Discussion
In a cubic crystal, the Burgers vector for a mixed dislocation line is \(\frac{a}{2}[110]\). The dislocation line lies along the [011] direction. The slip plane of the dislocation is
Answer: A
The slip plane is \((1\bar{1}1)\).
The slip plane of a dislocation contains both its Burgers vector and its line direction, so its normal is their cross product:
\[ \begin{aligned} [110] \times [011] &= (1\cdot1-0\cdot1,\ 0\cdot0-1\cdot1,\ 1\cdot1-1\cdot0) \\ &= [1\bar{1}1] \end{aligned} \]
In a cubic crystal the plane with normal \([1\bar{1}1]\) is \((1\bar{1}1)\). Check: \(1\cdot1 + 1\cdot(-1) + 0\cdot1 = 0\) and \(0\cdot1 + 1\cdot(-1) + 1\cdot1 = 0\), so both vectors lie in it. (111) contains neither vector, \((0\bar{1}1)\) contains [011] but not [110], and \((1\bar{1}0)\) contains neither.
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