JEE Main 2025 Jan 23 Shift 2, Mathematics Q18: Lines in Space

Take points and direction vectors:

$\vec{a}=(2,1,-3),\vec{p}=(1,2,-3),\vec{b}=(-1,-3,-5),\vec{q}=(2,4,-5).$

Cross product:

$\vec{p}\times \vec{q}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 12-3\\ 24-5\end{array}\right|=\hat{i}(2(-5)-(-3)(4))-\hat{j}(1(-5)-(-3)(2))+\hat{k}(1(4)-2(2))=2\hat{i}-\hat{j}+0\hat{k}.$

$|\vec{p}\times \vec{q}|=\sqrt{4+1}=\sqrt{5}.$

$\vec{b}-\vec{a}=(-3,-4,-2).$

$(\vec{b}-\vec{a})\cdot (\vec{p}\times \vec{q})=(-3)(2)+(-4)(-1)+(-2)(0)=-2.$

Shortest distance: $d=\frac{|(\vec{b}-\vec{a})\cdot (\vec{p}\times \vec{q})|}{|\vec{p}\times \vec{q}|}=\frac{2}{\sqrt{5}}$.

$d^2=\frac{4}{5}$. With $m=4$ and $n=5$ coprime, $m+n=9$.