Question

The shortest distance between the skew lines $\bar{r}=(i+2\bar{j}+3\bar{k})+t(i+3\bar{j}+2\bar{k})$ and $\bar{r}=(4i+5\bar{j}+6\bar{k})+t(2i+3\bar{j}+\bar{k})$ is

• A. $\sqrt6$
• B. 3
• C. $2\sqrt3$
• D. $\sqrt3$

Answer 1

Answer: (D)

$\sqrt3$

Answer 2

S.D = $\frac{\left[\overline{a} -\overline{c},\overline{b},\overline{d}\right]}{\left[\overline{b}\times\overline{d}\right]}$ $\bar{a} - \bar{c} = (\hat{i} + 2 \hat{j} + 3\hat{k} ) - ( 4 \hat{i} + 5\hat{j} + 6\hat{k})$ = $-3\hat{i} - 3\hat{j} -3 \hat{k} = -3(\hat{i} +\hat{j} +\hat{k} )$ $\bar{b} = \hat{i} + 3\hat{j} + \hat{k}$, $\bar{a} = 2\hat{i} +3\hat{j} +\hat{k}$ $\therefore$ $|\bar{a} \times \bar{b} | = \sqrt{9+9+9} = \sqrt{27} = 3 \sqrt{3}$ $\therefore$ S.D = $\frac{9}{3 \sqrt{3}} = \sqrt{3}$