Question

Statement 1: Lines $\overrightarrow{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j})$ and $\overrightarrow{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k})$ intersect.
Statement 2: If $\overrightarrow{b}\times \overrightarrow{d}=\hat{0}$, then lines $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ and $\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{d}$ do not intersect.

• A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.
• B. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.
• C. Statement 1 is true and Statement 2 is false.
• D. Statement 1 is false and Statement 2 is true.

Answer 1

The correct option is B Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

Statement 2 is true, that is if $\overrightarrow{a}$ and $\overrightarrow{b}$ are direction vectors of two lines then

If

$\overrightarrow{a}\times \overrightarrow{b}=0$, implies the lines are parallel and do not intersect.

However in 3D, it is not necessary for two lines to be parallel if they do not intersect.

Consider statement 1

We can clearly observe that

$3i-j$ is not parallel to $2i+3k$

Now for them to be intersecting

$\overrightarrow{r_1}=\overrightarrow{r_2}$

Hence

$i+j-k+\lambda (3i-j)=4i-k+\mu (2i+3k)$

Hence

$i(1+3\lambda )+j(1-\lambda )-k=i(4+2\mu )+k(1-3\mu )$

Comparing , we get

$1-\lambda =0$
Or

$\lambda =1$ and

$1-3\mu =1$
$\mu =0$

And

$1+3\lambda =4+2\mu$

If $\mu =0$

Then

$\lambda =1$

Hence for $\lambda =1$ and $\mu =0$, the lines intersect.

Hence the correct answer is B.