Question

The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Answer 1

The cartesian equation of the given line is
$x=ay+b,z=cy+d$

It can be re-written as

$\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}$

Thus, the given line passes through the point $\left(b,0,d\right)$ and its direction ratios are proportional to a, 1, c. It is also parallel to the vector $\overrightarrow{b}=a\hat{i}+\hat{j}+c\hat{k}$.

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 $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$.

Vector equation of the required line is
$$\begin{gathered} \overrightarrow{r}=\left(b\hat{i}+0\hat{j}+d\hat{k}\right)+\lambda \left(a\hat{i}+\hat{j}+c\hat{k}\right) \\ \mathrm{Here},\lambda \mathrm{is}\mathrm{a}\mathrm{parameter}. \end{gathered}$$