The equation of line passing through the point $(1,2,3)$ and perpendicular to the lines…

The equation of line passing through the point $(1,2,3)$ and perpendicular to the lines $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}$ and $\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$ is
  1. $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$
  2. $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$
  3. $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$
  4. $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$

Solution

Required line is perpendicular to the lines $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}$ and $\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$ $\therefore \quad$ Required line is parallel to vector $\overline{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 2 & -2 \\ 2 & -3 & 1 \end{array}\right|=-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}}$ $\therefore \quad$ The equation of the required line is $(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-4 \hat{i}-7 \hat{j}-13 \hat{k})$

Asked in: MHT CET 2023 (11 May Shift 2)

Practice more Line and Plane questions on Aicharya