91Ó°ÊÓ

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

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 ${\mathbf{d}}_1=⟨2,-4,5âŸ\copyright$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 ${\mathbf{d}}_2=⟨3,2,3âŸ\copyright$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 ${\mathbf{d}}_1=⟨2,-4,5âŸ\copyright$and ${\mathbf{d}}_2=⟨3,2,3âŸ\copyright$

The normal vector is:

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

$=⟨-22,9,16âŸ\copyright$

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

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

Therefore, the lines are skew.

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

$\text{distance}=\frac{\left|\mathbf{N}·\stackrel{→}{R_1{R}_2}\right|}{‖\mathbf{N}‖}$

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 $\stackrel{¯}{R_1{R}_2}$ is given by:

The distance from $P$ to the plane is:

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