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

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

Solution

We have $4 x-3 z+5=0$ and $y=2$ $\begin{aligned} & \therefore 4 x=3 z-5 \Rightarrow 4 x=3\left(z-\frac{5}{3}\right) \\ & \therefore \frac{4 x}{12}=\frac{3\left(z-\frac{5}{3}\right)}{12} \Rightarrow \frac{x}{3}=\frac{3\left(z-\frac{5}{3}\right)}{4} \end{aligned}$ Thus line passes through point $\left(0,2, \frac{5}{3}\right)$, and has direction ratios $3,0,4$. Hence required equation of line is $\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})$

Asked in: MHT CET 2021 (22 Sep Shift 2)

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