Distance and shortest distance

Three ways two lines in space can sit. Two lines either intersect, are parallel, or are skew — neither meeting nor parallel, lying in different planes (an overpass and the road beneath it). For lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$, the coplanarity test is

where: $\vec a_1,\vec a_2$ are position vectors of any point on each line and $\vec b_1,\vec b_2$ are their direction vectors. If this scalar triple product is zero the lines are coplanar (they intersect or are parallel); if it is non-zero, they are skew.

Distance of a point from a line. For point $P$ (position vector $\vec p$) and line $\vec r=\vec a+\lambda\vec b$, dropping a perpendicular from $P$ gives

where: $(\vec p-\vec a)$ joins a known point on the line to $P$, and $\vec b$ is the line's direction. The cross product's magnitude is the area of the parallelogram spanned by these two vectors; dividing by the base $|\vec b|$ leaves the height — exactly the perpendicular distance.

Shortest distance for parallel lines $\vec r=\vec a_1+\lambda\vec b$ and $\vec r=\vec a_2+\mu\vec b$ (same direction $\vec b$):

This is the point-from-line formula in disguise — any point on the second line measured against the first. For skew lines, no single perpendicular from a point works; instead there is a unique line segment perpendicular to both lines at once, and its length is the shortest distance — the subject of the derivation below. In Cartesian form, the same formulas apply once $\vec a_1,\vec a_2,\vec b_1,\vec b_2$ are read off as the given points and direction ratios, and every cross/dot product is expanded component-wise as a $3\times3$ determinant. It helps to notice that all three formulas — point-from-line, parallel-lines, and skew-lines — are variations on one theme: turn the gap between two geometric objects into a vector, use a cross product to build a perpendicular quantity out of the directions involved, and divide by a length to leave behind a pure distance. The skew-line case is the richest because two independent directions are in play, so the perpendicular direction $\vec b_1\times\vec b_2$ itself has to be constructed before any projection can happen; in the point-from-line and parallel-lines cases there is only one direction, $\vec b$, to work with throughout.

Find the shortest distance between the skew lines $\vec r = (\hat i+\hat j) + \lambda(2\hat i-\hat j+\hat k)$ and $\vec r = (2\hat i+\hat j-\hat k)+\mu(3\hat i-5\hat j+2\hat k)$.

Step 1. Read off the vectors: $\vec a_1=\hat i+\hat j$, $\vec b_1=2\hat i-\hat j+\hat k$, $\vec a_2=2\hat i+\hat j-\hat k$, $\vec b_2=3\hat i-5\hat j+2\hat k$, so $\vec a_2-\vec a_1=\hat i-\hat k$.

Step 2. Compute $\vec b_1\times\vec b_2$ by the determinant

Step 3. Dot this with $\vec a_2-\vec a_1$: $(\hat i-\hat k)\cdot(3\hat i-\hat j-7\hat k) = 3+0+7=10$. Since this is non-zero, the lines are confirmed skew.

Step 4. Find the magnitude $|\vec b_1\times\vec b_2|=|3\hat i-\hat j-7\hat k|=\sqrt{9+1+49}=\sqrt{59}$, and divide: $d=\left|\dfrac{10}{\sqrt{59}}\right|=\dfrac{10}{\sqrt{59}}$ units.

The shortest distance between the two skew lines is $d=\dfrac{10}{\sqrt{59}}$ units.

Competitive stretch

The same setup gives the point-from-line formula near-for-free: for $P(1,0,-3)$ and the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z}{-1}$ (through $A(-1,3,0)$, direction $\vec b=2\hat i+3\hat j-\hat k$), $(\vec p-\vec a)\times\vec b=12\hat i-4\hat j+12\hat k$, of magnitude $\sqrt{304}=4\sqrt{19}$, and $|\vec b|=\sqrt{14}$, giving $d=\dfrac{4\sqrt{19}}{\sqrt{14}}\approx 4.66$ units — the identical cross-product-over-magnitude pattern, just with one line and one external point instead of two lines.

• Using the parallel-lines formula (a plain cross product with a single $\vec b$) on lines that are actually skew, or the skew formula on lines that are parallel — check whether $\vec b_1$ and $\vec b_2$ are proportional before choosing.
• Sign or cofactor-order slips inside the $3\times3$ determinant when expanding $\vec b_1\times\vec b_2$ or $(\vec p-\vec a)\times\vec b$; one misplaced sign changes the final distance.
• Dropping the absolute value at the end — a distance cannot be negative, even though the scalar triple product inside can come out negative.
• Declaring lines "intersecting" from a rough sketch instead of formally testing $(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=0$.

Ask the tutor: if setting up the determinant for $\vec b_1\times\vec b_2$ feels error-prone, ask the AI tutor to walk through the cofactor expansion one row at a time with your own numbers.