The vector equation of a line whose Cartesian equations are $y=2,4 x-3 z+5=0$ is

The vector equation of a line whose Cartesian equations are $y=2,4 x-3 z+5=0$ is
  1. $\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \mathrm{k})+\lambda\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)$
  2. $\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \mathrm{k})+\lambda\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)$
  3. $\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \mathrm{k})$
  4. $\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \mathrm{k})$

Solution

Given cartesian equation of line is $\begin{aligned} & 4 x-3 z+5=0, y=2 \\ & \Rightarrow 4 x=3 z-5, y=2 \\ & \Rightarrow 4 x=3\left(z-\frac{5}{3}\right), y=2 \\ & \Rightarrow \frac{x}{3}=\frac{z-\frac{5}{3}}{4}, y=2 \end{aligned}$ $\therefore \quad$ The given line passes $\left(0,2, \frac{5}{3}\right)$ and the direction ratios are proportional to $3,0,4$. $\therefore \quad$ The vector equation is $\bar{r}=\left(2 \hat{j}+\frac{5}{3} \hat{k}\right)+\lambda(3 \hat{i}+4 \hat{k})$

Asked in: MHT CET 2024 (02 May Shift 1)

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