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Find the Vector Equation of the Plane Passing Through the Points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3). - Mathematics

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प्रश्न

Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).

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उत्तर

\[\text{ The required plane passes through the point} P(2, 5, -3) \text{ whose position vector is } \vec{a} =2 \hat{i} +5 \hat{j}  -3 \hat{k}  \text{ and is normal to the vector} \vec{n} \text{ given by } \]
\[ \vec{n} = \vec{PQ} ×  \vec{PR .} \]
\[ \text{ Clearly, } \vec{PQ} = \vec{OQ} - \vec{OP} = \left( - 2 \hat{i}  - 3 \hat{j}  + 5 \hat{k}  \right) - \left( 2 \hat{i}  + 5 \hat{j} - 3 \hat{k} \right) = - 4 \hat{i} - 8 \hat{j} + 8 \hat{k} \]
\[ \vec{PR} = \vec{OR} - \vec{OP} = \left( 5 \hat{i} + 3 \hat{j} - 3 \hat{k} \right) - \left( 2 \hat{i} + 5 \hat{j} - 3 \hat{k} \right) = 3 \hat{i} - 2 \hat{j}  - 0 \hat{k}  \]
\[ n^\to = \vec{PQ} \times \vec{PR} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 4 & - 8 & 8 \\ 3 & - 2 & 0\end{vmatrix} = 16 \hat{i} + 24 \hat{j} + 32 \hat{k} \]
\[\text{ The vector equation of the required plane is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \Rightarrow \vec{r} . \left( 16 \hat{i} + 24 \hat{j} + 32 \hat{k} \right) = \left( 2 \hat{i} +5 \hat{j} -3 \hat{k}  \right) . \left( 16 \hat{i} + 24 \hat{j} + 32 \hat{k} \right)\]
\[ \Rightarrow \vec{r} . \left[ 8 \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k}  \right) \right] = 32 + 120 - 96\]
\[ \Rightarrow \vec{r} . \left[ 8 \left( 2 \hat{i}  + 3 \hat{j} + 4 \hat{k}  \right) \right] = 56\]
\[ \Rightarrow \vec{r} . \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 7\]

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अध्याय 29: The Plane - Exercise 29.05 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Exercise 29.05 | Q 2 | पृष्ठ २३

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