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

$$\begin{gathered} \mathit{r}\mathbf{=}\mathbf{2}\mathit{j}\mathbf{+}\mathit{k}\mathbf{+}\mathbf{(}\mathbf{3}\mathit{i}\mathbf{-}\mathit{k}\mathbf{)}{\mathit{t}}_{\mathbf{1}}\mathbf{and} \\ \mathit{r}\mathbf{=}\mathbf{7}\mathit{i}\mathbf{+}\mathbf{2}\mathit{k}\mathbf{+}\mathbf{(}\mathbf{2}\mathit{i}\mathbf{-}\mathit{j}\mathbf{+}\mathit{k}\mathbf{)}{\mathit{t}}_{\mathbf{2}} \end{gathered}$$

Equation of two lines:

$$\begin{gathered} r=2j+k+(3i-k)t_1\mathrm{and} \\ r=7i+2k+(2i-j+k)t_2 \end{gathered}$$

To prove and find: Show that the given lines intersect and find the acute angle between.

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.

$(0,2,1)+(3,0,-1)t_1=(7,0,2)+(2,-1,1)t_2$

This gives equations.

$$\begin{gathered} 3t_1=7+t_2 \\ 2=-t_2 \\ 1-t_1=2+t_2 \end{gathered}$$

Substituting the second equation in the third one gives the following.

$$\begin{gathered} 1-t_1=2+(-2) \\ {t}_1=1 \end{gathered}$$

So, the point of intersection is given by substituting$t_1=1\mathrm{or}{t}_2=-2$

And both gives$r=3i+2j$

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

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

$$\begin{gathered} A\times B=AB\cos \theta \\ \cos \theta =\frac{A\times B}{AB} \end{gathered}$$

Now calculate the following values.

$$\begin{gathered} A=\left|A\right|=\sqrt{(3{)}^2+(0{)}^2+(-1{)}^2}=\sqrt{10} \\ B=\left|B\right|=\sqrt{(2{)}^2+(-1{)}^2+(1{)}^2}=\sqrt{6} \\ A\times B=\left(3\right)\left(2\right)+\left(0\right)\left(1\right)+(-1)\left(1\right)=6-1=5 \\ \cos \theta =\frac{A\times B}{AB}=\frac{5}{\sqrt{10\times 6}} \\ \theta =co{s}^{-1}\frac{5}{\sqrt{60}}=49.8 \end{gathered}$$

The angle between the lines is $\theta =49.8°$.