Question

Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line $\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}.$

Answer 1

We have

$\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$

It can be re-written as

$$\begin{gathered} \frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-3}{\left({\displaystyle \frac{3}{2}}\right)} \\ =\frac{x+2}{-2}=\frac{y+3}{14}=\frac{z-3}{3} \end{gathered}$$

This shows that the given line passes through the point $\left(-2,-3,3\right)$ and its direction ratios are proportional to $-$2, 14, 3.

Thus, the parallel vector is $\overrightarrow{b}=-2\hat{i}+14\hat{j}+3\hat{k}$.

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,
$$\begin{gathered} \overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{b}=-2\hat{i}+14\hat{j}+3\hat{k} \end{gathered}$$.

Vector equation of the required line is
$$\begin{gathered} \overrightarrow{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(-2\hat{i}+14\hat{j}+3\hat{k}\right)...\left(1\right) \\ \mathrm{Here},\lambda \mathrm{is}\mathrm{a}\mathrm{parameter}. \end{gathered}$$

Reducing (1) to cartesian form, we get

$$\begin{gathered} x\hat{i}+y\hat{j}+z\hat{k}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(-2\hat{i}+14\hat{j}+3\hat{k}\right)[\mathrm{Putting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\mathrm{in}(1\left)\right] \\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left(1-2\lambda \right)\hat{i}+\left(2+14\lambda \right)\hat{j}+\left(3+3\lambda \right)\hat{k} \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\hat{i},\hat{j}\mathrm{and}\hat{k},\mathrm{we}\mathrm{get} \\ x=1-2\lambda ,y=2+14\lambda ,z=3+3\lambda \\ \Rightarrow \frac{x-1}{-2}=\lambda ,\frac{y-2}{14}=\lambda ,\frac{z-3}{3}=\lambda \\ \Rightarrow \frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3}=\lambda \\ \mathrm{Hence},\mathrm{the}\mathrm{cartesian}\mathrm{form}\mathrm{of}\left(1\right)\mathrm{is} \\ \frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3} \end{gathered}$$