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Find the Vector Equation of the Line Through the Origin Which is Perpendicular to the Plane → R ⋅ ( ^ I + 2 ^ J + 3 ^ K ) = 3 . - Mathematics

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प्रश्न

Find the vector equation of the line through the origin which is perpendicular to the plane  \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j}  + 3 \hat{k}  \right) = 3 .\]

 
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उत्तर

\[\text{ The required line is perpendicular to the plane } \vec{r} . \left( \hat{i}  - 2 \hat{j}  + 3 \hat{k}  \right) = 3 . \]

\[\text{ Therefore, it is parallel to the normal }  \hat{i}  + 2 \hat{j}  + 3 \hat{k}  . \]

\[\text{ Thus, the required line passes through the point with position vector } \vec{a} = 0 \hat{i}  + 0 \hat{j} + 0 \hat{k}  \text{ and is parallel to the vector }  \vec{n} = \hat{i}  -2 \hat{j}  + 3 \hat{k}  . \]

\[\text{ So, its vector equation is } \]

\[ r^\to = \vec{a} + \lambda \hat{n}  \]

\[ \Rightarrow \vec{r} = 0 \hat{i}  + 0 \hat{j}  + 0 \hat{k}  + \lambda \left( \hat{i}  - 2 \hat{j}  + 3 \hat{k}  \right)\]

\[ \Rightarrow \vec{r} = \lambda \left( \hat{i}  - 2 \hat{j} + 3 \hat{k}  \right)\]

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अध्याय 29: The Plane - Exercise 29.11 [पृष्ठ ६१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Exercise 29.11 | Q 6 | पृष्ठ ६१

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