Show that the lines determined by

${\mathbf{r}}_1\left(t\right)=⟨7,3-4t,2+6t⟩$

${\mathbf{r}}_2\left(t\right)=⟨6-t,3+8t,9-5t⟩$

intersect, and then find an equation of the plane containing the two lines.

The two lines ${\mathbf{r}}_1\left(t\right)=⟨7,3-4t,2+6t⟩$and ${\mathbf{r}}_2\left(t\right)=⟨6-t,3+8t,9-5t⟩$

The goal is to demonstrate that lines intersect and to determine the equation of a plane containing two lines.

The direction vector of the line ${\mathbf{r}}_1\left(t\right)=⟨7,3-4t,2+6t⟩$is ${\mathbf{d}}_1=⟨0,-4,6⟩$and the direction vector of the line ${\mathbf{r}}_2\left(t\right)=⟨6-t,3+8t,9-5t⟩$is ${\mathbf{d}}_2=⟨-1,8,-5⟩$

The lines ${\mathbf{r}}_1\left(t\right)=⟨7,3-4t,2+6t⟩$and ${\mathbf{r}}_2\left(t\right)=⟨6-t,3+8t,9-5t⟩$are intersecting because the direction vectors ${\mathbf{d}}_1=⟨0,-4,6⟩$and ${\mathbf{d}}_2=⟨-1,8,-5⟩$are not the scalar multiple of each other.

The normal vector $\mathbf{N}$ is given by:

The point of intersection of the two lines ${\mathbf{r}}_1\left(t\right)=⟨7,3-4t,2+6t⟩$and ${\mathbf{r}}_2\left(v\right)=⟨6-v,3+8v,9-5v)$is given by:

$$\begin{gathered} 7=6-v\dots ..\left(1\right) \\ 3-4t=3+8v\dots ..\left(2\right) \\ 2+6t=9-5v\dots \dots \left(3\right) \end{gathered}$$

The equation $\left(1\right)$gives

$v=-1$

The equation (2) and $v=-1$gives:

$t=2$

The values $v=-1$and $t=2$satisfies the equation $\left(3\right)$

Therefore, point of intersection is:

$$\begin{gathered} P\left(\text{intersection}\right)=(7,3-4(2),2+6(2\left)\right) \\ =(7,-5,14) \end{gathered}$$

The point of intersection of two lines ${\mathbf{r}}_1\left(t\right)=⟨7,3-4t,2+6t⟩$and ${\mathbf{r}}_2\left(v\right)=⟨6-v,3+8v,9-5v⟩$is $(7,-5,14)$

The equation of the plane is:

The equation of the plane that contains the lines ${\mathbf{r}}_1\left(t\right)=⟨7,3-4t,2+6t⟩$and ${\mathbf{r}}_2\left(t\right)=⟨6-t,3+8t,9-5t⟩$ is $14x+3y+2z-111=0$