# Step-by-step Solution ## Step 1: Establish the coordinate system - Assume the cube side length \(a\). - the cube in a 3D system with: - \(B\) at the origin: \(B(, , )\) - \(A\) at \(( ,)\) \(C\) at \((a, , a)\) - \(D\) at \((a, , )\) - The other points are on the top face: - \(A(, a, a)\), etc. ## Step 2: Determine the points in the figure - \(A\) is at \((, a, a)\) - \(B\) at \((, , )\) - \(C\) at \((a, , a)\) - \(D\) at \((a, , )\) The problem shows the points \(A, C, D\) inside the cube, with lines connecting these points to \(A\) and \(B\), and to each other. ## Step 3: Find the directions ### 3.1. Direction \(\vec{A C}\) - \(A = (, a, a)\) - \(C = (a, , a)\) \[ \vec{A C} = C - A = (a - , - a, a - a) = (a, -a, ) \] **Index for \(\vec{A C}\):** \[ \hat{u}_{A C} = \frac{\vec{A C}}{|\vec{A C}|} \] \[ |\vec{A C}| = \sqrt{a^2 + (-a)^2 + ^2} = \sqrt{2a^2} = a\sqrt{2} \] \[ \Rightarrow \hat{u}_{A C} = \left(\frac{a}{a\sqrt{2}}, \frac{-a}{a\sqrt{2}}, \right) = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, \right) \] ### 3.2. Direction \(\vec{A D}\) - \(A = (, a, a)\) - \(D = (a, , )\) \[ \vec{A D} = D - A = (a - , - a, - a) = (a, -a, -a) \] \[ |\vec{A D}| = \sqrt{a^2 + (-a)^2 + (-a)^2} = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \] \[ \Rightarrow \hat{u}_{A D} = \left(\frac{a}{a\sqrt{3}}, \frac{-a}{a\sqrt{3}}, \frac{-a}{a\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \] ### 3.3. Direction \(\vec{B C}\) - \(B = (, , )\) - \(C = (a, , a)\) \[ \vec{B C} = C - B = (a, , a) \] \[ |\vec{B C}| = \sqrt{a^2 + + a^2} = a\sqrt{2} \] \[ \hat{u}_{B C} = \left(\frac{a}{a\sqrt{2}}, , \frac{a}{a\sqrt{2}}\right) = \left(\frac{1}{\sqrt{2}}, , \frac{1}{\sqrt{2}}\right) \] ## Step 4: Determine the planes Planes are defined by the points and their normal vectors, which are perpendicular to the directions of the lines. ### 4.1. Plane \(ABC\) - Points: \(A, B, C\) Compute vectors: \[ \vec{A B} = B - A = (, , ) - (, a, a) = (, -a, -a) \] \[ \vec{A C} = (a, -a, ) \] Normal vector: \[ \vec{N}_{ABC} = \vec{A B} \times \vec{A C} \] Cross product: \[ \begin{aligned} \vec{N}_{ABC} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ & -a & -a \\ a & -a & \end{vmatrix} \\ &= \mathbf{i}((-a)\cdot - (-a)(-a)) - \mathbf{j}( \cdot - (-a) \cdot a) + \mathbf{k}( \cdot (-a) - (-a) \cdot a) \\ &= \mathbf{i}( - a^2) - \mathbf{j}( + a^2) + \mathbf{k}( + a^2) \\ &= (-a^2, -a^2, a^2) \end{aligned} \] Simplify: \[ \vec{N}_{ABC} = a^2(-1, -1, 1) \] Equation of plane \(ABC\): \[ -1(x - x_) - 1(y - y_) + 1(z - z_) = \] Using point \(A(, a, a)\): \[ -1(x - ) - 1(y - a) + 1(z - a) = \] \[ - x - (y - a) + (z - a) = \] \[ - x - y + a + z - a = \] \[ - x - y + z = \] --- ### 4.2. Plane \(ABD\) - Points: \(A, B, D\) Vectors: \[ \vec{A B} = (, -a, -a) \] \[ \vec{A D} = (a, -a, -a) \] Normal vector: \[ \vec{N}_{ABD} = \vec{A B} \times \vec{A D} \] Calculations: \[ \begin{aligned} \vec{N}_{ABD} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ & -a & -a \\ a & -a & -a \end{vmatrix} \\ &= \mathbf{i}((-a)(-a) - (-a)(-a)) - \mathbf{j}( \cdot (-a) - (-a) \cdot a) + \mathbf{k}( \cdot (-a) - (-a) \cdot a) \\ &= \mathbf{i}(a^2 - a^2) - \mathbf{j}( + a^2) + \mathbf{k}( + a^2) \\ &= (, -a^2, a^2) \end{aligned} \] Equation of plane \(ABD\): Using point \(A(, a, a)\): \[ (x - ) - a^2(y - a) + a^2(z - a) = \] \[ - a^2 y + a^2 a + a^2 z - a^2 a = \] \[ - a^2 y + a^2 z = \] Divide through by \(a^2\): \[ - y + z = \Rightarrow y = z \] --- ### 4.3. Plane \(ACD\) - Points: \(A, C, D\) Vectors: \[ \vec{A C} = (a, -a, ) \] \[ \vec{A D} = (a, -a, -a) \] Normal vector: \[ \vec{N}_{ACD} = \vec{A C} \times \vec{A D} \] Calculate: \[ \begin{aligned} \vec{N}_{ACD} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & -a & \\ a & -a & -a \end{vmatrix} \\ &= \mathbf{i}((-a)(-a) - \cdot (-a)) - \mathbf{j}(a \cdot (-a) - \cdot a) + \mathbf{k}(a \cdot (-a) - (-a) \cdot a) \\ &= \mathbf{i}(a^2 - ) - \mathbf{j}(-a^2 - ) + \mathbf{k}(-a^2 - (-a^2)) \\ &= (a^2, a^2, ) \end{aligned} \] Equation of plane \(ACD\): Using point \(A(, a, a)\): \[ a^2(x - ) + a^2(y - a) + (z - a) = \] \[ a^2 x + a^2 y - a^3 = \] Divide through by \(a^2\): \[ x + y - a = \] ## Summary of plane equations: | Plane | Equation | |---------|--------------| | \(ABC\) | \(-x - y + z = \) | | \(ABD\) | \(y = z\) | | \(ACD\) | \(x + y = a\) | --- ## Final note: The indices for the directions are based on the normalized vectors: - \(\vec{A C}\): \(\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, \right)\) - \(\vec{A D}\): \(\left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)\) - \(\vec{B C}\): \(\left(\frac{1}{\sqrt{2}}, , \frac{1}{\sqrt{2}}\right)\) These vectors are normal to their respective planes and define the directions for the problem. --- **Note:** The above steps provide the detailed derivation of the directions and planes based on the given cube and points.

# Theory of Directions and Planes in Crystallography ## Introduction to Crystallography Crystallography is the study of crystal structures and their properties. It involves understanding the arrangement of atoms within a crystal and how this affects their physical and chemical properties. ## Directions in Crystals ### Definition In crystallography, a direction is represented by a vector that indicates the orientation of a straight line through the crystal lattice. The directions are typically expressed in terms of Miller indices. ### Miller Indices for Directions Miller indices for directions are denoted as \([u v w]\) and are derived from the coordinates of the vector connecting two points in the crystal lattice. The indices are obtained by: 1. Identifying the coordinates of the two points. 2. Calculating the vector. 3. Reducing the vector to the smallest integer values. ### Example: Direction \(\vec{A C}\) To find the direction from point \(A\) to point \(C\): - Compute the vector \(\vec{A C}\). - Normalize the vector by dividing each component by its length, yielding the direction indices. ## Planes in Crystals ### Definition Planes in crystallography represent two-dimensional surfaces that can be defined by the arrangement of atoms in the crystal lattice. ### Miller Indices for Planes Miller indices for planes are denoted as \((h k l)\) and are defined by the following steps: 1. Determine the intercepts of the plane with the axes. 2. Take the reciprocals of these intercepts. 3. Clear any fractions and reduce to the smallest integers. ### Example: Plane \(ABC\) To define the plane formed by points \(A\), \(B\), and \(C\): - Compute the vectors \(\vec{A B}\) and \(\vec{A C}\). - Calculate the normal vector using the cross product of these vectors. - Use the normal vector to write the equation of the plane. ## Summary Understanding directions and planes is crucial in crystallography for analyzing crystal structures. The use of Miller indices provides a standardized way to describe these orientations, facilitating communication about crystal properties and behaviors.