Question

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

 

Answer 1

\[ \text{ The equation of the plane passing through the line of intersection of the given planes is } \]

\[ \vec{r} . \left( \hat{i}  + 3 \hat{j}  \right) + 6 + \lambda \left( \vec{r} . \left( 3 \hat{i}  - \hat{j}  - 4 \hat{k}  \right) \right) = 0 \]

\[ \vec{r} . \left[ \left( 1 + 3\lambda \right) \hat{i}  + \left( 3 - \lambda \right) \hat{j}  - 4\lambda \hat{k} \right] + 6 = 0 . . . \left( 1 \right)\]

\[ \vec{r} . \left[ \left( 1 + 3\lambda \right) \hat{i} + \left( 3 - \lambda \right) \hat{j} - 4\lambda \hat{k} \right] = - 6\]

\[ \vec{r} . \left[ \left( - 1 - 3\lambda \right) \hat{i}  + \left( \lambda - 3 \right) \hat{j} + 4\lambda \hat{k}  \right] = 6\]

\[ \text{ Dividing both sides by} \sqrt{\left( - 1 - 3\lambda \right)^2 + \left( \lambda - 3 \right)^2 + 16 \lambda^2}, \text{ we get } \]

\[ \vec{r} . \frac{\left[ \left( - 1 - 3\lambda \right) \hat{i} + \left( \lambda - 3 \right) j + 4\lambda \hat{k}  \right]}{\sqrt{\left( - 1 - 3\lambda \right)^2 + \left( \lambda - 3 \right)^2 + 16 \lambda^2}} = \frac{6}{\sqrt{\left( - 1 - 3\lambda \right)^2 + \left( \lambda - 3 \right)^2 + 16 \lambda^2}}, \text{ which is the normal form of plane (1), where } \]

\[ \text { the perpendicular distance of plane (1) from the origin } =\frac{6}{\sqrt{\left( - 1 - 3\lambda \right)^2 + \left( \lambda - 3 \right)^2 + 16 \lambda^2}}\]

\[ \Rightarrow 1 = \frac{6}{\sqrt{\left( - 1 - 3\lambda \right)^2 + \left( \lambda - 3 \right)^2 + 16 \lambda^2}} (\text{ Given } )\]

\[ \Rightarrow \sqrt{\left( - 1 - 3\lambda \right)^2 + \left( \lambda - 3 \right)^2 + 16 \lambda^2} = 6\]

\[ \Rightarrow 1 + 9 \lambda^2 + 6\lambda + \lambda^2 + 9 - 6\lambda + 16 \lambda^2 = 36\]

\[ \Rightarrow 26 \lambda^2 - 26 = 0\]

\[ \Rightarrow \lambda^2 = 1\]

\[ \Rightarrow \lambda = 1 , - 1\]

\[\text{ Case 1: Substituting} \lambda = 1 \text{ in (1), we get } \]

\[ \vec{r} . \left[ 4 \hat{i}  + 2 \hat{j}  - 4 \hat{k}  \right] + 6 = 0\]

\[ \text{ Case 2: Substituting } \lambda = - \text{ 1 in (1), we get } \]

\[ \vec{r} . \left[ - 2 \hat{i}  + 4 \hat{j}  + 4 \hat{k} \right] + 6 = 0\]