Question

If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.

Answer 1

`\text{ The normal is passing through the points} \text{ A }(4, -1, 2) and \text{  B }(-10, 5, 4). So,  `

`\vec{n} = \vec{AB} = \vec{OB} - \vec{OA} =( \text{ -10 }\hat{ i } +\text{ 5 } \hat{ j } + 4\hat{  k }) - ( 4\hat{ i }- \hat{  j }+ \text{ 2 } \hat{ j } ) = - \text{ 14} \hat{ j } + \text{ 6} \hat{ j } + \text{2 }\hat{ j } `

`\text{ Since the plane passes through } (-10, 5, 4), \vec{a} = \text{- 10} \hat{ i } +\text{ 5 } \hat{ j } + \text{ 4} \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} = -\text{ 10 } \hat{ i } +\text{ 5 } \hat{ j } + \text{ 4 } \hat{ k } and \vec{n} = -\text{ 14 }\hat{ i } +\text{ 6 } \hat{ j } +\text{ 2 } \hat{ k } ,\text{ we get }`

`   \vec{r} . ( - \text{ 14 }\hat{ i } + \text{ 6 } \hat{ j } + \text{ 2 } \hat{ k }  ) = ( -\text{ 10  } \hat{ i } +\text{ 5 } \hat{ j } +\text{ 4 } \hat{ k } ) . ( - \text{ 14 } \hat{ i } + \text{ 6 } \hat{ j } + \text{ 2 } \hat { k }) `

`  ⇒ \vec{r} .( \text{- 14 } \hat{ i }  + \text{ 6 } \hat{ j }  +\text{ 2 } \hat{ k }  ) = 140 + 30 + 8  `

`  ⇒ \vec{r} .  (2  ( \text{ 7 } \hat{ i }  + \text{ 3 } \hat{ j }  + \hat{ k })  ) = 178  `

`  ⇒ \vec{r} .    ( \text{ 7 } \hat{ i }  + \text{ 3 } \hat{ j }  + \hat{ k }  ) = 89`

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

` ( \text{ x } \hat{ i } + \text{ y } \hat{ i } + \text{ z}\hat{k } ). \( \text{ 7 } \hat{ i }  + \text{ 3 } \hat{ j }  + \hat{ k }  ) = 89`

\[ \Rightarrow 7x - 3y - z = - 89\]

\[ \Rightarrow 7x - 3y - z + 89 = 0\]

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