Show that the lines determined by

${\mathbf{r}}_1\left(t\right)=⟨-2-5t,3+2t,4t⟩$and${\mathbf{r}}_2\left(t\right)=⟨8+15t,1-6t,3-12t⟩$

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

The lines determined by ${\mathbf{r}}_1\left(t\right)=⟨-2-5t,3+2t,4t⟩$and ${\mathbf{r}}_2\left(t\right)=⟨8+15t,1-6t,3-12t⟩$

The goal is to discover the plane equation that is determined by the given circumstances.

If the direction vectors of the provided lines are scalar multiples of each other, the lines are parallel.

The direction vector of the line ${\mathbf{r}}_1\left(t\right)=⟨-2-5t,3+2t,4t⟩$ is ${\mathbf{d}}_1=⟨-5,2,4⟩$ and direction vector of the line ${\mathbf{r}}_2\left(t\right)=⟨8+15t,1-6t,3-12t⟩$ is ${\mathbf{d}}_2=⟨15,-6,-12⟩$

The direction vector ${\mathbf{d}}_2=⟨15,-6,-12⟩$is:

${\mathbf{d}}_2=-3{\mathbf{d}}_1$

Because the direction vectors are scalar multiples of each other, the lines are parallel.

The point $Q_0=(-2,3,0)$is on the line ${\mathbf{r}}_i\left(t\right)=⟨-2-5t,3+2t,4t⟩$and the point $Q=(8,1,3)$is on the line ${\mathbf{r}}_2\left(t\right)=⟨8+15t,1-6t,3-12t⟩$

The vector parallel to the plane of interest is:

The normal vector to the plane is the cross product of $d_1=⟨-5,2,4⟩$and the vector $\overrightarrow{Q_{0}Q}$

The normal vector is:

$=⟨14,55,-10⟩$

The equation of the plane is:

The equation of the plane that contains the llne ${\mathbf{r}}_1\left(t\right)=⟨-2-5t,3+2t,4t⟩$and the line ${\mathbf{r}}_2\left(t\right)=⟨8+15t,1-6t,3-12t⟩\mid$is $14x+55y-10z-137=0$