Question

Let $A$ and $B$ be two points on the lines $\overrightarrow{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}-\hat{j}-\hat{k})+\mu (2\hat{i}+\hat{j}+2\hat{k})$ respectively, which are the nearest to each other, which of the following is/are CORRECT?

• A. Distance between $A$ and $B$ is $\frac{3\sqrt{2}}{2}$
• B. Equation of the plane $OAB$ (where O is the origin) is $\overrightarrow{r}.(\hat{i}+34\hat{j}+\hat{k})=0$
• C. Volume of the tetrahedron $OABC$ (where $O$ is the origin) and $C$ is $(1,0,1)$ is $\frac{1}{12}$
• D. Equation of the line of shortest distance is $\overrightarrow{r}=(\frac{17}{6}\hat{i}+\frac{1}{6}\hat{j}+\frac{17}{6}\hat{k})+t(\hat{i}-\hat{k})$ (where $t$ is a parameter)

Answer 1

Answer: (A), (C), (D)

The correct options are
A Distance between $A$ and $B$ is $\frac{3\sqrt{2}}{2}$
C Volume of the tetrahedron $OABC$ (where $O$ is the origin) and $C$ is $(1,0,1)$ is $\frac{1}{12}$
D Equation of the line of shortest distance is $\overrightarrow{r}=(\frac{17}{6}\hat{i}+\frac{1}{6}\hat{j}+\frac{17}{6}\hat{k})+t(\hat{i}-\hat{k})$ (where $t$ is a parameter)

$\begin{array}{c}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{n_1}\\ \overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{n_2}\end{array}]$

$AB$ will be line of shortest distance.

dr's of $AB$ $=\overrightarrow{n_1}\times \overrightarrow{n_2}=-3\hat{i}+0\hat{j}+3\hat{k}$

dr's of $AB$ is $(-1,0,1)$

$AB=\frac{\left|[\overrightarrow{b}-\overrightarrow{a}\overrightarrow{n_1}\overrightarrow{n_2}]\right|}{|\overrightarrow{n_1}\times \overrightarrow{n_2}|}=\frac{3}{\sqrt{2}}$

Let $A$ is $(1+\lambda ,2-\lambda ,1+\lambda )$ and $B$ is $(2+2\mu ,\mu -1,2\mu -1)$

dr's of $AB$ is $2\mu -\lambda +1,\mu +\lambda -3,-\lambda +2\mu -2$

$\therefore \frac{2\mu -\lambda +1}{-1}=\frac{\mu +\lambda -3}{0}=\frac{-\lambda +2\mu -2}{1}$

$\mu +\lambda =3,2\lambda -4\mu =-1$

$\lambda =\frac{11}{6},\mu =\frac{7}{6}$

$A(\frac{17}{6},\frac{1}{6},\frac{17}{6})$, $B(\frac{26}{6},\frac{1}{6},\frac{8}{6})$

Length of $AB=\sqrt{{\left(\frac{9}{6}\right)}^2+{(-\frac{9}{6})}^2}=\frac{3}{\sqrt{2}}$

Normal vector to the plane $OAB$ is -

$=\frac{1}{36}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 17& 1& 17\\ 26& 1& 8\end{array}\right|=\frac{1}{36}(-9,306,-9)=-\frac{1}{4}(1,-34,1)$

Equation of plane is $\overrightarrow{r}\cdot (\hat{i}-34\hat{j}+\hat{k})=0$



Volume of tetrahedron
$=\frac{1}{6}\left|\begin{array}{ccc}1& 0& 1\\ \frac{17}{6}& \frac{1}{6}& \frac{17}{6}\\ \frac{26}{6}& \frac{1}{6}& \frac{8}{6}\end{array}\right|$

$=\frac{18}{216}=\frac{1}{12}$