Solution:

Given two unit vectors $\vec{a}$ and $\vec{b}$, and the vectors $\vec{a} + 2\vec{b}$ and $5\vec{a} - 4\vec{b}$ are perpendicular, we use the condition for perpendicular vectors: $$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$$ Expanding the dot product: $$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = \vec{a} \cdot 5\vec{a} + \vec{a} \cdot (-4\vec{b}) + 2\vec{b} \cdot 5\vec{a} + 2\vec{b} \cdot (-4\vec{b})$$ Using properties of dot products and knowing $\vec{a}$ and $\vec{b}$ are unit vectors ($\vec{a} \cdot \vec{a} = 1$ and $\vec{b} \cdot \vec{b} = 1$): $$5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$$ Simplifying: $$5(1) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8(1) = 0$$ $$5 - 8 + 6(\vec{a} \cdot \vec{b}) = 0$$ $$-3 + 6(\vec{a} \cdot \vec{b}) = 0$$ $$6(\vec{a} \cdot \vec{b}) = 3$$ $$\vec{a} \cdot \vec{b} = \frac{1}{2}$$ The dot product $\vec{a} \cdot \vec{b} = \cos \theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$: $$\cos \theta = \frac{1}{2}$$ Therefore, the angle $\theta$ is: $$\theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ$$
Final Answer:
$$\boxed{60^\circ}$$

Solution:

Given vectors:

• $\vec{a} = \hat{i} - \hat{j}$
• $\vec{b} = \hat{i} + \hat{j} + \hat{k}$
• And $(\vec{a} \times \vec{c}) + \vec{b} = 0$
• $\vec{a} \cdot \vec{c} = 4$

From $(\vec{a} \times \vec{c}) + \vec{b} = 0$, we get: $$\vec{a} \times \vec{c} = -\vec{b}$$ Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$. The cross product $\vec{a} \times \vec{c}$ is: $$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix}$$ Expanding this determinant: $$\vec{a} \times \vec{c} = (z \hat{i} + z \hat{j} + (x + y) \hat{k})$$ Setting $\vec{a} \times \vec{c} = -\vec{b}$, we get: $$z = -1, \quad z = -1, \quad x + y = -1$$ Therefore: $$x + y = -1$$ Now, from $\vec{a} \cdot \vec{c} = 4$: $$\vec{a} \cdot \vec{c} = 1 \cdot x + (-1) \cdot y = 4$$ Simplifying: $$x - y = 4$$ Solving the system of equations: $$x + y = -1$$ $$x - y = 4$$ Adding the two equations: $$2x = 3 \quad \Rightarrow \quad x = \frac{3}{2}$$ Substituting into $x + y = -1$: $$\frac{3}{2} + y = -1 \quad \Rightarrow \quad y = -\frac{5}{2}$$ Now, $\vec{c} = \frac{3}{2} \hat{i} - \frac{5}{2} \hat{j} - \hat{k}$. To find $|\vec{c}|^2$, we compute: $$|\vec{c}|^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-1)^2 = \frac{9}{4} + \frac{25}{4} + 1$$ $$|\vec{c}|^2 = \frac{9 + 25 + 4}{4} = \frac{38}{4} = 9.5$$
Final Answer:
$$\boxed{9.5}$$