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

Answer 1

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

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 $\frac{1}{\sqrt{3}}$

Line of intersection of planes $x+y+z+1=0=2x-y+z+3$ is

Eliminating $y$ gives $3x+2z+4=0$

$\Rightarrow x=\frac{-2z-4}{3}$ ----(1)

Substituting above $x$ in $x+y+z+1=0$ gives $3y+z-1=0$

$\Rightarrow z=-3y+1$ ----(2)

From (1) and (2), line equation is $\frac{3x+4}{-2}=-3y+1=z$

$\Rightarrow \frac{x-(-4/3)}{-2/3}=\frac{y-\left(1/3\right)}{-1/3}=\frac{z}{1}$

Given line is $\frac{x-1}{\alpha }=\frac{y+1}{-1}=\frac{z}{1}$

Shortest distance between above two skew lines is

$\frac{(\overrightarrow{b}-\overrightarrow{a})\cdot (\overrightarrow{c}\times \overrightarrow{d})}{|\overrightarrow{c}\times \overrightarrow{d}|}$

where one line is passing through $\overrightarrow{a}$ and parallel to $\overrightarrow{c}$ and other one passes through $\overrightarrow{b}$ and parallel to

$\overrightarrow{d}$

$\overrightarrow{a}=(1,-1,0)$

$\overrightarrow{b}=(-\frac{4}{3},\frac{1}{3},0)$

$\overrightarrow{c}=(\alpha ,-1,1)$

$\overrightarrow{d}=(-\frac{2}{3},-\frac{1}{3},1)$

$\therefore \frac{(\overrightarrow{b}-\overrightarrow{a})\cdot (\overrightarrow{c}\times \overrightarrow{d})}{|\overrightarrow{c}\times \overrightarrow{d}|}=\left|\frac{\left|\begin{array}{ccc}7/3& -4/3& 0\\ \alpha & -1& 1\\ -2/3& -1/3& 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & -1& 1\\ -2/3& -1/3& 1\end{array}\right|}\right|=\frac{1}{\sqrt{3}}$

Expanding determinants and simplifying gives

$48{\alpha }^2-48\alpha +12=10{\alpha }^2+16\alpha +12$

$\Rightarrow 38{\alpha }^2=64\alpha$

$\Rightarrow \alpha =\frac{64}{38}=\frac{32}{19}$

Hence, option A.