Question

If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda )=(z-3),\lambda \in R$ is $\frac{1}{\sqrt{38}},$ then the integral value of $\lambda$ is equal to:

• A. $2$
• B. $5$
• C. $3$
• D. $-1$

Answer 1

The correct option is C $3$

The given lines are
$3(x-1)=6(y-2)=2(z-1)$
$\Rightarrow \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$
$\therefore \overrightarrow{r}=(\hat{i}+2\hat{j}+\hat{k})+t(2\hat{i}+\hat{j}+3\hat{k})\cdots \left(i\right)$

and $4(x-2)=2(y-\lambda )=(z-3)$
$\Rightarrow \frac{x-2}{1}=\frac{y-\lambda }{2}=\frac{z-3}{4}$
$\therefore \overrightarrow{r}=(2\hat{i}+\lambda \hat{j}+3\hat{k})+s(\hat{i}+2\hat{j}+4\hat{k})\cdots \left(ii\right)$

$\therefore {\overrightarrow{a}}_1=\hat{i}+2\hat{j}+\hat{k},{\overrightarrow{a}}_2=2\hat{i}+\lambda \hat{j}+3\hat{k}$
$\therefore {\overrightarrow{a}}_1-{\overrightarrow{a}}_2=-\hat{i}+(2-\lambda )\hat{j}-2\hat{k}$
${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=(2\hat{i}+\hat{j}+3\hat{k})\times (\hat{i}+2\hat{j}+4\hat{k})$
$=-2\hat{i}-5\hat{j}+3\hat{k}$
$\therefore S.D.=\left|\frac{({\overrightarrow{a}}_1-{\overrightarrow{a}}_2).({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}\right|=\frac{1}{\sqrt{38}}$
$\therefore |5\lambda -14|=1$
$\therefore \lambda =3$$\left(\lambda \text{is integer}\right)$