Determine the angle between $$r$$ and $$s$$:
$$$ r:\dfrac{x+1}{3}=\dfrac{y-2}{-1}=\dfrac{z+1}{3} \qquad s:\left\{ \begin{array}{l} x=1-k \\ y=2+k \\ z=k \end{array} \right. $$$
See development and solution
Development:
We start by looking for governing vectors of $$r$$ and $$s$$:
- a governing vector of $$r$$ is $$\vec{u}= (3, -1, 3)$$.
- a governing vector of $$s$$ is $$\vec{v}=(-1, 1, 1)$$.
We can already apply the formula:
$$$ \begin{array}{rl} \cos(\alpha)=&\dfrac{|u_1 v_1+u_2 v_2+u_3 v_3|} {\sqrt{u_1^2+u_2^2+u_3^2}\sqrt{v_1^2+v_2^2+v_3^2}}= \dfrac{|3\cdot(-1)+(-1)\cdot1+3\cdot1|} {\sqrt{3^2+(-1)^2+3^2}\sqrt{(-1)^2+1^2+1^2}} \\ =& \dfrac{1}{\sqrt{19}\sqrt{3}}=0.13245 \end{array}$$$
Therefore,
$$$\alpha=\arccos(0.13245)=82.39^\circ$$$
Solution:
$$\alpha=82.39^\circ$$