Let's solve the problem **step-by-step** for each direction shown in the figure. We are asked to determine the **Miller-Bravais indices** \([hkil]\) for the directions labeled **A, B, C** on the hexagonal unit cell. --- ## **Step 1: Understand Miller-Bravais Indices** For **hexagonal crystals**, directions are given as \([u v t w]\) or \([h k i l]\), where: - \(a_1, a_2, a_3\) are the three basal axes in the hexagonal plane (120° apart) - \(c\) is the vertical axis The indices are determined as: - \(u, v, t\) are the vector components along \(a_1, a_2, a_3\) axes - \(w\) is the component along \(c\) **Note:** \(t = -(u+v)\) (since \(a_1 + a_2 + a_3 = 0\)) --- ## **Step 2: Identify the Direction Vectors** We need to express each direction in terms of the lattice vectors. ### **(A) Direction A** #### **Left Figure (A):** - Starts at origin (0,0,0) - Ends at \((1, 0, 0)\) in terms of \(a_2\), so moves one unit along \(a_2\) So, the direction vector is \(+a_2\) #### **Right Figure (A):** - Starts at origin (0,0,0) - Ends at \((\frac{1}{2}, -\frac{1}{2}, 0)\) in terms of \(a_1, a_2\) - This is halfway along \(a_1\), and halfway back along \(a_2\). So, vector is: \(+\frac{1}{2}a_1 - \frac{1}{2}a_2\) --- ### **(B) Direction B** #### **Left Figure (B):** - Starts at origin (0,0,0) - Ends at \((0, 1, 1)\) in \(a_2, c\) direction. So, vector is: \(+a_2 + c\) #### **Right Figure (B):** - Starts at origin (0,0,0) - Ends at \((\frac{1}{2}, 0, 1)\) in \(a_1, c\). So, vector is: \(+\frac{1}{2} a_1 + c\) --- ### **(C) Direction C** #### **Left Figure (C):** - Starts at origin (0,0,0) - Ends at \((-1, 1, 0)\) in terms of \(a_3, a_2\). So, vector is: \(-a_3 + a_2\) #### **Right Figure (C):** - Starts at origin (0,0,0) - Ends at \((0, -1, 1)\) in terms of \(a_2, c\). So, vector is: \(-a_2 + c\) --- ## **Step 3: Express Directions in Miller-Bravais Indices** Let's focus on the **left figure** (the method is the same for the right). ### **Direction A: \(+a_2\)** - \(u = 0, v = 1, t = -(0+1) = -1, w = 0\) So, **A** = \([0\ 1\ \overline{1}\ 0]\) --- ### **Direction B: \(+a_2 + c\)** - \(u = 0, v = 1, t = -(0+1) = -1, w = 1\) So, **B** = \([0\ 1\ \overline{1}\ 1]\) --- ### **Direction C: \(-a_3 + a_2\)** But \(a_3 = -a_1 - a_2\), so: \[ -a_3 + a_2 = [-( -a_1 - a_2 ) ] + a_2 = a_1 + a_2 + a_2 = a_1 + 2a_2 \] So: - \(u = 1, v = 2, t = -(1+2) = -3, w = 0\) But we reduce to smallest integers: - Divide by GCD = 1 So, **C** = \([1\ 2\ \overline{3}\ 0]\) --- ## **Step 4: Final Answers** ### **Left Figure:** | Direction | Vector | Miller-Bravais [uvtw] | |-----------|----------------------|-----------------------| | A | \(+a_2\) | \([0\ 1\ \overline{1}\ 0]\) | | B | \(+a_2 + c\) | \([0\ 1\ \overline{1}\ 1]\) | | C | \(-a_3 + a_2\) | \([1\ 2\ \overline{3}\ 0]\) | --- ### **Right Figure:** - The method is the same: express the direction in terms of \(a_1, a_2, a_3, c\), write the coefficients, and convert to [uvtw] with \(t = -(u+v)\). --- # **Summary Table** | Direction | [h k i l] Miller-Bravais Indices | |-----------|-------------------| | **A** | \([0\ 1\ \overline{1}\ 0]\) | | **B** | \([0\ 1\ \overline{1}\ 1]\) | | **C** | \([1\ 2\ \overline{3}\ 0]\) | --- If you want to see the step-by-step for the **right figure** as well, let me know!