Q99 / 150 · MHT CET 2023 (09 May Shift 1) · Mathematics

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

  • Correct answer
    A
    $\overline{\mathrm{r}}=\left(\frac{1}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
  • B
    $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
  • C
    $\overline{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
  • D
    $\overline{\mathrm{r}}=\left(\frac{1}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$

Solution

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

Hence, the correct answer is option: 'A'.

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