The Cartesian equation of a line is $2 x-2=3 y+1=6 z-2$, then the vector equation of the line is

The Cartesian equation of a line is $2 x-2=3 y+1=6 z-2$, then the vector equation of the line is
  1. $\bar{r}=\left(\hat{i}-\frac{\hat{j}}{3}+\frac{\hat{k}}{3}\right)+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})$
  2. $\overline{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{3}\right)+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)$
  3. $\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$
  4. $\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)$

Solution

Given Cartesian equation of the line is $\begin{aligned} & 2 x-2=3 y+1=6 z-2 \\ & \Rightarrow 2(x-1)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{3}\right) \\ & \Rightarrow \frac{x-1}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{3}}{\frac{1}{6}} \\ & \Rightarrow \frac{x-1}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{3}}{1} \end{aligned}$ $\therefore \quad$ The given line passes through $\left(1,-\frac{1}{3}, \frac{1}{3}\right)$ and has direction ratios proportional to $3,2,1$. $\therefore \quad$ Vector equation is $\overline{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{\hat{\mathrm{j}}}{3}+\frac{\hat{\mathrm{k}}}{3}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

Asked in: MHT CET 2024 (04 May Shift 2)

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