Question

Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).

Answer 1

We know that the vector equation of a line passing through the points with position vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\overrightarrow{r}=\overrightarrow{a}+\lambda \left(\overrightarrow{b}-\overrightarrow{a}\right)$, where $\lambda$ is a scalar.

Here,
$$\begin{gathered} \overrightarrow{a}=\hat{i}+2\hat{j}-\hat{k} \\ \overrightarrow{b}=2\hat{i}+\hat{j}+\hat{k} \end{gathered}$$

Vector equation of the required line is
$$\begin{gathered} \overrightarrow{r}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\lambda \left\{\left(2\hat{i}+\hat{j}+\hat{k}\right)-\left(\hat{i}+2\hat{j}-\hat{k}\right)\right\} \\ \Rightarrow \overrightarrow{r}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+2\hat{k}\right)...\left(1\right) \\ \mathrm{Here},\lambda \mathrm{is}\mathrm{a}\mathrm{parameter}. \end{gathered}$$

Reducing (1) to cartesian form, we get

$$\begin{gathered} x\hat{i}+y\hat{j}+z\hat{k}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+2\hat{k}\right)[\mathrm{Putting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}in(1\left)\right] \\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left(1+\lambda \right)\hat{i}+\left(2-\lambda \right)\hat{j}+\left(-1+2\lambda \right)\hat{k} \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\hat{i},\hat{j}\mathrm{and}\hat{k},\mathrm{we}\mathrm{get} \\ x=1+\lambda ,y=2-\lambda ,z=-1+2\lambda \\ \Rightarrow x-1=\lambda ,\frac{y-2}{-1}=\lambda ,\frac{z+1}{2}=\lambda \\ \Rightarrow \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2}=\lambda \\ \mathrm{Hence},\mathrm{the}\mathrm{cartesian}\mathrm{form}\mathrm{of}\left(1\right)\mathrm{is} \\ \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2} \end{gathered}$$