Question

Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.

Answer 1

The normal is passing through the points A (-1, 2, 3) and B (3, -5, 6). So,

` \vec{n} = \vec{AB} = \vec{OB} - \vec{OA} =( \text{ 3 } \hat{ i } -\text{ 5 } \hat{ j  } + \text{ 6 } \hat{ k } ) - \(  - \hat{ i } + \text{ 2 } \hat{ j } + 3 { k} ) = \text{ 4 } \hat{ i } - \text{ 7 } \hat{ j } + \text{ 3 } \hat{ k } ` 

` \text{ Mid-point of  AB}=( \frac{- 1 + 3}{2}, \frac{2 - 5}{2}, \frac{3 + 6}{2} )=( 1, \frac{- 3}{2}, \frac{9}{2} `

`  \text{ Since the plane passes through}( 1, \frac{- 3}{2}, \frac{9}{2} ), \vec{a} =\hat{ i } - \frac{3}{2}\hat{ j } + \frac{9}{2} \hat{ k } `

`  \text{ We know that the vector equation of the plane passing through a point   } \vec{a} \text{ and normal to} \vec{n} \text{ is}  `

\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \] `     

` \text { Substituting }\vec{a} = \hat{ i } - \hat{ j } + \hat{ k }\text{ and }\vec{n} =\text{ 4 } \hat{ i}- \text{ 7 } \hat{ j } +\text{ 3 } \hat{ k } ,\text{  we get } `

` \vec{r} . ( =\text{ 4 } \hat{ i}- \text{ 7 } \hat{ j } +\text{ 3 } \hat{ k }) = (\hat{ i} - \frac{3}{2} \hat{ j } + \frac{9}{2}\hat{ k } ) . (\text{ 4 } \hat{ i}- \text{ 7 } + \text{ 3 } \hat{ k }) `

`  ⇒ \vec{r} . ( =\text{ 4 } \hat{ i}- \text{ 7 } \hat{ j } +\text{ 3 } \hat{ k }) = 28  `

\[\]

 `  \text{ Substituting }\vec{r} = \text{ x } \hat{ i} +\text{ y } \hat{ j }+\text{ z} \hat{ k}\text{  in the vector equation, we get } `

` (\text{ x } \hat{ i} +\text{ y } \hat{ j }+\text{ z} \hat{ k} ). (\text{ 4 } \hat{ i}- \text{ 7 } \hat{ j } +\text{ 3 } \hat{ k }) = 28  `

\[ \Rightarrow 4x - 7y + 3z = 28\]

\[ \Rightarrow 4x - 7y + 3x - 28 = 0\]

\[\]

\[\]