Shortest Distance between two lines

How to find shortest distance between two lines in space.

The shortest distance d between two lines

$\dpi{120} \mathbf{\overrightarrow{r_{1}}=\overrightarrow{a_{1}}+s\overrightarrow{b_{1}}}$ and

$\dpi{120} \mathbf{\overrightarrow{r_{2}}=\overrightarrow{a_{2}}+s\overrightarrow{b_{2}}}$

is given by the following formula.

$\dpi{150} \mathbf{d = \left |\frac{\left ( \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right ).\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )}{\left | \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right |} \right |}$

Example1: Find shortest distance between the lines

r =〈4,-1,0〉 + s〈1,2-3〉 and r =〈1,-1,2〉 + t〈2,4,-5〉 .

Solution: Here we have a1 = (4,-1,0) , a2 = (1,-1,2) and b1 = (1,2,-3) , b2 = (2,4,-5)

Step1) a2-a1 = (1,-1,2)-(4,-1,0) =

Step2)

$\dpi{120} \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}}=\begin{vmatrix} i & j & k\\ 1& 2& -3\\ 2 & 4 & -5 \end{vmatrix}$

= i(-10+12) –j(-5+6) +k(4-4)

= 2i-j+0k

Step3) $\dpi{120} \left |\overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right |=\sqrt{2^{2}+(-1)^{2}+0^{2}}=\sqrt{5}$

Step4) Plug in all the values into formula

$\dpi{120} d = \left |\frac{\left ( \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right ).\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )}{\left | \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right |} \right | =\left |\frac{\left \langle -3,0,2 \right \rangle.\left \langle 2,-1,0 \right \rangle}{\sqrt{5}} \right |$

$\dpi{120} =\left |\frac{-6+0+0}{\sqrt{5}} \right |= \frac{6}{\sqrt{5}}$

Example2.By computing the shortest distance determine whether the following pair of lines intersect or not.

$\dpi{120} {\color{Red} \frac{x-1}{2}=\frac{y+1}{3}=z }$ and $\dpi{120} {\color{Red} \frac{x+1}{5}=\frac{y-2}{1},z=2 }$

Solution: Given equation can be rewritten in proper form as,

$\dpi{120} \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-0}{1}$ and $\dpi{120} \frac{x+1}{5}=\frac{y-2}{1}=\frac{z-2}{0}$

Here we have a1 = (1,-1,0) , a2 =(-1,2,2) , b1= (2,3,1) and b2=(5,1,0)

Step1) a2 -a1 = (-1,2,2)-(1,-1,0) = (-2,3,2)

Step2)

$\dpi{120} \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}}=\begin{vmatrix} i & j & k\\ 2& 3& 1\\ 5 & 1 & 0 \end{vmatrix}$

= i(0-1) –j(0-5)+k(2-15)

= -i+5j-13k

Step3) $\dpi{120} \left (\overrightarrow{ a_{2}}-\overrightarrow{a_{1}} \right ).\left ( \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right )$

= (-2,3,2).(-1,5,-13)

= 2+15-26= -9 ≠ 0

If the shortest distance between two lines is 0 , then the lines intersect each other.

Here we got that distance is not 0 so the two lines doesn’t intersect each other.

Shortest distance between two Parallel lines:

Lets assume two parallel lines with same direction vector b are given by the equations

$\dpi{120} \mathbf{\overrightarrow{r_{1}}=\overrightarrow{a_{1}}+s\overrightarrow{b}}$ and

$\dpi{120} \mathbf{\overrightarrow{r_{2}}=\overrightarrow{a_{2}}+t\overrightarrow{b}}$

Then the shortest distance between them is found by the following formula.

$\dpi{150} \mathbf{d = \left |\frac{\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times b}{\left | \overrightarrow{b} \right |} \right |}$

Example3: Find shortest distance between the lines

r = 〈1,2,3〉 + s 〈2,3,4〉 and r =〈2,4,5〉+ t 〈4,6,8〉 .

Solution: Rewriting second equation we get ,

r = 〈2,4,5〉 + 2t 〈2,3,4〉

We see that both lines have same direction vector , therefore both are parallel lines. Here we have a1 = (1,2,3) , a2 = (2,4,5) and b= (2,3,4).

Step1)

a2 -a1 = (2,4,5)-(1,2,3) = 〈1,2,2〉

Step2)

$\dpi{120} \left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times\overrightarrow{ b}=\begin{vmatrix} i & j & k\\ 1& 2& 2\\ 2 & 3 & 4 \end{vmatrix}$

= i(8-6) –j(4-4)+k(3-4)

= 2i-0j-k

Step3)

$\dpi{120} \left |\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times \overrightarrow{b} \right |=\sqrt{2^{2}+0+(-1)^{2}}=\sqrt{5}$

$\dpi{120} \left | b \right |=\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}$

Step 4) Plug in the values into the formula.

$\dpi{120} \mathbf{d = \left |\frac{\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times b}{\left | \overrightarrow{b} \right |} \right |}=\frac{\sqrt{5}}{\sqrt{29}}$