If the square of the shortest distance between the lines $\dfrac{x-2}{1}=\dfrac{y-1}{2}=\dfrac{z+3}{-3}$ and $\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{-5}$ is $\dfrac{m}{n}$, where $m,n$ are coprime numbers, then $m+n$ is equal to:
Step-by-Step Solution
Key Concept: Use $S_d=\dfrac{|(\vec{b}-\vec{a})\cdot(\vec{p}\times\vec{q})|}{|\vec{p}\times\vec{q}|}$ where $\vec{p},\vec{q}$ are direction vectors and $\vec{b}-\vec{a}$ connects points on the two lines.
$\vec{p}=(1,2,-3)$, $\vec{q}=(2,4,-5)$, $\vec{b}-\vec{a}=(-3,-4,-2)$.
$\vec{p}\times\vec{q}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&-3\\2&4&-5\end{vmatrix}=\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4)=2\hat{i}-\hat{j}$.
$|\vec{p}\times\vec{q}|=\sqrt{5}$.
$(\vec{b}-\vec{a})\cdot(\vec{p}\times\vec{q})=(-3)(2)+(-4)(-1)+(-2)(0)=-6+4=−2$.
$S_d=\dfrac{|-2|}{\sqrt{5}}=\dfrac{2}{\sqrt{5}}$, $S_d^2=\dfrac{4}{5}$.
$m=4,n=5$ (coprime), $m+n=9$.
Correct Answer: 2