3D Coordinate Geometry - Skew Lines

There are three possible types of relations that two different lines can have in a three-dimensional space. They can be

1. parallel, when their direction vectors are parallel and the two lines never meet;
2. meeting at a single point, when their direction vectors are not parallel and the two lines intersect;
3. skew, which means that they never meet and are not parallel.

All three of these relations can be found in a cuboid. In the cuboid shown in the diagram below, edges $\overline{AB}$ and $\overline{CD}$ are parallel. Edges $\overline{AB}$ and $\overline{BC}$ intersect at a single point $B.$ Edges $\overline{AB}$ and $\overline{EH}$ are skew, since they are not parallel and never meet.

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Find all edges that are skew to $\overline{AE}$ in the cuboid shown below.

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We should find all the lines that do not meet with $\overline{AE}$ and are not parallel to $\overline{AE},$ which are edges $\overline{CD},$ $\overline{GH},$ $\overline{BC},$ and $\overline{FG}.$ Observe that any edge in a cuboid is skew to four other edges. $_\square$

Find all edges that are skew to $\overline{AB}$ in the pentagonal prism shown below.

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We should find all the edges that do not meet with $\overline{AB}$ and are not parallel to $\overline{AB}.$ The edges that meet with $\overline{AB}$ are $\overline{AE},$ $\overline{AV},$ $\overline{BW},$ and $\overline{BC}.$ The edge $\overline{VW}$ is parallel to $\overline{AB}.$ Therefore all the edges except for these are skew to $\overline{AB},$ which are edges $\overline{VZ},$ $\overline{EZ},$ $\overline{ED},$ $\overline{ZY},$ $\overline{DY},$ $\overline{DC},$ $\overline{YX},$ $\overline{CX},$ and $\overline{XW}.$ $_\square$

Determine the relation between the following two lines:

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

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The direction vectors of the two lines are $\vec{d_1}=(2,3,-5)$ and $\vec{d_2}=(1,1,3).$ Since their direction vectors are not parallel, the two lines either intersect at a single point or are skew to each other.

Now let's find out if the two lines meet. Equating the first equation to $t$ gives

$$\begin{aligned} \frac{x-1}{2}=\frac{y}{3}=\frac{z+2}{-5}&=t\\ \Rightarrow x&=2t+1\\ y&=3t\\ z&=-5t-2, \end{aligned}$$

so any point on the first line can be expressed as $(2t+1,3t,-5t-2).$ Plugging this into the second equation gives

$$2t+1=3t-4=\frac{-5t-2}{3}.$$

Observe that there is no such real number $t$ that satisfies this equation. Then the two lines do not meet, so they are skew (because they are not parallel, either, as proved earlier). $_\square$

The (shortest) distance between a pair of skew lines can be found by obtaining the length of the line segment that meets perpendicularly with both lines, which is $d$ in the figure below.

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Find the distance between the following pair of skew lines:

$$x=-y+2=-z+2\quad \text{and}\quad x-2=-y+1=z+1.$$

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Let $x=-y+2=-z+2$ be $l_1$ and $x-2=-y+1=z+1$ be $l_2.$ We should find the length of $\overline{AB},$ which is the line segment that meets perpendicularly with both $l_1$ and $l_2.$ Equating the equation of $l_1$ with $t$ gives

$$\begin{aligned} x=-y+2=-z+2&=t\\ \Rightarrow x&=t\\ y&=-t+2\\ z&=-t+2. \end{aligned}$$

Hence point $A$ can be expressed as $(t,-t+2,-t+2)$ for some real number $t.$ Applying the same method for $l_2$ gives

$$\begin{aligned} x-2=-y+1=z+1&=s\\ x&=s+2\\ y&=-s+1\\ z&=s-1. \end{aligned}$$

Thus point $B$ can be expressed as $(s+2,-s+1,s-1)$ for some real number $s.$ Then we have

$$\begin{aligned} \overrightarrow{AB}&=(s+2,-s+1,s-1)-(t,-t+2,-t+2)\\ &=(s-t+2,-s+t-1,s+t-3). \end{aligned}$$

Now, let $\vec{d_1}$ denote the direction vector of $l_1,$ and $\vec{d_2}$ be that of $l_2.$ Then we have $\vec{d_1}=(1,-1,-1)$ and $\vec{d_2}=(1,-1,1).$ Since $\overrightarrow{AB}$ should be perpendicular to both $\vec{d_1}$ and $\vec{d_2},$ it must be true that $\overrightarrow{AB}\cdot\vec{d_1}=\overrightarrow{AB}\cdot\vec{d_2}=0.$ Hence we have

$$\begin{aligned} \overrightarrow{AB}\cdot\vec{d_1}&=0\\ (s-t+2,-s+t-1,s+t-3)\cdot(1,-1,-1)&=0\\ (s-t+2)-(-s+t-1)-(s+t-3)&=0\\ \Rightarrow s-3t+6&=0 \qquad(1)\\ \overrightarrow{AB}\cdot\vec{d_2}&=0\\ (s-t+2,-s+t-1,s+t-3)\cdot(1,-1,1)&=0\\ (s-t+2)-(-s+t-1)+(s+t-3)&=0\\ \Rightarrow 3s-t&=0. \qquad(2) \end{aligned}$$

Solving the simultaneous equations (1) and (2) gives

$$\begin{array}{c}&&t=\frac{9}{4}, &&s=\frac{3}{4}.\end{array}$$

Therefore $\overrightarrow{AB}=\left(\frac{1}{2},\frac{1}{2},0\right),$ and the distance between the two skew lines is

$$\begin{aligned} d&=\lvert\overrightarrow{AB}\rvert\\ &=\left\lvert \left(\frac{1}{2},\frac{1}{2},0\right) \right\rvert\\ &=\frac{\sqrt{2}}{2}.\ _\square \end{aligned}$$