Question

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

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

Short answer

The Acute angle between the lines is:$\theta =49.8$

Step-by-step solution

Given Information:

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.

Concept used

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.

To prove that the lines intersect

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

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$r=3i+2j$

To find the acute angle between the lines

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

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 $\theta =49.8°$.