Question

Prove that the line $\overrightarrow{r}=\left(\hat{i}+\hat{j}-\hat{k}\right)+\lambda \left(3\hat{i}-\hat{j}\right)\mathrm{and}\overrightarrow{r}=\left(4\hat{i}-\hat{k}\right)+\mu \left(2\hat{i}+3\hat{k}\right)$ intersect and find their point of intersection.

Answer 1

The position vectors of two arbitrary points on the given lines are
$$\begin{gathered} \left(\hat{i}+\hat{j}-\hat{k}\right)+\lambda \left(3\hat{i}-\hat{j}\right)=\left(1+3\lambda \right)\hat{i}+\left(1-\lambda \right)\hat{j}-\hat{k} \\ \left(4\hat{i}-\hat{k}\right)+\mu \left(2\hat{i}+3\hat{k}\right)=\left(4+2\mu \right)\hat{i}+0\hat{j}+\left(3\mu -1\right)\hat{k} \end{gathered}$$

If the lines intersect, then they have a common point. So, for some values of $\lambda \mathrm{and}\mu$, we must have

$\left(1+3\lambda \right)\hat{i}+\left(1-\lambda \right)\hat{j}-\hat{k}=\left(4+2\mu \right)\hat{i}+0\hat{j}+\left(3\mu -1\right)\hat{k}$

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

$$\begin{gathered} 1+3\lambda =4+2\mu ...\left(1\right) \\ 1-\lambda =0...\left(2\right) \\ 3\mathrm{\mu }-1=-1...\left(3\right) \end{gathered}$$

Solving (2) and (3), we get
$$\begin{gathered} \lambda =1 \\ \mu =0 \end{gathered}$$

Substituting the values $\lambda =1\mathrm{and}\mu =0$ in (1), we get

$$\begin{gathered} \mathrm{LHS}=1+3\lambda \\ =1+3\left(1\right) \\ =4 \\ \mathrm{RHS}=4+2\mu \\ =4+2\left(0\right) \\ =4 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Since}\lambda =1\mathrm{and}\mu =0\mathrm{satisfy}\left(3\right),\mathrm{the}\mathrm{given}\mathrm{lines}\mathrm{intersect}. \end{gathered}$$

Substituting $\mu =0$ in the second line, we get $\overrightarrow{r}=4\hat{i}+0\hat{j}-\hat{k}$ as the position vector of the point of intersection.

Thus, the coordinates of the point of intersection are (4, 0, $-$1).