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प्रश्न
Find the vector equation of the plane passing through three points with position vectors \[\hat{i} + \hat{j} - 2 \hat{k} , 2 \hat{i} - \hat{j} + \hat{k} \text{ and } \hat{i} + 2 \hat{j} + \hat{k} .\] Also, find the coordinates of the point of intersection of this plane and the line \[\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + \lambda\left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right) .\]
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उत्तर
\[\text{ Let A(1, 1 , -2),B(2, -1, 1) and C(1, 2, 1) be the points represented by the given position vectors } .\]
\[\text{ The required plane passes through the point A (1, 1, -1) whose position vector is } \vec{a} = \hat{i} + \hat{j} - 2 \hat{k} \text{ and is normal to the vector } \vec{n} \text{ given by } \]
\[ \vec{n} = \vec{AB} \times \vec{AC .} \]
\[\text{ Clearly } , \vec{AB} = \vec{OB} - \vec{OA} = \left( 2 \hat{i} - \hat{j} + \hat{k} \right) - \left( \hat{i} + \hat{j} - 2 \hat{k} \right) = \hat{i} - 2 \hat{j} + 3 \hat{k} \]
\[ \vec{AC} = \vec{OC} - \vec{OA} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) - \left( \hat{i} + \hat{j} -2 \hat{k} \right) = 0 \hat{i} + \hat{j} + 3 \hat{k} \]
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 0 & 1 & 3\end{vmatrix} = - 9 \hat{i} - 3 \hat{j} + \hat{k} \]
\[\text{ The vector equation of the required plane is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \Rightarrow \vec{r} . \left( - 9 \hat{i} - 3 \hat{j} + \hat{k} \right) = \left( \hat{i} + \hat{j} -2 \hat{k} \right) . \left( - 9 \hat{i} - 3 \hat{j} + \hat{k} \right)\]
\[ \Rightarrow \vec{r} . \left( - 9 \hat{i} - 3 \hat{j} + \hat{k} \right) = - 9 - 3 - 2\]
\[ \Rightarrow \vec{r} . \left[ - \left( 9 \hat{i} + 3 \hat{j} - \hat{k} \right) \right] = - 14\]
\[ \Rightarrow \vec{r} . \left( 9 \hat{i} + 3 \hat{j} - \hat{k} \right) = 14\]
\[\text{ To find the point of intersection of this plane } \]
\[\text{ The given equation of the line is } \]
\[ \vec{r} = \left( 3 \hat{i} - \hat{j} - \hat{k} \right) + \lambda \left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right)\]
\[ \Rightarrow \vec{r} = \left( 3 + 2\lambda \right) \hat{i} + \left( - 1 - 2\lambda \right) \hat{j} + \left( - 1 + \lambda \right) \hat{k} \]
\[\text{ The coordinates of any point on this line are in the form of } \left( 3 + 2\lambda \right) \hat{i} + \left( - 1 - 2\lambda \right) \hat{j} + \left( - 1 + \lambda \right) \hat{k} \text{ or } \left( 3 + 2\lambda, - 1 - 2\lambda, - 1 + \lambda \right)\]
\[\text{ Since this point lies on the plane } \vec{r} .\left( 9 \hat{i} + 3 \hat{j} - \hat{k} \right) = 14, \]
\[\left[ \left( 3 + 2\lambda \right) \hat{i} + \left( - 1 - 2\lambda \right) \hat{j} + \left( - 1 + \lambda \right) \hat{k} \right] . \left( 9 \hat{i} + 3 \hat{j} - \hat{k} \right) = 14\]
\[ \Rightarrow 27 + 18\lambda - 3 - 6\lambda + 1 - \lambda = 14\]
\[ \Rightarrow 11\lambda = - 11\]
\[ \Rightarrow \lambda = - 1\]
\[\text{ So, the coordinates of the point are } \]
\[\left( 3 + 2\lambda, - 1 - 2\lambda, - 1 + \lambda \right)\]
\[ = \left( 3 - 2, - 1 + 2, - 1 - 1 \right)\]
\[ = \left( 1, 1, - 2 \right)\]
