Discussion Forum

The shortest distance between the lines  $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$  and 

$\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}$ is 

Answer:

Explanation:

 Given, lines are   

$\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$  and 

$\frac{x-10}{3}=\frac{y-304}{8}=\frac{4-z}{5}$ 

 The vector  form of given lines are

  $r= 7\hat{i}-4\hat{j}+6\hat{k} +\lambda (3\hat{i}-16\hat{j}+7\hat{k})$   and

$r=10\hat{i}+30\hat{j}+4\hat{k} +\mu (3\hat{i}+8\hat{j}-5\hat{k})$

 On comparing these equations with

$r=a_{1}+\lambda b_{1}$  and   $r=a_{2}+\mu b_{2}$, we get 

$\overrightarrow{a_{1}}=7\hat{i}-4\hat{j}+6\hat{k}$

$\overrightarrow{a_{2}}=10\hat{i}+30\hat{j}+4\hat{k}$

$\overrightarrow{b_{1}}=3\hat{i}-16\hat{j}+7\hat{k}$

 and $\overrightarrow{b_{2}}=3\hat{i}+8\hat{j}-5\hat{k}$

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

 = $|\frac{(3\hat{i}+34\hat{j}-2\hat{k}).(24\hat{i}+36\hat{j}+72\hat{k})}{84}|$

 = $|\frac{72+1224-144}{84}|=|\frac{1152}{84}|=\frac{288}{41}$  units