Shortest distance between two lines

Take two lines in vector form, $\vec r = \vec{a_1} + \lambda \vec{b_1}$ and $\vec r = \vec{a_2} + \mu \vec{b_2}$, where $\vec{a_1}, \vec{a_2}$ locate one point on each line and $\vec{b_1}, \vec{b_2}$ are their direction vectors. Comparing the two lines always lands in exactly one of three cases: they intersect, they are parallel, or they are skew.

Case 1 — skew or otherwise non-parallel lines. Whenever $\vec{b_1}$ and $\vec{b_2}$ are not parallel, $\vec{b_1}\times\vec{b_2}$ is a nonzero vector perpendicular to both lines — the direction of the common perpendicular. Dotting it with $\vec{a_2}-\vec{a_1}$, the vector joining a point on line 1 to a point on line 2, projects that separation onto this common-perpendicular direction, giving the shortest distance directly:

where: $d$ is the shortest distance, $\vec{a_2}-\vec{a_1}$ is the vector between the two given points, and $\vec{b_1}\times\vec{b_2}$ is the cross product of the two direction vectors. The numerator, $(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})$, is a scalar triple product, and the modulus bars around the whole fraction guarantee the answer is a non-negative length regardless of the sign the triple product happens to produce.

Case 2 — parallel lines. If $\vec{b_1}$ and $\vec{b_2}$ point along the same (or exactly opposite) direction, then $\vec{b_1}\times\vec{b_2}=\vec 0$ and Case 1's formula divides by zero. Parallel lines never converge to a common perpendicular pinned by a cross product in that way; instead the distance is the length of the component of $\vec{a_2}-\vec{a_1}$ perpendicular to the common direction $\vec b$:

where: $\vec b$ is the common direction of both lines (either $\vec{b_1}$ or $\vec{b_2}$, since they are parallel), and the rest of the notation matches Case 1.

Reading $d=0$. If the Case-1 formula evaluates to $d=0$, the numerator scalar triple product is zero, which means $\vec{a_2}-\vec{a_1}$, $\vec{b_1}$ and $\vec{b_2}$ are coplanar. Combined with the lines not being parallel, this forces them to actually meet at a point — the lines intersect. In Cartesian form, with points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and direction ratios $(a_1,b_1,c_1)$, $(a_2,b_2,c_2)$, the Case-1 formula becomes a ratio of a $3\times3$ determinant to the same cross-product magnitude:

Problem. Find the shortest distance between the lines $\vec r = (\hat\imath+2\hat\jmath+\hat k) + \lambda(\hat\imath-\hat\jmath+\hat k)$ and $\vec r = (2\hat\imath-\hat\jmath-\hat k) + \mu(2\hat\imath+\hat\jmath+2\hat k)$.

Step 1. Identify $\vec{a_1}=\hat\imath+2\hat\jmath+\hat k$, $\vec{b_1}=\hat\imath-\hat\jmath+\hat k$, $\vec{a_2}=2\hat\imath-\hat\jmath-\hat k$, $\vec{b_2}=2\hat\imath+\hat\jmath+2\hat k$. The direction ratios $(1,-1,1)$ and $(2,1,2)$ are not proportional, so the lines are not parallel and this is a genuine skew-or-intersecting check.

Step 2. Compute the separation: $\vec{a_2}-\vec{a_1} = (2-1)\hat\imath + (-1-2)\hat\jmath + (-1-1)\hat k = \hat\imath - 3\hat\jmath - 2\hat k$.

Step 3. Compute the cross product:

Step 4. Dot the two results: $(1)(-3) + (-3)(0) + (-2)(3) = -3 + 0 - 6 = -9$.

Step 5. Compute the magnitude $|\vec{b_1}\times\vec{b_2}| = \sqrt{(-3)^2+0^2+3^2} = \sqrt{18} = 3\sqrt2$, then apply the formula:

The shortest distance is $\dfrac{3}{\sqrt2}$ units, equivalently $\dfrac{3\sqrt2}{2}$ units.

Competitive stretch

Problem. Check whether $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{2}=\dfrac{z-1}{-1}$ intersect.

Step 1. Parametrize: line 1 is $x=1+2\lambda,\ y=-1+3\lambda,\ z=4\lambda$; line 2 is $x=3+\mu,\ y=2+2\mu,\ z=1-\mu$. Equating $x$ and $y$ gives $2\lambda-\mu=2$ and $3\lambda-2\mu=3$, which solve to $\lambda=1, \mu=0$.

Step 2. Check the $z$-equation with these values: line 1 gives $z=4(1)=4$, line 2 gives $z=1-0=1$. Since $4 \neq 1$, no common $\lambda,\mu$ satisfies all three coordinates, so the lines do not intersect; since their direction ratios $(2,3,4)$ and $(1,2,-1)$ are not proportional either, they are skew, and the Case-1 formula above would be applied next to find how far apart they actually are.

• Using the skew-line formula on parallel lines. If $\vec{b_1}\times\vec{b_2}=\vec 0$, the Case-1 formula divides by zero — always check for parallel direction ratios first and switch to the Case-2 formula if needed.
• Sign slip in the cross-product expansion, especially the middle ($\hat\jmath$) term, which carries a built-in minus sign that is easy to drop.
• Forgetting the final modulus. The scalar triple product in the numerator can be negative; the reported shortest distance is always taken as non-negative.
• Assuming a nonzero $d$ automatically means "skew" without checking parallelism. A nonzero distance only rules out intersection — confirm the direction ratios are not proportional before calling the lines skew rather than parallel-and-separated.

Ask the tutor: Want to work through another shortest-distance problem, or verify whether a given pair of lines is parallel, skew, or intersecting? Ask the AI tutor to walk through the calculation with you.