Mathematics - Vector Question with Solution | TestHub

The given lines are $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda )=(z-3)$

And, they can be written as $L_1:\frac{(x-1)}{2}=\frac{(y-2)}{1}=\frac{(z-1)}{3}$ and 

$L_2:\frac{(x-2)}{1}=\frac{y-\lambda }{2}=\frac{z-3}{4}.$

The lines are in the form $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n},$ where $\left(x_1, y_1, z_1\right)$ is a point on the line and $l, m, n$ are the direction ratios of the vector parallel to the line.

Thus, the position vectors of the points on the line are respectively $\overrightarrow{a_1}=\hat{i}+2\hat{j}+\hat{k} \& \overrightarrow{a_2}=2\hat{i}+\lambda \hat{j}+3\hat{k}$ and direction of the lines are along the vectors ${\overrightarrow{r}}_1=2\hat{i}+\hat{j}+3\hat{k}$ and $\overrightarrow{r_2}=\hat{i}+2\hat{j}+4\hat{k}.$

Thus, we have $\overrightarrow{a}=\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}+\left(\lambda -2\right)\hat{j}+2\hat{k}.$

Shortest distance $=$ Projection of $\overrightarrow{a}$ on $\overrightarrow{r_1}\times \overrightarrow{r_2}$

$=\frac{\left|\overrightarrow{a}·\left(\overrightarrow{r_1}\times \overrightarrow{r_2}\right)\right|}{\left|\overrightarrow{r_1}\times \overrightarrow{r_2}\right|}$

Now $\overrightarrow{r_1}\times \overrightarrow{r_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 3\\ 1& 2& 4\end{array}\right|$

$\Rightarrow \overrightarrow{r_1}\times \overrightarrow{r_2}=\hat{i}\left(4-6\right)-\hat{j}\left(8-3\right)+\hat{k}\left(4-1\right)$

$\Rightarrow \overrightarrow{r_1}\times \overrightarrow{r_2}=-2\hat{i}-5\hat{j}+3\hat{k}$

$\Rightarrow \left|\overrightarrow{r_1}\times \overrightarrow{r_2}\right|=\sqrt{{\left(-2\right)}^2+{\left(-5\right)}^2+3^2}=\sqrt{4+25+9}=\sqrt{38}$

And, $\left|\overrightarrow{a}·\left(\overrightarrow{r_1}\times {\overrightarrow{r}}_2\right)\right|=\left|\begin{array}{ccc}1& \lambda -2& 2\\ 2& 1& 3\\ 1& 2& 4\end{array}\right|$

$\Rightarrow \left|\overrightarrow{a}·\left(\overrightarrow{r_1}\times {\overrightarrow{r}}_2\right)\right|=\left|1\left(4-6\right)-\left(\lambda -2\right)\left(8-3\right)+2\left(4-1\right)\right|$

$\Rightarrow \left|\overrightarrow{a}·\left(\overrightarrow{r_1}\times {\overrightarrow{r}}_2\right)\right|=\left|-2-5\lambda +10+6\right|$

$\Rightarrow \left|\overrightarrow{a}·\left(\overrightarrow{r_1}\times {\overrightarrow{r}}_2\right)\right|=\left|14-5\lambda \right|$

Given, the shortest distance is $\frac{1}{\sqrt{38}},$ hence, we have

$\frac{1}{\sqrt{38}}=\frac{\left|14-5\lambda \right|}{\sqrt{38}}$

$\Rightarrow \left|14-5\lambda \right|=1$

$\Rightarrow 14-5\lambda =1$ or $14-5\lambda =-1$

$\Rightarrow \lambda =\frac{13}{5}$ or $3$

$\therefore$ Integral value is $\lambda =3.$