Question

Find the distance between the skew lines with parametric equations x = 2 + t, y = 2 + 6t, z = 2t, and x = 2 + 2s, y = 4 + 14s, z = −1 + 5s.

Answer 1

we are given

skew lines as

L1: x = 2 + t, y = 2 + 6t, z = 2t

L2: x = 2 + 2s, y = 4 + 14s, z = −1 + 5s

Firstly, we will find directional vector

`u=(1,6,2)`

`v=(2,14,5)`

now, we will find cross product

`n=uXv`

`n=(1,6,2)X(2,14,5)`

`n=(2,-1,2)`

we are given two points as

P: (2,2,0) and Q:(2,4,-1)

The dot product of orthogonal vectors is zero

`n*PR=0`

`(2,-1,2)*((x-2) , (y-2) , (z-0))=0`

`2(x-2)-1(y-2)+2(z-0)=0`

`2x-4-y+2+2z=0`

`2x-y+2z-2=0`

now, we can find distance from point Q

`d=(|2*2-4+2*-1-2|)/(√(2^2+(-1)^2+(2)^2) )`

`d=(4)/(3)`..........Answer