Question

The shortest distance between the lines $\overrightarrow{r}=3\hat{i}+5\hat{j}+7\hat{k}+\lambda (\hat{i}+2\hat{j}+\hat{k})and\overrightarrow{r}=-\hat{i}-\hat{j}-\hat{k}+\mu (7\hat{i}-6\hat{j}+\hat{k})is$

• A. $\frac{16}{5\sqrt{5}}$
• B. $\frac{26}{5\sqrt{5}}$
• C. $\frac{36}{5\sqrt{5}}$
• D. $\frac{46}{5\sqrt{5}}$

Answer 1

The correct option is D $\frac{46}{5\sqrt{5}}$

$(\overrightarrow{b_1}\times \overrightarrow{b_2})$=$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 7& -6& 1\end{array}\right|=\frac{8\hat{i}+6\hat{j}-20\hat{k}}{\sqrt{64+36+400}}=\frac{8\hat{i}+6\hat{j}-20\hat{k}}{10\sqrt{5}}$

$\overrightarrow{a_2}-\overrightarrow{a_1}=4\hat{i}+6\hat{j}+8\hat{k}$

$\therefore$ shortest distance $=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})|}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$

$=\frac{|32+36-160|}{10\sqrt{5}}$

$=\frac{92}{10\sqrt{5}}=\frac{46}{5\sqrt{5}}$