Line and Plane — Maths STD 12 Science — Question

$\begin{aligned} & 4 x=3 z-5=3\left(z-\frac{5}{3}\right) \\ & \therefore \frac{4 x}{12}=\frac{3\left(z-\frac{5}{3}\right)}{12} \\ & \therefore \frac{x}{3}=\frac{z-\frac{5}{3}}{4}\end{aligned}$

$\therefore$ the cartesian equations of the line are

$\frac{x}{3}=\frac{z-\frac{5}{3}}{4}, y=2$

This line passes through the point $\mathrm{A}\left(0,2, \frac{5}{3}\right)$ position vector is $\bar{a}=2 \hat{j}+\frac{5}{3} \hat{k}$

Also the line has direction ratio 3, 0, 4.

If $\bar{b}$ is a vector parallel to the line, then $\bar{b}=3 \hat{i}+4 \hat{k}$

The vector equation of the line passing through $\mathrm{A}(\bar{a})$ and parallel to $b$ is $\bar{r}=\bar{a}+\lambda b$

where $\lambda$ is $\bar{a}$ scalar,

∴ the vector equation of the required line is

$\bar{r}=\left(2 \hat{j}+\frac{5}{3} \hat{k}\right)+\lambda(3 \hat{i}+4 \hat{k})$