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. $\overline{\mathrm{r}}=(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}-4 \hat{\mathrm{k}})$
  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. $\overline{\mathrm{r}}=(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})$
  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 lines $y-2=0$ and $4 x-3 z+5=0$ $\begin{aligned} & \therefore 4 x=3 z-5=3\left[z-\left(\frac{5}{3}\right)\right] \\ & \therefore \frac{4 x}{12}=\frac{3\left[z-\left(\frac{5}{3}\right)\right]}{12} \Rightarrow \frac{x}{3}=\frac{z-\left(\frac{5}{3}\right)}{4}, y=2 \end{aligned}$ Thus line passes through the point $\left(0,2, \frac{5}{3}\right)$ i.e. a point having position vector $2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}$ Also direction ratios of a line are $3,0,4$ Hence required vector equation is $\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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