Show that the lines determined by

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

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

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

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

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

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

$$\begin{gathered} \mathbf{N}={\mathbf{d}}_1\times {\mathbf{d}}_2 \\ =⟨-1,-4,4⟩\times ⟨-1,2,1⟩ \end{gathered}$$

$$\begin{gathered} =⟨-12,-3,-6⟩ \\ =⟨4,1,2⟩ \end{gathered}$$

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

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

The equation $\left(1\right)$ and $\left(2\right)$ gives

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

Therefore, point of intersection is:

The point of intersection of two lines ${\mathbf{r}}_1\left(t\right)=⟨3-t,4-4t,-3+4t⟩$ and ${\mathbf{r}}_2\left(v\right)=⟨5-v,-6+2v,-2+v)$ is $(2,0,1)$