The vector equation of the line $2 x+4=3 y+1=6 z-3$ is

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

Solution

The equation of line is $\begin{aligned} & 2 x+4=3 y+1=6 z-3 \\ & \Rightarrow 2(x+2)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{2}\right) \\ & \Rightarrow \frac{x+2}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{2}}{\frac{1}{6}} \\ & \Rightarrow \frac{x+2}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{2}}{1} \end{aligned}$ $\therefore \quad$ The given line passes through $\left(-2,-\frac{1}{3}, \frac{1}{2}\right)$ and has direction ratios proportional to $3,2,1$. $\therefore \quad$ Vector equation of the line is $\overline{\mathrm{r}}=\left(-2 \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

Asked in: MHT CET 2023 (11 May Shift 1)

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