#### Question

Find the shortest distance between the lines `(x-1)/2=(y-2)/3=(z-3)/4 and (x-2)/3=(y-4)/4=(z-5)/5`

#### Solution

The lines are

`(X-1)/2=(Y-2)/3=(Z-3)/4`................(1)

`(x-2)/3=(y-4)/4=(z-5)/5`.................(2)

Here

`x_1=1,y_1=2,z_1=3 and a_1=2,b_1=3,c_1=4`

`x_2=2,y_2=4,z_2=5 and a_2=3,b_2=4,c_2=5`

Shortest distance between the lines is

`d=|[x_2-x_1,y_2-y_1,z_2-z_1],[a_1,b_1,c_1],[a_2,b_2,c_2]|/sqrt((b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2)`

Now `|[x_2-x_1,y_2-y_1,z_2-z_1],[a_1,b_1,c_1],[a_2,b_2,c_2]|=|[1,2,2],[2,3,4],[3,4,5]|`

`=1(15-16)-2(10-12)+2(8-9)=-1+4-2=1`

and `(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2=(15-16)^2+(12-10)^2+(8-9)^2`

`=1+4+1=6`

Hence, the shortest distance between the lines (i) and (ii) is=`|1/sqrt6|=1/sqrt6 units`