Question

Show that x-1/3 = y+1/2 = z-1/5 and x+2/4 = y-1/3 = z+1/-2 are skew lines,Find the shortest distance between them.

Answer 1

Answer:

Shortest Distance between the lines is $\frac{107}{\sqrt{1038}}$

Step-by-step explanation:

$\textrm{skew lines are the lines which do not intersect and which are not parallel}$

$\textrm{Given two lines are}L_{1}=\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}L_{2}=\frac{y-1}{3}=\frac{y-1}{3}=\frac{z+1}{-2}$$\textrm{For two lines to be parallel,there direction cosines should be proportional to each other}$$\textrm{for}L_{1}\textrm{direction cosines are 3,5,4}$

$\textrm{for} L_{2}\textrm{direction cosines are 4,3,-2}$

given direction cosines are not Proportional to each other,Hence given pair of lines are not parallel to each other

$\textrm{For Intersection part}$

$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=t\textrm{ for some t}$

$\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=s\textrm { for some s}$

$\textrm{Therefore for}L_{1}\textrm{x=3t+1,y=2t-1,z=5t+1}$

$\textrm{Similarly for}L_{2}\textrm{x=4s-2,y=3s+1,z=-2s-1}$

$\textrm{Therefore 3t+1=4s-2...(i)2t-1=3s+1...(ii)5t+1=-2s-1 ...(iii)}$

$\textrm{Finding value of s and t from eq.(i)\ and\ eq.(ii) and putting in eq.(iii) we get,}$

$\textrm{we got t=-17 and s=12}$

$\textrm{by putting the value of s and t in eq.(iii)}i.e\ 5(-17)\neq-2(12)-1=-85\neq-25$

$\textrm{Hence,the line don't intersect each other.Therefore given pair lines are not Skew Lines}$

$\textrm{Now Calculating the Distance D}$

$\textrm{D=}\dfrac{\begin{vmatrix}x_{2}-x_{1}& y_{2}-y_{1}& z_{2}-z_{1} \\a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}$

$\textrm{Here}a_{1}=3,b_{1}=2,c_{1}=5\ and\ a_{2}=4,b_{2}=3,c_{2}=-2$

$x_{1}=1,x_{2}=-2,y_{1}=-1,y_{2}=1,z_{1}=1,z_{2}=-1$

$\dfrac{\begin{vmatrix}-3&2&-2\\3&2&5\\4&3&-2\end{vmatrix}}{\sqrt{(-4-15)^{2}+(20+6)^{2}+(9-8)^{2}}}$

$\frac{107}{\sqrt1038}$