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Calculate the distance between the lines $$L_1:x=4+2t,y=3+2t,z=3+6t$$ $$L_2: x=-2+3s ,y=3+4s ,z=5+9s$$

I tried subtraction $L_1$ from $L_2$ then multiplying the resting vector by the $t$'s and $s$'s original values and trying to find value for $t$ to or $s,$ but I found $t=\frac{19}{12} s$ and I don't know how to keep solving this.

Vim
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Nathalie Olcese
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  • is there a typo in the last equation $$z=5=9s$$? – Dr. Sonnhard Graubner Sep 16 '17 at 14:52
  • what do you think is a reasonable definition of "the distance between two lines"? – Long Sep 16 '17 at 14:57
  • @Dr.SonnhardGraubner yes its a typo it should be z=5+9s – Nathalie Olcese Sep 16 '17 at 15:41
  • A few very similar questions: https://math.stackexchange.com/questions/210848/finding-the-shortest-distance-between-two-lines, https://math.stackexchange.com/questions/2310599/find-the-shortest-distance-between-two-lines-in-3d, https://math.stackexchange.com/questions/813497/find-the-distance-between-two-lines-given-in-parametric-form-should-be-easy, https://math.stackexchange.com/questions/1188423/find-the-distance-between-two-lines, https://math.stackexchange.com/questions/1709200/distance-between-two-skew-lines, https://math.stackexchange.com/questions/1554820 – David K Sep 16 '17 at 16:06
  • You can take a look at the link and follow the directions to find the distance between two lines in vector forms. Distance between two non parallel lines – Satish Ramanathan Sep 16 '17 at 14:58

3 Answers3

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The distance between the lines

$L_1:x=4+2t,y=3+2t,z=3+6t$ and $L_2: x=-2+3s ,y=3+4s ,z=5+9s$ is

$d=2 \sqrt{10}$

Indeed consider the function which gives the distance between a generic point of the first line and a generic point of the second line

$f(t,s)=\sqrt{(2 t-4 s)^2+(-3 s+2 t+6)^2+(-9 s+6 t-2)^2}$

and set to zero the partial derivatives $\partial f_t=0,\partial f_s=0$

$\left\{ \begin{array}{l} 4 (2 t-4 s)+4 (-3 s+2 t+6)+12 (-9 s+6 t-2)=0 \\ -8 (2 t-4 s)-6 (-3 s+2 t+6)+18 (-9 s+6 t-2)=0 \\ \end{array} \right. $

$\left\{ \begin{array}{l} 17 s-11 t=0 \\ 53 s-34 t=0 \\ \end{array} \right.$

which has one solution

$t= 0,\;s=0$

we have

$f(0,0)=2 \sqrt{10}$

Hope it helps

Raffaele
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A general solution is :
take a point on $l_1 :A=(4+2t,3+2t,3+6t)$
take a point on $L_2: B=(-2+3s ,3+4s ,9s)$

find vector $\overline{AB}=B-A=\\(-2+3s ,3+4s ,9s)-(4+2t,3+2t,3+6t)$

then solve the system of equation $$\begin{cases}\overline{AB}.V_{l_1}=0\\\overline{AB}.V_{l_2}=0\end{cases}\\ \begin{cases}\overline{AB}.(2,2,6)=0\\\overline{AB}.(3,4,9)=0\end{cases}$$ you will have two unknown and two linear equation ,here . $|\overline{AB}|$ is the minimum distance between two lines.

Khosrotash
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$ \hat a = 4\hat i + 3\hat j+3\hat k$

$ \hat b = -2\hat i + 3\hat j+5\hat k$

$\hat t =2\hat i + 2\hat j+6\hat k$

$\hat s = 3\hat i + 4\hat j+9\hat k$

$L1: \hat a+t\hat t$

$L2: \hat b+s\hat s$

So the distance $$d = \dfrac{\left|(\hat a - \hat b).(\hat t \times \hat s)\right|}{|\hat t \times \hat s|}$$

$\hat t \times \hat s = -6\hat i + 2\hat k$

$ \hat a - \hat b = 6\hat i - 2\hat k$

$\left|(\hat a - \hat b).(\hat t \times \hat s)\right| = 40$

$|\hat t \times \hat s| = \sqrt{40}$

Thus the distance $= \frac{40}{\sqrt{40}}$

$d = \sqrt{40} = 2\sqrt{10}$

Satish Ramanathan
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