Question

Find the shortest distance between the line whose equations are $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$ and $\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu (4\hat{i}+6\hat{j}+8\hat{k})$

• A. $\sqrt{\frac{5}{29}}$
• B. $\sqrt{\frac{6}{29}}$
• C. $\sqrt{\frac{7}{29}}$
• D. $\sqrt{\frac{8}{29}}$

Answer 1

The correct option is A $\sqrt{\frac{5}{29}}$

given lines

$\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$

$\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+2\mu (2\hat{i}+3\hat{j}+4\hat{k})$

both lines are parallel

position vector of line

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{c}=2\hat{i}+4\hat{j}+5\hat{k}$

normal vector

$\overrightarrow{b}=2\hat{i}+3\hat{j}+4\hat{k}$

$SD=\left|\frac{(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right|$

$SD=\left|\frac{(2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k}\left)\right)\times (2\hat{i}+3\hat{j}+4\hat{k})}{\sqrt{2^2+3^2+4^2}}\right|$

$SD=\left|\frac{(2\hat{i}+4\hat{j}+5\hat{k}-\hat{i}-2\hat{j}-3\hat{k})\times (2\hat{i}+3\hat{j}+4\hat{k})}{\sqrt{4+9+16}}\right|$

$SD=\left|\frac{(\hat{i}+2\hat{j}+2\hat{k})\times (2\hat{i}+3\hat{j}+4\hat{k})}{\sqrt{29}}\right|$

$SD=\frac{\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 3& 4\end{array}\right|\right|}{\sqrt{29}}$

$SD=\frac{\left|\hat{i}\right(8-6)-\hat{j}(4-4)+\hat{k}(3-4\left)\right|}{\sqrt{29}}$

$SD=\frac{|2\hat{i}-\hat{k}|}{\sqrt{29}}$

$SD=\frac{\sqrt{4+1}}{\sqrt{29}}$

$SD=\sqrt{\frac{5}{29}}$