Question

The shortest distance $\left(\text{in units}\right)$ between the lines whose equations are $\overrightarrow{r}=3\hat{i}+5\hat{j}+7\hat{k}+\lambda (\hat{i}+2\hat{j}+\hat{k})$ and $\overrightarrow{r}=-\hat{i}-\hat{j}+\hat{k}+s(7\hat{i}-6\hat{j}+\hat{k})$ is:

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

Answer 1

The correct option is B $\frac{26}{5\sqrt{5}}$

We know, shortest distance $=\frac{|(\overrightarrow{a}-\overrightarrow{c})\cdot (\overrightarrow{b}\times \overrightarrow{d})|}{|\overrightarrow{b}\times \overrightarrow{d}|}$
From the given lines, $\overrightarrow{r}=3\hat{i}+5\hat{j}+7\hat{k}+\lambda (\hat{i}+2\hat{j}+\hat{k})$ and $\overrightarrow{r}=-\hat{i}-\hat{j}+\hat{k}+s(7\hat{i}-6\hat{j}+\hat{k})$
we have $\overrightarrow{a}=3\hat{i}+5\hat{j}+7\hat{k};\overrightarrow{c}=-\hat{i}-\hat{j}+\hat{k};\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k};\overrightarrow{d}=7\hat{i}-6\hat{j}+\hat{k}$
On solving we have $(\overrightarrow{a}-\overrightarrow{c})=4\hat{i}+6\hat{j}+6\hat{k}$ and $\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 7& -6& 1\end{array}\right|=8\hat{i}+6\hat{j}-20\hat{k}\Rightarrow |\overrightarrow{b}\times \overrightarrow{d}|=\sqrt{500}$
Hence shortest distance $=\frac{52}{10\sqrt{5}}=\frac{26}{5\sqrt{5}}$