JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 30

Shortest distance $\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$ with $\vec a_2-\vec a_1=(1,-1,-1)$.

Cross product. $\vec b_1\times\vec b_2=(1,-a,0)\times(1,-1,1)=(-a,-1,a-1)$; dot product $-a+1-(a-1)=2-2a$; $|\vec b_1\times\vec b_2|^2=2a^2-2a+2$.

Equation. $\dfrac{4(1-a)^2}{2a^2-2a+2}=\dfrac23$ gives $2a^2-5a+2=0$, so $a=2$ or $\frac12$.

Integral value $a=2$.

Answer: 2