Question

Find the vector equation of the plane passing through the intersection of the planes

\[\vec{r} \cdot \left( \hat{ i } + \hat{ j }+ \hat{ k }\right) = \text{ 6 and }\vec{r} \cdot \left( \text{ 2  } \hat{ i} +\text{  3 } \hat{  j } + \text{ 4 } \hat{ k } \right) = - 5\] and the point (1, 1, 1).

Answer 1

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

` \vec{r} . \left(\hat{  i } + \hat{  j } +\hat{  k } \right) - 6 + \lambda \left( \vec{r} . \left(   2  \hat{  i } + 3 \hat{  j } + 4 \hat{  k} \right) + 5 \right) = 0 `

` \vec{r} . \left[ \left( 1 + 2\lambda \right) \hat{  i } + \left( 1 + 3\lambda \right) \hat{  j } + \left( 1 + 4\lambda \right) \hat{  k } \right] - 6 + 5\lambda = 0 . . . \left( 1 \right)` 
`  \text{ This passes through }\hat{  i } +\hat{  j } + \hat{  k } . \text{ So  },`

`( \hat{  i }+\hat{  j } +\hat{  k } \right) . \left[ \left( 1 + 2\lambda \right) \hat{  i } + \left( 1 + 3\lambda \right)\hat{  j } + \left( 1 + 4\lambda \right) \hat{  k }\right] - 6 + 5\lambda = 0`

\[ \Rightarrow 1 + 2\lambda + 1 + 3\lambda + 1 + 4\lambda - 6 + 5\lambda = 0\]
\[ \Rightarrow 14\lambda - 3 = 0\]
\[ \Rightarrow \lambda = \frac{3}{14}\]
\[\text{ Substituting this in (1), we get }\]
` \vec{r} . \left[ \left( 1 + 2 \left( \frac{3}{14} \right) \right) \hat{  i } + \left( 1 + 3 \left( \frac{3}{14} \right) \right)\hat{  j } + \left( 1 + 4 \left( \frac{3}{14} \right) \right)\hat{   k } \right] - 6 + 5 \left( \frac{3}{14} \right) = 0`

 

` \Rightarrow \vec{r} . \left( 20\hat{  i } + 23\hat{  j } + 26 \hat{  k }\right) = 69 `