Question

Find the equation of the plane that contains the line of intersection of the planes  \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j}  + 3 \hat{k}  \right) - 4 = 0 \text{ and }  \vec{r} \cdot \left( 2 \hat{i}  + \hat{j} - \hat{k}  \right) + 5 = 0\] and which is perpendicular  to the plane \[\vec{r} \cdot \left( 5 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right) + 8 = 0 .\]

  

Answer 1

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

\[ \vec{r} . \left( \hat{i}  + 2 \hat{j}  + 3 \hat{k}  \right) - 4 + \lambda \left[ \vec{r} . \left( 2 \hat{i} + \hat{j}  - \hat{k}  \right) + 5 \right] = 0 \]

\[ \vec{r} . \left[ \left( 1 + 2\lambda \right) \hat{i}  + \left( 2 + \lambda \right) \hat{j}  + \left( 3 - \lambda \right) \hat{ k} \right] - 4 + 5\lambda = 0 . . . \left( 1 \right)\]

\[\text{ This plane is perpendicular to } \vec{r} . \left( 5 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right) + 8 = 0 . \text{ So} ,\]

\[5 \left( 1 + 2\lambda \right) + 3 \left( 2 + \lambda \right) - 6 \left( 3 - \lambda \right) = \text{ 0 }(\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]

\[ \Rightarrow 5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0\]

\[ \Rightarrow 19\lambda - 7 = 0\]

\[ \Rightarrow \lambda = \frac{7}{19}\]

\[\text{  Substituting this in (1), we get} \]

\[ \vec{r} . \left[ \left( 1 + 2 \left( \frac{7}{19} \right) \right) \hat{i}  + \left( 2 + \frac{7}{19} \right) \hat{j}  + \left( 3 - \frac{7}{19} \right) \hat{k}  \right] - 4 + 5 \left( \frac{7}{19} \right) = 0\]

\[ \Rightarrow \vec{r} . \left( 33 \hat{i}  + 45 \hat{j}  + 50 \hat{k}  \right) - 41 = 0\]

\[ \Rightarrow \left( x \hat{i}  + y \hat{j}  + z \hat{k}  \right) . \left( 33 \hat{i}  + 45 \hat{j}  + 50 \hat{k } \right) - 41 = 0\]

\[ \Rightarrow 33x + 45y + 50z - 41 = 0\]