Question

The correct option about the lines $\overrightarrow{r}=\hat{i}-\hat{j}-\hat{k}+s(\hat{i}+2\hat{j}-2\hat{k})$ and $\overrightarrow{r}=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-2\hat{k})$ is

• A. intersect
• B. don't intersect
• C. skew lines
• D. Cannot be determined

Answer 1

The correct option is C skew lines

Given: $\overrightarrow{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k})$ and $\overrightarrow{r}=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-2\hat{k})$
We have two lines are $\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}$and$\overrightarrow{r}=\overrightarrow{c}+t\overrightarrow{d}$
$\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$
$\overrightarrow{c}=\hat{i}-2\hat{j}+3\hat{k}$
$\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\overrightarrow{d}=-\hat{i}+\hat{j}-2\hat{k}$
To check the type of lines.
Lets find the value of
$\Rightarrow [\overrightarrow{a}-\overrightarrow{c}\overrightarrow{b}\overrightarrow{d}]$
$\Rightarrow \left[\right(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})\hat{i}+2\hat{j}-2\hat{k}-\hat{i}+\hat{j}-2\hat{k}]$
$\Rightarrow [\hat{j}-4\hat{k}\hat{i}+2\hat{j}-2\hat{k}-\hat{i}+\hat{j}-2\hat{k}]$
$\Rightarrow \left|\begin{array}{ccc}0& 1& -4\\ 1& 2& -2\\ -1& 1& -2\end{array}\right|=-1(-2+4)-1(-2+8)$
$\Rightarrow -2-6=-8$
Hence,
$[\overrightarrow{a}-\overrightarrow{c}\overrightarrow{b}\overrightarrow{d}]\ne 0\Rightarrow$ Given lines are non coplanar.
$\therefore$These are skew lines.