Discussion Forum

The shortest distance between  the lines x=y+2=6z-6 and x+1=2y=-12z is 

Answer:

Explanation:

The line are   $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$

 and   $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$

 Here,   $\overrightarrow{a_{1}}=-2\hat{j}+\hat{k},\overrightarrow{b_{1}}=6\hat{i}+6\hat{j}+\hat{k},\overrightarrow{a_{2}}=-\hat{i}$

   $\overrightarrow{b_{2}}=12\hat{i}+6\hat{j}-\hat{k}$

 $\overrightarrow{b_{1}}\times \overrightarrow{b_{2}}=\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\6& 6&1\\12&6&-1 \end{bmatrix}=-12\hat{i}+18\hat{j}-36\hat{k}$.

 Shortest distance =   $\frac{|(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}).(\overrightarrow{b_{1}}-\overrightarrow{b_{2}})|}{|\overrightarrow{b_{1}}\times\overrightarrow{b_{2}}|}$

=$\frac{|(-\hat{i}+2\hat{j}-\hat{k}).(-12\hat{i}+18\hat{j}-36\hat{k})|}{\sqrt{(-12)^{2}+(18)^{2}+(-36)^{2}}}$

 =  $\frac{|12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2$