Question

Find the equation of the plane through the line of intersection of the planes $\overrightarrow{r}·\left(\hat{i}+3\hat{j}\right)+6=0\mathrm{and}\overrightarrow{r}·\left(3\hat{i}-\hat{j}-4\hat{k}\right)=0,$ which is at a unit distance from the origin.

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ \overrightarrow{r}.\left(\hat{i}+3\hat{j}\right)+6+\lambda \left(\overrightarrow{r}.\left(3\hat{i}-\hat{j}-4\hat{k}\right)\right)=0 \\ \overrightarrow{r}.\left[\left(1+3\lambda \right)\hat{i}+\left(3-\lambda \right)\hat{j}-4\lambda \hat{k}\right]+6=0...\left(1\right) \\ \overrightarrow{r}.\left[\left(1+3\lambda \right)\hat{i}+\left(3-\lambda \right)\hat{j}-4\lambda \hat{k}\right]=-6 \\ \overrightarrow{r}.\left[\left(-1-3\lambda \right)\hat{i}+\left(\lambda -3\right)\hat{j}+4\lambda \hat{k}\right]=6 \\ \text{Dividing both sides by}\sqrt{{\left(-1-3\lambda \right)}^2+{\left(\lambda -3\right)}^2+16{\lambda }^2}\text{, we get} \\ \overrightarrow{r}.\frac{\left[\left(-1-3\lambda \right)\hat{i}+\left(\lambda -3\right)\hat{j}+4\lambda \hat{k}\right]}{\sqrt{{\left(-1-3\lambda \right)}^2+{\left(\lambda -3\right)}^2+16{\lambda }^2}}=\frac{6}{\sqrt{{\left(-1-3\lambda \right)}^2+{\left(\lambda -3\right)}^2+16{\lambda }^2}},\text{which is the normal form of plane (1), where} \\ \text{the perpendicular distance of plane (1) from the origin =}\frac{6}{\sqrt{{\left(-1-3\lambda \right)}^2+{\left(\lambda -3\right)}^2+16{\lambda }^2}} \\ \Rightarrow 1=\frac{6}{\sqrt{{\left(-1-3\lambda \right)}^2+{\left(\lambda -3\right)}^2+16{\lambda }^2}}\text{(Given)} \\ \Rightarrow \sqrt{{\left(-1-3\lambda \right)}^2+{\left(\lambda -3\right)}^2+16{\lambda }^2}=6 \\ \Rightarrow 1+9{\lambda }^2+6\lambda +{\lambda }^2+9-6\lambda +16{\lambda }^2=36 \\ \Rightarrow 26{\lambda }^2-26=0 \\ \Rightarrow {\lambda }^2=1 \\ \Rightarrow \lambda =1,-1 \\ \text{Case 1: Substituting}\lambda =1\text{in (1), we get} \\ \overrightarrow{r}.\left[4\hat{i}+2\hat{j}-4\hat{k}\right]+6=0 \\ \text{Case 2: Substituting}\lambda =-1\text{in (1), we get} \\ \overrightarrow{r}.\left[-2\hat{i}+4\hat{j}+4\hat{k}\right]+6=0 \end{gathered}$$