= Solution
For FCC <aluminium>, consider $\{111\}\langle110\rangle$ slip. Normalize the tensile direction and one candidate system:
$$
\mathbf l_0=\frac{(1,2,4)}{\sqrt{21}},\qquad
\mathbf n=\frac{(-1,1,1)}{\sqrt3},\qquad
\mathbf s=\frac{(1,0,1)}{\sqrt2}.
$$
Here $\mathbf n\cdot\mathbf s=0$. Comparing all twelve systems gives the largest <Schmid factor> on $(\overline111)[101]$:
$$
m=|\mathbf l_0\cdot\mathbf n|\,|\mathbf l_0\cdot\mathbf s|
=\frac{25}{21\sqrt6}=0.48601.
$$
<Schmid's law> therefore gives
$$
\boxed{\sigma_y=\frac{0.78}{0.48601}=1.605\,\mathrm{MPa}.}
$$
During ideal single slip, the tensile direction expressed in crystal coordinates rotates \b[towards $[101]$], along the great-circle path joining the original axis to that slip direction. It reaches a symmetry boundary where another <slip system> becomes equally favored. This <single-slip tensile-axis rotation> concerns the crystal relative to the externally held loading direction, rather than a rotation of the laboratory loading apparatus.
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