Answer
The angle between the two vectors $v$ and $w$ is $ \theta =\cos^{-1} \dfrac{3}{3 \sqrt {14}}=74.5^{\circ}$ and $v \cdot w=3$.
Work Step by Step
Let us consider two vectors $v=pi+qj+rk$ and $w=xi+yj+zk$. If $\theta$ is the angle between the two vectors, then we have: $ \cos \theta =\dfrac{v \cdot w}{|v||w|}$
In our case: $v \cdot w = (2i+2j-k) \cdot (i+2j+3k)=2+4-3=3$
$ \cos \theta =\dfrac{3}{3 \sqrt {14}}$
or, $ \theta =\cos^{-1} \dfrac{3}{3 \sqrt {14}}$
Therefore, the angle between the two vectors $v$ and $w$ is $ \theta =\cos^{-1} \dfrac{3}{3 \sqrt {14}}=74.5^{\circ}$ and $v \cdot w=3$.
