Question

Show that the lines with the equations

$\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z-2}{5}$and$\frac{x-4}{3}=\frac{y+1}{2}=\frac{z}{3}$

are skew, find the equations of the parallel planes containing the lines, and find the distance between the lines.

Short answer

The distance between parallel to the plane is $\frac{178}{\sqrt{821}}$

Step-by-step solution

Given information

The lines with the equations $\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z-2}{5}$and $\frac{x-4}{3}=\frac{y+1}{2}=\frac{z}{3}$

Calculation

The goal is to demonstrate that lines are skew and to determine the equation of the parallel planes that contain the lines as well as the distance between them.

The line $\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z-2}{5}$gives the direction vector ${\mathbf{d}}_1=⟨2,-4,5⟩$and the point $Q_0=(-1,3,2)$

The line $\frac{x-4}{3}=\frac{y+1}{2}=\frac{z}{3}$gives the direction vector ${\mathbf{d}}_2=⟨3,2,3⟩$and the point $Q_1=(4,-1,0)$

The direction vectors do not have a scalar multiple. As a result, the lines are either crossing or skew.

The normal vector to the plane is the cross product of ${\mathbf{d}}_1=⟨2,-4,5⟩$and ${\mathbf{d}}_2=⟨3,2,3⟩$

The normal vector is:

$\mathbf{N}=\left|\begin{array}{c}ijk\\ 2-45\\ 323\end{array}\right|$

$=⟨-22,9,16⟩$

The equation of the plane from the line $\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z-2}{5}$is:

$$\begin{gathered} -22(x+1)+9(y-3)+16(z-2)=0 \\ -22x+9y+16z-22-27-32=0\text{(Simplify)} \\ -22x+9y+16z-81=0\text{(Combine like terms)} \end{gathered}$$

The equation of the plane from the line $\frac{x-4}{3}=\frac{y+1}{2}=\frac{z}{3}$ is:

$$\begin{gathered} -22(x-4)+9(y+1)+16(z-0)=0 \\ -22x+9y+16z+88+9+0=0\text{(Simplify)} \\ -22x+9y+16z+97=0\text{(Combine like terms)} \end{gathered}$$

Therefore, the lines are skew.

Calculation

The distance between two parallel planes can be calculated as follows:

$\text{distance}=\frac{\left|\mathbf{N}·\overrightarrow{R_1{R}_2}\right|}{\Vert \mathbf{N}\Vert }$

The point $R_1=(0,9,0)$ is lies on the plane $-22x+9y+16z-81=0$ and the point $R_2=\left(0,-\frac{97}{9},0\right)$ lies on the plane $-22x+9y+16z+97=0$

The vector $\overline{R_1{R}_2}$ is given by:

The distance from $P$ to the plane is:

Therefore, the distance between parallel to the plane is $\frac{178}{\sqrt{821}}$