Correct me if im wrong, but this is what i know so far

The cross product of two skew lines is the perpendicular vector between both those lines.

The perpendicular vector intersecting two skew lines is the shortest distance between the two lines.

What I would like to know, is what are the xyz coordinates of the two points in which the perpendicular line intersects the two original lines.

I found this How to find shortest distance between two skew lines in 3D? which shows the distance, but not the coordinates, and I do not see where the coordinate would be calculated in it. As in, it doesnt seem to find the two coordinates, and then get the distance between those two, but instead directly calculates the distance?

The question is, how to find these xyz coordinates in a general way (so it can be programmed)?

  • $\begingroup$ Perhaps this answer and the comments beneath it will help. $\endgroup$ – Ted Shifrin Jul 13 '15 at 14:05
  • $\begingroup$ I added an example (and found a sign bug in the process :-) $\endgroup$ – mvw Jul 13 '15 at 19:06

First the equations for all vectors $x$ on line $g$ and all vectors $y$ on line $h$: $$ \begin{align} g: x &= a + \lambda b \quad \\ h: y &= c + \mu d \end{align} \quad (*) $$ The difference vector between two of those vectors is $$ D = y - x = c - a + \mu d - \lambda b $$ The length of $D$ squared is: \begin{align} q(\lambda, \mu) &= D \cdot D \\ &= (c - a)^2 + (\mu d-\lambda b)^2 + 2 (c-a)\cdot(\mu d - \lambda b) \\ &= (c - a)^2 + \mu^2 d^2 + \lambda^2 b^2 - 2 \mu \lambda (d\cdot b) + 2 \mu ((c-a)\cdot d) - 2 \lambda ((c-a)\cdot b) \\ \end{align} The gradient of $q$ is: $$ q_\lambda = 2\lambda b^2 - 2\mu (d\cdot b)-2((c-a)\cdot b) \\ q_\mu = 2\mu d^2 - 2\lambda (d\cdot b)+2((c-a)\cdot d) $$ It should vanish for local extrema of $q$ which leads to the system $$ A u = v $$ with $$ A = \left( \begin{matrix} b^2 & - d\cdot b \\ - d \cdot b & d^2 \end{matrix} \right) \quad u = \left( \begin{matrix} \lambda \\ \mu \end{matrix} \right) \quad v = \left( \begin{matrix} (c-a) \cdot b \\ -(c-a) \cdot d \end{matrix} \right) $$ A unique solution exists for $$ 0 \ne \mbox{det A} = b^2 d^2 - (d \cdot b)^2 $$ Note that the dot operator stands for the scalar product. That solution is $$ u = A^{-1} v $$ with $$ A^{-1} = \frac{1}{\mbox{det } A} \left( \begin{matrix} d^2 & d\cdot b \\ d \cdot b & b^2 \end{matrix} \right) $$ Inserting the found values for $\lambda$ and $\mu$ into equations $(*)$ will provide you the two vectors, whose points are closest to each other.


minimal distance

$$ a = (2,0,0), b=(1,1,1), c=(0,1,-1), d=(-1,0,-1) $$ leads to $$ \lambda = 1, \mu = -2.5, x_\min = (3,1,1), y_\min=(2.5,1,1.5), D=(-0.5,0,0.5) $$

The image renders the $x$ values in green, the $y$ values in purple, and $D$ in red.


Let $L_1, L_2$ be two skew lines in $\mathbb{R}^3.$ Assume $P_j$ are points in $L_j$ and $\mathbf{v}_j$ the direction vector of $L_j,$ respectively, where $j=1,2.$ Put $\mathbf{v}_3=\mathbf{v}_1\times \mathbf{v}_2.$ Then the two feet $Q_1, Q_2$ of common perpendicular to $L_1,L_2$ are given by \begin{gather*} \begin{aligned} &Q_1=P_1+\frac{\overrightarrow{P_1P_2}\cdot(\mathbf{v}_2\times \mathbf{v}_3)}{\mathbf{v}_3\cdot \mathbf{v}_3}\mathbf{v}_1,\\ &Q_2=P_2+\frac{\overrightarrow{P_1P_2}\cdot(\mathbf{v}_1\times \mathbf{v}_3)}{\mathbf{v}_3\cdot \mathbf{v}_3}\mathbf{v}_2. \end{aligned} \end{gather*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.