Question

Find shortest distance between lines $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+\hat{j}+4\hat{k})$ and

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

• A. $2$
• B. $\frac{1}{6}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\sqrt{6}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{6}}$

Given eq of lines

$\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$----(1)

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

position vector and normal vector of line (1)

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

$\overrightarrow{n_1}=2\hat{i}+3\hat{j}+4\hat{k}$

position vector and normal vector of line (2)

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

$\overrightarrow{n_2}=3\hat{i}+4\hat{j}+5\hat{k}$

shortest between two skews line

$SD=\frac{\overrightarrow{AC}\cdot (\overrightarrow{n_1}\times \overrightarrow{n_2})}{|\overrightarrow{n_1}\times \overrightarrow{n_2}|}$

$AC=\overrightarrow{c}-\overrightarrow{a}$

$AC=2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})$

$AC=2\hat{i}+4\hat{j}+5\hat{k}-\hat{i}-2\hat{j}-3\hat{k}$

$AC=\hat{i}+2\hat{j}+2\hat{k}$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=-\hat{i}+2\hat{j}-\hat{k}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{(-1{)}^2+2^2+(-1{)}^2}$

$|\overrightarrow{n_1}\times \overrightarrow{n_2}|=\sqrt{6}$

putting AC $\overrightarrow{n_1}\times \overrightarrow{n_2},|\overrightarrow{n_1}\times \overrightarrow{n_2}|$ in formula

$SD=\frac{\overrightarrow{AC}\cdot (\overrightarrow{n_1}\times \overrightarrow{n_2})}{|\overrightarrow{n_1}\times \overrightarrow{n_2}|}$

$SD=\frac{(\hat{i}+2\hat{j}+2\hat{k})\cdot (-\hat{i}+2\hat{j}-\hat{k})}{\sqrt{6}}$

$SD=\frac{-1+4-2}{\sqrt{6}}$

$SD=\frac{1}{\sqrt{6}}$