Answer
$109.47^{\circ}$
Work Step by Step
The dot product of two vectors is:
$\cos \theta=\dfrac{\vec{a} \cdot \vec{b}}{|a| |b|}\\ =\dfrac{(i+j+k) (i-j-k)}{|i+j+k| |i-j-k|}\\=\dfrac{1}{\sqrt 3 \sqrt 3}\\=\dfrac{1}{3}$
Now, $\theta = \cos^{-1} \dfrac{1}{3} \approx 109.47^{\circ}$
