Question

If the shortest distance between the lines $\frac{x-1}{\alpha }=\frac{y+1}{-1}=\frac{z}{1},(\alpha \ne -1)$ and $x+y+z+1=0=2x–y+z+3$ is $\frac{1}{\sqrt{3}},$ then a value of $\alpha$ is :

• A. $\frac{32}{19}$
• B. $\frac{19}{32}$
• C. $-\frac{16}{19}$
• D. $-\frac{19}{16}$

Answer 1

The correct option is A $\frac{32}{19}$

Family of plane is,
$x+y+z+1+k(2x–y+z+3)=0(1+2k)x+(1-k)y+(1+k)z+(1+3k)=0$
If given family of plane is parallel to the line then
$(1+2k)\alpha -(1-k)+(1+k)=0\Rightarrow \alpha =\frac{-2k}{1+2k}\text{Perpendicular distance between}\frac{x-1}{\alpha }=\frac{y+1}{-1}=\frac{z}{1}\&(1+2k)x+(1-k)y+(1+k)z+(1+3k)=0\text{is}\frac{|(1+2k)\left(1\right)+(1-k)(-1)+(1+k)\left(0\right)+(1+3k)|}{\sqrt{(1+2k{)}^2+(1-k{)}^2+(1+k{)}^2}}=\frac{1}{\sqrt{3}}\Rightarrow k=0,-\frac{32}{102}\Rightarrow \alpha =0,\frac{32}{19}$