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Find-vector-equation-plane-passing-through-points-3-4-2-7-0-6-perpendicular-plane-2x-5y-15-0 - Mathematics

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प्रश्न

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j}  - 2 \hat{k}  + \lambda\left( \hat{i}  - \hat{j}  + \hat{k}  \right) .\]

 

उत्तर

\[\text{ The equation of any plane passing through (3, 4, 2) is } \]

\[a \left( x - 3 \right) + b \left( y - 4 \right) + c \left( z - 2 \right) = 0 . . . \left( 1 \right)\]

\[\text{ It is given that (1) is passing through (7, 0, 6). So, } \]

\[a \left( 7 - 3 \right) + b \left( 0 - 4 \right) + c \left( 6 - 2 \right) = 0 \]

\[ \Rightarrow 4a - 4b + 4c = 0\]

\[ \Rightarrow a - b + c = 0 . . . \left( 2 \right)\]

\[\text{ It is given that (1) is perpendicular to the plane 2x - 5y + 0z + 15z = 0 . So, } \]

\[2a - 5b + 0c = 0 . . . \left( 3 \right)\]

\[\text{ Solving (1), (2) and (3), we get } \]

\[\begin{vmatrix}x - 3 & y - 4 & z - 2 \\ 1 & - 1 & 1 \\ 2 & - 5 & 0\end{vmatrix} = 0\]

\[ \Rightarrow 5 \left( x - 3 \right) + 2 \left( y - 4 \right) - 3 \left( z - 2 \right) = 0\]

\[ \Rightarrow 5x + 2y - 3z = 17\]

\[\text{ Or }  \vec{r} .\left( 5 \hat{i}  + 2 \hat{j}  - 3 \hat{k}  \right)= 17\]

\[\text{ Showing that the plane contains the line } \]

\[\text{ The line }  \vec{r} = \left( \hat{i} + 3 \hat{j} - 2 \hat{k}  \right) + \lambda \left( \hat{i} - \hat{j}  + \hat{k}  \right) \text{ passes through a point whose positon vector is } \vec{a} = \hat{i}  + 3 \hat{j} - 2 \hat{k}  \text{ and is parallel to the vector } \vec{b} = \hat{i} - \hat{j}  + \hat{k} . \]

\[\text{ If the plane }  \vec{r} .\left( 5 \hat{i}  + 2 \hat{j}  - 3 \hat{k}  \right)=17 \text{ contains the given line, then } \]

\[(1) \text{ it should pass through the point } \hat{i}  + 3 \hat{j} - 2 \hat{k}  \]

\[(2) \text{ it should be parallel to the line } \]

\[\text{ Now } ,\left( \hat{i} + 3 \hat{j}  - 2 \hat{k}  \right).\left( 5 \hat{i}  + 2 \hat{j}  - 3 \hat{k}  \right)= 5 + 6 + 6 = 17\]

\[\text{ So, the plane passes through the point } \hat{i}+ 3 \hat{j} - 2 \hat{k}  . \]

\[\text{ The normal vector to the given plane is } \vec{n} = \hat{i}  - \hat{j}  + \hat{k .} \]

\[\text{ We observe that } \]

\[ \vec{b} . \vec{n} = \left( \hat{i}  - \hat{j} + \hat{k}  \right) . \left( 5 \hat{i} + 2 \hat{j}  - 3 \hat{k}  \right) = 5 - 2 - 3 = 0\]

\[\text{ Therefore, the plane is parallel to the line. } \]

\[\text{ Hence, the given plane contains the given line } .\]

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पाठ 29: The Plane - Exercise 29.13 [पृष्ठ ७४]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 29 The Plane
Exercise 29.13 | Q 8 | पृष्ठ ७४

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