The shortest distance between the lines: x-1/alpha= y+1/-1 =​​z/1,(alpha not equal to -1) x+y+z+1=0= 2x-y+z+3 is 1/rt3, then alpha=? [32/19]

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \frac{\mathrm{x}-1}{\mathrm{\alpha }}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}}{1} \\ \mathrm{So} \mathrm{we} \mathrm{have} \\ \left({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1\right)=\left(1,-1,0\right)\to \mathrm{Point} \mathrm{through} \mathrm{which} \mathrm{line} \mathrm{passes} \\ \mathrm{Direction} \mathrm{ratios} \mathrm{are} \\ \left({\mathrm{a}}_1,{\mathrm{b}}_1,{\mathrm{c}}_1\right)=\left(\mathrm{\alpha },-1,1\right) \\ \mathrm{and} \\ \mathrm{x}+\mathrm{y}+\mathrm{z}+1=0=2\mathrm{x}-\mathrm{y}+\mathrm{z}+3 \\ \mathrm{i}.\mathrm{e}. \\ \mathrm{x}+\mathrm{y}=-\mathrm{z}-1...\left(\mathrm{i}\right) \\ 2\mathrm{x}-\mathrm{y}+\mathrm{z}+3=0 \\ \Rightarrow 2\mathrm{x}-\mathrm{y}=-\mathrm{z}-3...\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)+\left(\mathrm{ii}\right) \\ \mathrm{x}+\mathrm{y}+2\mathrm{x}-\mathrm{y}=-\mathrm{z}-1-\mathrm{z}-3 \\ \Rightarrow 3\mathrm{x}=-2\mathrm{z}-4 \left[\mathrm{x}=\frac{-2\mathrm{z}-4}{3}\right] \\ \Rightarrow \frac{3\mathrm{x}+4}{-2}=\mathrm{z} \\ \frac{-2\mathrm{z}-4}{3}+\mathrm{y}=-\mathrm{z}-1 \\ \Rightarrow \mathrm{y}=\frac{2\mathrm{z}+4}{3}-\mathrm{z}-1 \\ \Rightarrow \mathrm{y}=\frac{2\mathrm{z}+4-3\mathrm{z}-3}{3} \\ \Rightarrow 3\mathrm{y}=-\mathrm{z}+1 \\ \Rightarrow 3\mathrm{y}-1=-\mathrm{z} \\ \Rightarrow \mathrm{z}=\frac{3\mathrm{y}-1}{-1} \\ \mathrm{Hence} \mathrm{we} \mathrm{have} \\ \frac{3\mathrm{x}+4}{-2}=\frac{3\mathrm{y}-1}{-1}=\frac{\mathrm{z}}{1} \\ \mathrm{i}.\mathrm{e}. \\ \frac{\mathrm{x}+{\displaystyle \frac{4}{3}}}{-{\displaystyle \frac{2}{3}}}=\frac{\mathrm{y}-{\displaystyle \frac{1}{3}}}{-{\displaystyle \frac{1}{3}}}=\frac{\mathrm{z}}{1} \\ \mathrm{So} \\ \left({\mathrm{x}}_2,{\mathrm{y}}_2,{\mathrm{z}}_2\right)=\left(-\frac{4}{3},\frac{1}{3},0\right) \\ \mathrm{Direction} \mathrm{ratios} \mathrm{is} \mathrm{given} \mathrm{by} \\ \left({\mathrm{a}}_2,{\mathrm{b}}_2,{\mathrm{c}}_2\right)=\left(-\frac{2}{3},-\frac{1}{3},1\right) \\ \mathrm{Shortest} \mathrm{distance} \mathrm{is} \mathrm{given} \mathrm{by} \end{gathered}$$


$$\begin{gathered} \mathrm{Shortest} \mathrm{distance}=\frac{1}{\sqrt{3}} \\ \Rightarrow \frac{\left|\begin{array}{ccc}-{\displaystyle \frac{4}{3}}-1& {\displaystyle \frac{1}{3}}+1& 0\\ \mathrm{\alpha }& -1& 1\\ -{\displaystyle \frac{2}{3}}& -{\displaystyle \frac{1}{3}}& 1\end{array}\right|}{\sqrt{{\left(-1+{\displaystyle \frac{1}{3}}\right)}^2+{\left(-{\displaystyle \frac{2}{3}}-\mathrm{\alpha }\right)}^2+{\left(-{\displaystyle \frac{\mathrm{\alpha }}{3}}-{\displaystyle \frac{2}{3}}\right)}^2}}=\frac{1}{\sqrt{3}} \\ \Rightarrow \left|\begin{array}{ccc}-\frac{7}{3}& \frac{4}{3}& 0\\ \mathrm{\alpha }& -1& 1\\ -\frac{2}{3}& -\frac{1}{3}& 1\end{array}\right|=\frac{1}{\sqrt{3}}\sqrt{{\left(-\frac{2}{3}\right)}^2+{\left(-\frac{2}{3}-\mathrm{\alpha }\right)}^2+{\left(-\frac{\mathrm{\alpha }}{3}-\frac{2}{3}\right)}^2} \\ \Rightarrow -\frac{7}{3}\left(-1+\frac{1}{3}\right)-\frac{4}{3}\left(\mathrm{\alpha }+\frac{2}{3}\right)=\frac{1}{\sqrt{3}}\sqrt{{\left(-\frac{2}{3}\right)}^2+{\left(\frac{2}{3}+\mathrm{\alpha }\right)}^2+{\left(\frac{\mathrm{\alpha }}{3}+\frac{2}{3}\right)}^2} \\ \Rightarrow -\frac{7}{3}\left(\frac{2}{3}\right)-\frac{4}{3}\left(\frac{3\mathrm{\alpha }+2}{3}\right)=\frac{1}{\sqrt{3}}\sqrt{{\left(-\frac{2}{3}\right)}^2+{\left(\frac{2}{3}+\mathrm{\alpha }\right)}^2+{\left(\frac{\mathrm{\alpha }}{3}+\frac{2}{3}\right)}^2} \\ \Rightarrow -\frac{15}{9}-\frac{12\mathrm{\alpha }}{9}-\frac{8}{9}=\frac{1}{\sqrt{3}}\sqrt{{\left(-\frac{2}{3}\right)}^2+{\left(\frac{2}{3}+\mathrm{\alpha }\right)}^2+{\left(\frac{\mathrm{\alpha }}{3}+\frac{2}{3}\right)}^2} \\ \Rightarrow \frac{-23-12\mathrm{\alpha }}{9}=\frac{1}{\sqrt{3}}\sqrt{{\left(-\frac{2}{3}\right)}^2+{\left(\frac{2+3\mathrm{\alpha }}{3}\right)}^2+{\left(\frac{\mathrm{\alpha }+2}{3}\right)}^2} \\ \mathrm{Square} \mathrm{both} \mathrm{sides} \mathrm{and} \mathrm{solve} \mathrm{for} \mathrm{\alpha } \mathrm{to} \mathrm{get} \mathrm{your} \mathrm{answer} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards