Question

Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector $\hat{i}-2\hat{j}+3\hat{k}.$ Reduce the corresponding equation in cartesian from.

Answer 1

We know that the vector equation of a line passing through a point with position vector $\overrightarrow{a}$ and parallel to the vector $\overrightarrow{b}$ is $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$.

Here,
$$\begin{gathered} \overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{b}=\hat{i}-2\hat{j}+3\hat{k} \end{gathered}$$

Vector equation of the required line is
$$\begin{gathered} \overrightarrow{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}-2\hat{j}+3\hat{k}\right)...\left(1\right) \\ \mathrm{Here},\lambda \mathrm{is}\mathrm{a}\mathrm{parameter}. \end{gathered}$$

Reducing (1) to cartesian form, we get

$$\begin{gathered} x\hat{i}+y\hat{j}+z\hat{k}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}-2\hat{j}+3\hat{k}\right)[\mathrm{Putting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\mathrm{in}(1\left)\right] \\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left(1+\lambda \right)\hat{i}+\left(2-2\lambda \right)\hat{j}+\left(3+3\lambda \right)\hat{k} \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\hat{i},\hat{j}\mathrm{and}\hat{k},\mathrm{we}\mathrm{get} \\ x=1+\lambda ,y=2-2\lambda ,z=3+3\lambda \\ \Rightarrow x-1=\lambda ,\frac{y-2}{-2}=\lambda ,\frac{z-3}{3}=\lambda \\ \Rightarrow \frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{3}=\lambda \\ \mathrm{Hence},\mathrm{the}\mathrm{cartesian}\mathrm{form}\mathrm{of}\left(1\right)\hspace{0.17em}\mathrm{is} \\ \frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{3} \end{gathered}$$