find the indices of the direction that result from the intersection of each of the following pairs of planes…

Explanation:

Step1: Recall vector - cross - product rule for finding intersection direction

The direction of the intersection of two planes with Miller indices $(h_1,k_1,l_1)$ and $(h_2,k_2,l_2)$ in a cubic crystal can be found using the vector cross - product $\vec{u}\times\vec{v}$, where $\vec{u}=(h_1,k_1,l_1)$ and $\vec{v}=(h_2,k_2,l_2)$. The cross - product formula is $\vec{u}\times\vec{v}=\left(k_1l_2 - k_2l_1,l_1h_2 - l_2h_1,h_1k_2 - h_2k_1\right)$.

Step2: Solve part (a)

For the planes $(1\ 0\ 0)$ and $(0\ 1\ 0)$, using the cross - product formula: [ \begin{align*} \vec{u}\times\vec{v}&=(0\times0 - 1\times0,0\times0 - 0\times1,1\times1 - 0\times0)\ &=(0,0,1) \end{align*} ] The direction indices are $[0\ 0\ 1]$.

Step3: Solve part (b)

For the planes $(1\ 1\ 1)$ and $(1\ 1\ - 1)$, [ \begin{align*} \vec{u}\times\vec{v}&=(1\times(-1)-1\times1,1\times1 - (-1)\times1,1\times1 - 1\times1)\ &=(-2,2,0) \end{align*} ] Dividing by the greatest common divisor (2), the direction indices are $[-1\ 1\ 0]$.

Step4: Solve part (c)

For the planes $(1\ 0\ - 1)$ and $(0\ 0\ 1)$, [ \begin{align*} \vec{u}\times\vec{v}&=(0\times1 - 0\times(-1),(-1)\times0 - 1\times1,1\times0 - 0\times0)\ &=(0,-1,0) \end{align*} ] The direction indices are $[0\ - 1\ 0]$.

Answer:

(a) $[0\ 0\ 1]$ (b) $[-1\ 1\ 0]$ (c) $[0\ - 1\ 0]$