Question

The lines $\overrightarrow{r}=\hat{i}+\hat{j}-\hat{k}+s(3\hat{i}-\hat{j})$ and $\overrightarrow{r}=4\hat{i}-\hat{k}+t(2\hat{i}+3\hat{k})$

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

Answer 1

The correct option is A intersect

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