Question

If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP.

Answer 1

`\text{  The normal is passing through the points O (0, 0, 0) and P (1, 2, -3). So, } `

`  \vec{n} = \vec{OP} =( \hat{i } + 2  \hat{j } - 3 \hat{k } ) - ( 0 \hat{i }+ 0 \hat{j }+ 0 \hat{k }) = \hat{i }+ 2 \hat{j }- 3  \hat{k }`

`  \text{ Since the plane passes through P(1, 2, -3),} \vec{a} = \hat{i  }+ 2 \hat{j }- 3 \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} =2   \hat{i }+ 3   \hat{j } - \hat{k } \text{ and }\vec{n} = \hat{i }+ 2 \hat{j }- 3 \hat{k }\text{  in the relation, we get }` 

`  \vec{r} . ( \hat{i }+ 2    \hat{j }- 3    \hat{k }) = ( \hat{i }+ 2   \hat{j  } - 3  \hat{k }) . ( \hat{i }+ 2   \hat{j } - 3   \hat{k }) `

` ⇒ \vec{r} .  (\hat{i }+ 2    \hat{j }- 3    \hat{k })= 1 + 4 + 9    `

` ⇒ \vec{r} .  (\hat{i }+ 2    \hat{j }- 3    \hat{k })= 14   `

` ⇒ \vec{r} .  (\hat{i }+ 2    \hat{j }- 3    \hat{k })= 14   `

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

`( x  \hat{i  }+ y   \hat{j } + z   \hat{k } ) . ( \hat{i } + 2    \hat{j } - 3   \hat{k } ) = 14 `

\[ \Rightarrow x + 2y - 3z = 14\]

\[\]

\[\]