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प्रश्न
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.
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उत्तर
`\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 `
\[ \Rightarrow x + 2y - 3z = 14\]
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