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

$d=\frac{|2*2-4+2*-1-2|}{\sqrt{2^2+(-1)^2+(2)^2} }$

$d=\frac{4}{3}$..........Answer