Question

For the given lines $\overrightarrow{r}=2\hat{i}+2\hat{j}-\hat{k}+\lambda (\hat{i}-2\hat{j}-\hat{k})$ and $\overrightarrow{r}=3\hat{i}-2\hat{j}+\hat{k}+\mu (-\hat{i}-\hat{j}+\hat{k}),$ which of the following is correct ?

• A. both lines are parallel
• B. both lines are intersecting
• C. both lines are coincident
• D. both lines are skew

Answer 1

The correct option is D both lines are skew

Given lines:
$\overrightarrow{r}=2\hat{i}+2\hat{j}-\hat{k}+\lambda (\hat{i}-2\hat{j}-\hat{k})=\overrightarrow{a}+\lambda \overrightarrow{b}$ and $\overrightarrow{r}=3\hat{i}-2\hat{j}+\hat{k}+\mu (-\hat{i}-\hat{j}+\hat{k})=\overrightarrow{c}+\mu \overrightarrow{d}$
As $\overrightarrow{b}$ is not parallel to $\overrightarrow{d}$
So, lines are either intersecting or skew.
Now any point on line $L_1=(2+\lambda ,2-2\lambda ,-1-\lambda )$ and on $L_2=(3-\mu ,-2-\mu ,1+\mu )$
For point of intersection , $L_1=L_2$
$\Rightarrow 2+\lambda =3-\mu \cdots \left(i\right)2-2\lambda =-2-\mu \cdots \left(ii\right)-1-\lambda =1+\mu \cdots \left(iii\right)$
Adding $\left(i\right)$ and $\left(iii\right):$ we get $,1=4$
which is not true
So, there is no point of intersection.
$\therefore$ lines are skew.