Answer
$$ \theta =\cos^{-1}\frac{8}{3 \sqrt{21} }$$
Work Step by Step
Given $$v=(1,3,-1,4),w=(-1,1,-2,1)$$
Let $\theta$ be an angle between the vectors $v$ and $w .$
We know that
$$\cos \theta=\frac{\langle v,w\rangle}{\|v\|\|w\|}$$ .
Since, we have
\begin{aligned}
\langle v, w\rangle&= 1\cdot(-1)+(3) \cdot 1+(-1) \cdot(-2)+4 \cdot (1)\\
&=8\\
\|v\|&=\sqrt{\langle v, v\rangle}\\
&=\sqrt{1 \cdot 1+3 \cdot 3+(-1) \cdot(-1)+4 \cdot 4}\\
&=\sqrt{1+9+1+16}\\
&=\sqrt{27}\\
&=3 \sqrt{3}\\
\|w\|&=\sqrt{\langle w,w\rangle}\\
&=\sqrt{(-1) \cdot(-1)+1 \cdot 1+(-2) \cdot(-2)+1 \cdot 1}\\
&=\sqrt{1+1+4+1}\\
&=\sqrt{7}
\end{aligned} and
So we get:
$$\cos \theta=\frac{\langle v,w\rangle}{\|v\|\|w\|}=\frac{8}{3 \sqrt{3} \sqrt{7}}=\frac{8}{3 \sqrt{21} }$$
The angle between $v$ and $w$ is $\cos^{-1}\frac{8}{3 \sqrt{21} }$
