The shortest distance between the lines $\frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-5}{8}$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\sqrt{\frac{5}{29}}$
Given lines
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$
$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-5}{8}$
A point on first line is $A(1,2,3)$ and its direction vector is
$\vec a=(2,3,4)$
A point on second line is $B(2,4,5)$ and its direction vector is
$\vec b=(4,6,8)=2(2,3,4)$
Hence the lines are parallel
Shortest distance between two parallel lines is
$\frac{|(\vec{AB}\times\vec a)|}{|\vec a|}$
$\vec{AB}=(1,2,2)$
$\vec{AB}\times\vec a= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 2 & 2\\ 2 & 3 & 4 \end{vmatrix} =(2,0,-1)$
$|\vec{AB}\times\vec a|=\sqrt{5}$
$|\vec a|=\sqrt{29}$
Shortest distance $=\frac{\sqrt{5}}{\sqrt{29}}=\sqrt{\frac{5}{29}}$
The shortest distance between the given lines is $\sqrt{\frac{5}{29}}$.