt{2}}{6} + \frac{1}{3\sqrt{2}}\right) \hat{j} + \left(-\frac{1}{2} + \frac{1}{3} \sqrt{2} - \frac{\sqrt{2}}{3}\right) \hat{k}$$ Simplify each component: $$i: \frac{1}{3} \sqrt{2}$$ $$j: -\frac{\sqrt{2}}{6} + \frac{1}{3\sqrt{2}} = -\frac{\sqrt{2}}{6} + \frac{\sqrt{2}}{6} = 0$$ $$k: -\frac{1}{2} + \frac{1}{3} \sqrt{2} - \frac{\sqrt{2}}{3} = -\frac{1}{2} + 0 = -\frac{1}{2}$$ So, $$\vec{C}_1 + \vec{D} = \frac{1}{3} \sqrt{2} \hat{i} + 0 \hat{j} - \frac{1}{2} \hat{k}$$ - For $\vec{C}_2 + \vec{D}$: $$\left(\frac{1}{2} - \frac{1}{3} \sqrt{2} - \frac{1}{2}\right) \hat{i} + \left(\frac{\sqrt{2}}{6} + \frac{1}{3\sqrt{2}}\right) \hat{j} + \left(-\frac{1}{2} - \frac{1}{3} \sqrt{2} - \frac{\sqrt{2}}{3}\right) \hat{k}$$ Simplify each component: $$i: -\frac{1}{3} \sqrt{2}$$ $$j: \frac{\sqrt{2}}{6} + \frac{\sqrt{2}}{6} = \frac{\sqrt{2}}{3}$$ $$k: -\frac{1}{2} - \frac{1}{3} \sqrt{2} - \frac{\sqrt{2}}{3} = -\frac{1}{2} - \frac{2}{3} \sqrt{2}$$ 14. **Final answers:** $$\boxed{\vec{C}_1 + \vec{D} = \frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}}$$ $$\boxed{\vec{C}_2 + \vec{D} = -\frac{\sqrt{2}}{3} \hat{i} + \frac{\sqrt{2}}{3} \hat{j} + \left(-\frac{1}{2} - \frac{2}{3} \sqrt{2}\right) \hat{k}}$$