Question

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

Answer 1

The position vectors of two arbitrary points on the given lines are

$$\begin{gathered} 3\hat{i}+2\hat{j}-4\hat{k}+\lambda \left(\hat{i}+2\hat{j}+2\hat{k}\right)=\left(3+\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(2\lambda -4\right)\hat{k} \\ 5\hat{i}-2\hat{j}+\mu \left(3\hat{i}+2\hat{j}+6\hat{k}\right)=\left(5+3\mu \right)\hat{i}+\left(-2+2\mu \right)\hat{j}+6\mu \hat{k} \end{gathered}$$

If the lines intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have
$\left(3+\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(2\lambda -4\right)\hat{k}=\left(5+3\mu \right)\hat{i}+\left(-2+2\mu \right)\hat{j}+6\mu \hat{k}$

Equating the coefficients of $\hat{i},\hat{j}\mathrm{and}\hat{k}$, we get

$$\begin{gathered} 3+\lambda =5+3\mu ...\left(1\right) \\ 2+2\lambda =-2+2\mu ...\left(2\right) \\ 2\lambda -4=6\mu ...\left(3\right) \end{gathered}$$

Solving (1) and (2), we get
$$\begin{gathered} \lambda =-4 \\ \mu =-2 \end{gathered}$$.

Substituting the values $\lambda =-4\mathrm{and}\mu =-2$ in (3), we get

$$\begin{gathered} \mathrm{LHS}=2\lambda -4 \\ =2\left(-4\right)-4 \\ =-12 \\ \mathrm{RHS}=6\mu \\ =6\left(-2\right) \\ =-12 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Since}\lambda =-4\mathrm{and}\mu =-2\mathrm{satisfy}\left(3\right),\mathrm{the}\mathrm{lines}\mathrm{intersect}. \end{gathered}$$

Substituting $\mu =-2$ in the second line, we get $\overrightarrow{r}=5\hat{i}-2\hat{j}-6\hat{i}-4\hat{j}-12\hat{k}=-\hat{i}-6\hat{j}-12\hat{k}$ as the position vector of the point of intersection.

Thus, the coordinates of the point of intersection are ($-$1, $-$6, $-$12).