Question

Find the shortest distance between the skew – lines
$\overline{r}=(2+\lambda )\hat{i}-(1+2\lambda )\hat{j}+(1+3\lambda )\hat{k}and\overline{r}=(\hat{i}+2\hat{j}+3\hat{k})+\mu (3\hat{i}-2\hat{j}-\hat{k})$

• A. $\frac{\sqrt{3}}{2}$
• B. $\sqrt{\frac{3}{2}}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\sqrt{6}$

Answer 1

The correct option is B $\sqrt{\frac{3}{2}}$

$\overline{r}=(2i-j+k)+\lambda (i-2j-3k)=\overline{a}+\lambda \overline{b}$
$\overline{r}=(i+2j+3k)+\mu (3i-2j-k)=\overline{c}+\mu \overline{d}$
$S.D=\frac{|[\overline{a}-\overline{c}\overline{b}\overline{d}]|}{|b\times d|}=\frac{|[i-3j-2ki-2j-2k3i-2j-k]|}{|(i-2j-3k)\times (3i-2j-k)|}=\frac{\left|\begin{array}{ccc}1& -3& -2\\ 1& -2& -3\\ 3& -2& -1\end{array}\right|}{\left|\begin{array}{ccc}i& j& k\\ 1& -2& -3\\ 3& -2& -1\end{array}\right|}=\frac{12}{|-4(i+2j+k)|}=\frac{3}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{2}}$