Question

Find the vector equation of a line passing through the point with position vector $\hat{i}-2\hat{j}-3\hat{k}$and parallel to the line joining the points with position vectors $\hat{i}-\hat{j}+4\hat{k}\mathrm{and}2\hat{i}+\hat{j}+2\hat{k}.$ Also, find the cartesian equivalent of this equation.

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,
$$\begin{gathered} \overrightarrow{a}=\hat{i}-2\hat{j}-3\hat{k} \\ \overrightarrow{b}=\left(2\hat{i}+\hat{j}+2\hat{k}\right)-\left(\hat{i}-\hat{j}+4\hat{k}\right)=\hat{i}+2\hat{j}-2\hat{k} \end{gathered}$$

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

Reducing (1) to cartesian form, we get

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