A vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and…

A vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})=2$ is
  1. $-2 \hat{i}+7 \hat{j}+13 \hat{k}$
  2. $2 \hat{i}-7 \hat{j}+13 \hat{k}$
  3. $-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
  4. $\hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

Solution

The line of intersection of the planes $\overline{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=1$ and $\overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})=2$ is perpendicular to each of the normal vectors $\overline{n_1}=3 \hat{i}-\hat{j}+\hat{k}$ and $\overline{n_2}=\hat{i}+4 \hat{j}-2 \hat{k}$. $\therefore \quad$ The line is parallel to the vector $\overline{\mathrm{n}}_1 \times \overline{\mathrm{n}}_2$ $\begin{aligned} \therefore \quad \overline{\mathrm{n}}_1 \times \overline{\mathrm{n}}_2 & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{array}\right| \\ & =-2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+13 \hat{\mathrm{k}} \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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