The vector equation of the line $\frac{x+3}{2}=\frac{2 y-3}{5} ; z=-1$ is

The vector equation of the line $\frac{x+3}{2}=\frac{2 y-3}{5} ; z=-1$ is
  1. $\bar{r}=\left(3 \hat{\imath}-\frac{3}{2} \hat{\jmath}-\hat{k}\right)+\lambda(4 \hat{\imath}+5 \hat{\jmath})$
  2. $\bar{r}=\left(-3 \hat{\imath}+\frac{3}{2} \hat{\jmath}-\hat{k}\right)+\lambda(4 \hat{\imath}+5 \hat{\jmath})$
  3. $\bar{r}=\left(-3 \hat{\imath}+\frac{3}{2} \hat{\jmath}+\hat{k}\right)+\lambda(4 \hat{\imath}+5 \hat{\jmath})$
  4. $\bar{r}=\left(3 \hat{\imath}+\frac{3}{2} \hat{\jmath}-\hat{k}\right)+\lambda\left(4 \hat{\imath}+\frac{5}{2} \hat{\jmath}\right)$

Solution

Equation of line is $\frac{x+3}{2}=\frac{2 y-3}{5} ; z=-1$ $\therefore \frac{x+3}{2}=\frac{2\left(y-\frac{3}{2}\right)}{5} ; z=-1 \quad \Rightarrow \frac{x+3}{2}=\frac{y-\frac{3}{2}}{\left(\frac{5}{2}\right)} ; z=-1$ This line passes through point $\left(-3, \frac{3}{2},-1\right)$ and d.r.s. are $2, \frac{5}{2}, 0$ i.e. $4,5,0$ Hence vector equation of given line is $\overline{\mathrm{r}}=\left(-3 \overline{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)+\lambda(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}})$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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