Question

Show that the lines
$\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})$
$\overrightarrow{r}=5\hat{i}-2\hat{j}+\mu (3\hat{i}+2\hat{j}+6\hat{k})$
are intersecting. Hence find their point of intersection.

Answer 1

Let the lines are $M:\overrightarrow{r}=3\overrightarrow{i}+2\overrightarrow{j}-4\overrightarrow{k}+\lambda (\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k})$ and $N:\overrightarrow{r}=5\overrightarrow{i}-2\overrightarrow{j}+\mu (3\overrightarrow{i}+2\overrightarrow{j}+6\overrightarrow{k})$

Coordinates of any random point on $M$ are $P(3+\lambda ,2+2\lambda ,-4+2\lambda )$ and on $N$ are $Q(5+3\mu ,-2+2\mu ,6\mu )$
If the lines $M$ and $N$ intersect then, they must have a common point on them i.e., $P$ and $Q$ must coincide for some values of $\lambda$ and $\mu$
Now, $3+\lambda =5+3\mu$ ----- (1)
$2+2\lambda =-2+2\mu$----- (2
$-4+2\lambda =6\mu$-----(3)
Solving 1 and 2,
we get $\lambda =-4,\mu =-2$
Substitute the values in equation 3,
$-4+2(-4)=6(-2)$
$-4=-12+8$
$-4=-4$
So, the given lines intersect each other
Now, point of intersection is $P(-1,-6,-12).$