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प्रश्न
Find the equation of the plane through the point \[2 \hat{i} + \hat{j} - \hat{k} \] and passing through the line of intersection of the planes \[\vec{r} \cdot \left( \hat{i} + 3 \hat{j} - \hat{k} \right) = 0 \text{ and } \vec{r} \cdot \left( \hat{j} + 2 \hat{k} \right) = 0 .\]
उत्तर
\[\text{ The equation of the plane passing through the line of intersection of the given planes is } \]
\[ \vec{r} . \left( \hat{i} + 3 \hat{j} - \hat{k} \right) + \lambda \left( \vec{r} . \left( \hat{j} + 2 \hat{k} \right) \right) = 0 \]
\[ \vec{r} . \left[ \hat{i} + \left( 3 + \lambda \right) \hat{j} + \left( - 1 + 2\lambda \right) \hat{k} \right] = 0 . . . \left( 1 \right)\]
\[\text { This passes through } 2 \hat{i} + \hat{j} - \hat{k} . \text{ So } ,\]
\[\left( 2 \hat{i} + \hat{j} - \hat{k} \right) \left[ \hat{i} + \left( 3 + \lambda \right) \hat{j} + \left( - 1 + 2\lambda \right) \hat{k}\right] = 0\]
\[ \Rightarrow 2 + 3 + \lambda + 1 - 2\lambda = 0\]
\[ \Rightarrow \lambda = 6\]
\[\text{ Substituting this in (1), we get } \]
\[ \vec{r} . \left[ \hat{i} + \left( 3 + 6 \right) \hat{j} + \left( - 1 + 12 \right) \hat{k} \right] = 0\]
\[ \Rightarrow \vec{r} . \left( \hat{i} + 9 \hat{j} + 11 \hat{k} \right) = 0\]
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