Question

The correct statement about the lines $\overrightarrow{r}=\hat{i}+2\hat{j}+s(3\hat{i}-\hat{j}+\hat{k})$ and $\overrightarrow{r}=10\hat{i}-\hat{j}+\alpha \hat{k}+t(-6\hat{i}+2\hat{j}-2\hat{k})$ is/are:

• A. They are skew lines for any value of $\alpha$
• B. They are coincident for $\alpha =3$
• C. They are parallel for $\alpha =2$
• D. They have more than one common point for $\alpha =1$

Answer 1

Answer: (B), (C)

They are parallel for $\alpha =2$

$\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}$ and $\overrightarrow{r}=\overrightarrow{c}+t\overrightarrow{d}$
Here $\overrightarrow{a}=\hat{i}+2\hat{j},\overrightarrow{b}=3\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{c}=10\hat{i}-\hat{j}+\alpha \hat{k},\overrightarrow{d}=-6\hat{i}+2\hat{j}-2\hat{k}$
Clearly, $\overrightarrow{b}$ and $\overrightarrow{d}$ are parallel. Hence given lines are parallel.
Now for points of intersection,
$\overrightarrow{a}+s\overrightarrow{b}=\overrightarrow{c}+t\overrightarrow{d}$
$\Rightarrow \hat{i}+2\hat{j}+s(3\hat{i}-\hat{j}+\hat{k})=10\hat{i}-\hat{j}+\alpha \hat{k}+t(-6\hat{i}+2\hat{j}-2\hat{k})$
On comparing the coefficients of $\hat{i},\hat{j},\hat{k}$ on either sides, we have equations $s+2t=3$ and $s+2t=\alpha$
$\therefore \alpha =3\Rightarrow$Collinear and $\alpha \in R-\left\{3\right\}\Rightarrow$ Parallel but not collinear.