If cartesian equation of the line is $x-1=2 y+3=3-z$, then its vector equation is

If cartesian equation of the line is $x-1=2 y+3=3-z$, then its vector equation is
  1. $\bar{r}=(\hat{\imath}-3 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+\hat{\jmath}-2)$
  2. $\bar{r}=(-\hat{\imath}-3 \hat{\jmath}+3 \hat{k})+\lambda\left(\hat{\imath}+\frac{1}{2} \hat{\jmath}-\hat{k}\right)$
  3. $\bar{r}=\left(-\hat{\imath}+\frac{3}{2} \hat{\jmath}-3 \hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})$
  4. $\bar{r}=\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+3 \hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})$

Solution

Given Cartesian equation is $\begin{aligned} x-1 &=2 y+3=3-z \\ \therefore \frac{x-1}{1} &=\frac{2\left(y+\frac{3}{2}\right)}{1}=\frac{-(z-3)}{1} \Rightarrow \frac{x-1}{1}=\frac{y+\frac{3}{2}}{\left(\frac{1}{2}\right)}=\frac{z-3}{-1} \end{aligned}$ $\begin{array}{l} \therefore 1, \frac{1}{2},-1 \quad \text { i.e. } 2,1,-2 \text { are d.r. of given line. } \\ \therefore \quad \mathrm{A}\left(1,-\frac{3}{2}, 3\right) \text { lies on given line. } \\ \therefore \quad \text { vector equation is } \\ \qquad \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \end{array}$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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