Question

The vector equations of two lines $L_1$ and $L_2$ are given by
$L_2:r=(2\hat{i}+9\hat{j}+13\hat{k})+\lambda (\hat{i}+2\hat{j}+3\hat{k})$ and $L_2:r=(-3\hat{i}+7\hat{j}+p\hat{k})+\mu (-\hat{i}+2\hat{j}-3\hat{k})$. Then, the lines $L_1$ and $L_2$ are

• A. skew lines for all $p\in R$
• B. intersecting for all $p\in R$ and the point of intersection is $(-1,3,4)$
• C. intersecting lines for $p=-2$
• D. intersecting for all real $p\in R$

Answer 1

The correct option is C intersecting lines for $p=-2$

The vector along $L_1=\overline{b}=\hat{i}+2\hat{j}+3\hat{k}$
The vector along $L_2=\overline{d}=-\hat{i}+2\hat{j}-3\hat{k}$
Now, the two lines are not parallel to each other. They can be either skew or intersecting. If they are intersecting, the shortest distance between the two will be zero.
$\overline{b}\times \overline{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ -1& 2& -3\end{array}\right|=-12\hat{i}+4\hat{k}$
The vector joining two points, one on each line is given by $\overline{r}=5\hat{i}+2\hat{j}+(13-p)\hat{k}$
If the lines are intersecting, $\overline{r}.(\overline{b}\times \overline{d})=0$
$\Rightarrow -60+4(13-p)=0$
$\Rightarrow p=-2$
Hence, option C is correct.