Answer
$60^{\circ}$
Work Step by Step
We calculate the vectors as follows:
$\vec{AB}=⟨1-0,0-0,0-1⟩=⟨1,0,-1⟩$
$\vec{AD}=⟨0-0,1-0,0-1⟩=⟨0,1,-1⟩$
Next, compute the lengths:
$||\vec{AB}||=\sqrt {1^{2}+0^{2}+(-1)^{2}}=\sqrt 2$
$||\vec{AD}||=\sqrt {0^{2}+1^{2}+(-1)^{2}}=\sqrt 2$
Now, calculate the dot product:
$\vec{AB}\cdot\vec{AC}=⟨1,0,-1⟩\cdot⟨0,1,-1⟩$
$=1\times0+0\times1+(-1)\times(-1)=1$
But $\vec{AB}\cdot\vec{AD}=||\vec{AB}||\,||\vec{AD}||\cos\theta$, where $\theta$ is the angle between $\overline{AB}$ and $\overline{AD}$.
$\implies 1=\sqrt {2}\times\sqrt {2}\times\cos\theta$
Or $\theta=\cos^{-1}(\frac{1}{2})=60^{\circ}$
