Show that the given lines intersect and find the acute angle between them.

$$\begin{gathered} \mathbf{r}\mathbf{=}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{t}}_{\mathbf{1}}\mathbf{and} \\ \mathbf{r}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{)}{\mathbf{t}}_{\mathbf{2}} \end{gathered}$$

Equation of two lines are as follows,

$$\begin{gathered} r=(5,-2,0)+(1,-1-1)t_1\mathrm{and} \\ r=(4,-4,-1)+(0,3,2)t_2 \end{gathered}$$

Equating both the line equations and substituting the value for the variable gives the point of intersection. Finding the angle between the two given lines using formula for gives the acute angle between the lines.

$(5,-2,0)+(1,-1-1)t_1=(4,-4,-1)+(0,3,2)t_2$

This gives 3 equations

$$\begin{gathered} 5+t_1=4\to t_1=-1 \\ -2-t_1=4+3t_2 \\ -t_1=-1+2t_2 \end{gathered}$$

Substitute the second equation in the third one.

$-(-1)=-1+2t_1\to -2t_2=-2\to t_2$

So, the point of intersection is given by substituting $t_1=-1\mathrm{or}{t}_2=1$And both gives $r=4i-j+1k.$

The two vectors are as follows

$$\begin{gathered} A=i-j-k \\ B=3j+2k \end{gathered}$$

Let the angle between the A vectors B or$\theta$

$$\begin{gathered} A\cdot B=AB \\ \cos \theta \cos \theta =(A\cdot B)/AB \end{gathered}$$

$$\begin{gathered} A=\left|A\right|=\sqrt{(1{)}^2+(-1{)}^2+(-1{)}^2}=\sqrt{3} \\ B=\left|B\right|=\sqrt{(0{)}^2+(3{)}^2+(2{)}^2}=\sqrt{13} \\ A\times B=\left(1\right)\left(0\right)+(-1)\left(3\right)+(-1)\left(2\right)=-3-2=-5 \\ \cos \theta =\frac{A\times B}{AB}=\frac{-5}{\sqrt{13\times 3}} \\ \theta =co{s}^{-1}\frac{5}{\sqrt{39}}=143.2 \end{gathered}$$

But the acute angle is$180-\theta =180-143.2=36.8$

The angle between the lines is $\theta =36.8^0$.