Answer
Angle $\approx 35^{\circ}$
Work Step by Step
We calculate the vectors as follows:
$\vec{AB}=⟨1-0,0-0,0-1⟩=⟨1,0,-1⟩$
$\vec{AC}=⟨1-0,1-0,0-1⟩=⟨1,1,-1⟩$
How, we calculate the lengths:
$||\vec{AB}||=\sqrt {1^{2}+0^{2}+(-1)^{2}}=\sqrt 2$
$||\vec{AC}||=\sqrt {1^{2}+1^{2}+(-1)^{2}}=\sqrt 3$
Next, compute the dot product:
$\vec{AB}\cdot\vec{AC}=⟨1,0,-1⟩\cdot⟨1,1,-1⟩$
$=1\times1+0\times1+(-1)\times(-1)=2$
But $\vec{AB}\cdot\vec{AC}=||\vec{AB}||\,||\vec{AC}||\cos\theta$, where $\theta$ is the angle between $\overline{AB}$ and $\overline{AC}$.
$\implies 2=\sqrt {2}\times\sqrt {3}\times\cos\theta$
Or,
$\theta=\cos^{-1}(\frac{2}{\sqrt {2}\times\sqrt {3}})\approx 35^{\circ}$
