Question

The lines $\overrightarrow{r}=\left(\hat{i}-\hat{j}\right)+l(2\hat{i}+\hat{k})$ and $\overrightarrow{r}=\left(2\hat{i}-\hat{j}\right)+m(\hat{i}+\hat{j}-\hat{k})$.

• A. do not intersect at any values of $l$ and $m$
• B. intersect when $l=1$and $m=2$
• C. intersect when $l=2$ and $m=\frac{1}{2}$
• D. intersect at any values of $l$ and $m$

Answer 1

The correct option is A

do not intersect at any values of $l$ and $m$

Explanation for correct option

Given: $\overrightarrow{r}=\left(\hat{i}-\hat{j}\right)+l(2\hat{i}+\hat{k})$ and $\overrightarrow{r}=\left(2\hat{i}-\hat{j}\right)+m(\hat{i}+\hat{j}-\hat{k})$

$\overrightarrow{r}=\left(\hat{i}-\hat{j}\right)+l(2\hat{i}+\hat{k})$ and $\overrightarrow{r}=\left(2\hat{i}-\hat{j}\right)+m(\hat{i}+\hat{j}-\hat{k})$

let these lines

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ and $\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{d}$

Shortest distance between the lines $D=\frac{\left|\left(\overrightarrow{a}-\overrightarrow{c}\right)·\left(\overrightarrow{b}\times \overrightarrow{d}\right)\right|}{\left|\overrightarrow{b}\times \overrightarrow{d}\right|}$

cross product of $(a_1\hat{i}+b_1\hat{j}+c_1\hat{k})$and $(a_2\hat{i}+b_2\hat{j}+c_2\hat{k})$ is $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=\hat{i}(b_1{c}_2-c_1{b}_2)-\hat{j}(a_1{c}_2-c_1{a}_2)+\hat{k}(a_1{b}_2-b_1{a}_2)$

cross product of $(2\hat{i}+\hat{k})$and $(\hat{i}+\hat{j}-\hat{k})$ is $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& 1\\ 1& 1& -1\end{array}\right|=\hat{i}(0\times 1-1\times 1)-\hat{j}(2\times 1-1)+\hat{k}(2\times 1-0\times 1)$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& 1\\ 1& 1& -1\end{array}\right|=-\hat{i}-\hat{j}+\widehat{2k}$

$$\begin{gathered} \left(D\right)=\frac{\left|\left(\hat{i}-\hat{j}-2\hat{i}+\hat{j}\right)·\left((2\hat{i}+\widehat{k)}\times (\hat{i}+\hat{j}-\hat{k})\right)\right|}{\left|(2\hat{i}+\widehat{k)}\times (\hat{i}+\hat{j}-\hat{k})\right|} \\ \left(D\right)=\frac{\left|\left(-\hat{i}\right)·\left(-\hat{i}-\hat{j}+2\hat{k}\right)\right|}{\left|\left(-\hat{i}-\hat{j}+2\hat{k}\right)\right|} \\ \left(D\right)=\frac{1}{\sqrt{1^2+1^2+2^2}} \\ \left(D\right)=\frac{1}{\sqrt{6}} \end{gathered}$$

Since the shortest distance is not equal to zero, both the lines do not intersect.

Hence, option (A) is correct.