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प्रश्न
Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.
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उत्तर
` \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{ 5 }\hat{i }+ \text{ 2 }\hat{j } - \text{ 4 }\hat{k } \text{ and }\vec{n} = \text{ 2 }\hat{i } + \text{ 3 }\hat{j } - \hat{k } (\text{ because the direction ratios of } \vec{n} \text{ are 2, 3, -1)}, \text{ we get } `
` \vec{r} . (\text{ 2 }\hat{i} + \text{ 3 }\hat{ j } - \hat{ k})= \text{ 5 }\hat{ i }+ \text{ 2 }\hat{ j } - \text{ 4 }\hat{ k } \text{ and }\vec{n}. ( 2 \hat{i }+3 \hat{j }- \hat{k } )`
` ⇒ \vec{r} . (\text{ 2 }\hat{ i } + \text{ 3 }\hat{ j } - \hat{ k}) = 10 + 6 + 4 `
` ⇒ \vec{r} . (\text{ 2 }\hat{ i } + \text{ 3 }\hat{ j } - \hat{ k}) = 20 `
` \text { For Cartesian form, we need to substitute } \vec{r} = \vec{r} . (\text{ x }\hat{ i } + \text{ y }\hat{ j } - \text{ z }\hat{ k}) \text{ in this equation. Then, we get } `
` (\text{ x }\hat{i } + \text{ y }\hat{j } + \text{ z }\hat{k }) .(\text{ 2 }\hat{i } + \text{ 3 }\hat{ j } - \hat{k})= 20 `
\[ \Rightarrow 2x + 3y - z = 20\]
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