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Question
Determine the value of λ for which the following planes are perpendicular to each other.
3x − 6y − 2z = 7 and 2x + y − λz = 5
Solution
` \text{ We know that the planes } a_1 x + b_1 y + c_1 z + d_1 = 0 \text{ and } a_2 x + b_2 y + c_2 z + d_2 = 0 \text{ are perperndicular to each other only if} `
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0\]
\[\text{ The given planes are } 3x - 6y - 2z = 7 \text{ and }2x + y - \lambda z = 5 . \]
\[ \Rightarrow a_1 = 3; b_1 = - 6; c_1 = - 2; a_2 = 2; b_2 = 1; c_2 = - \lambda\]
\[\text{ The given planes are perpendicular} .\]
\[ \Rightarrow a_1 a_2 + b_1 b_2 + c_1 c_2 = 0\]
\[ \Rightarrow \left( 3 \right) \left( 2 \right) + \left( - 6 \right) \left( 1 \right) + \left( - 2 \right) \left( - \lambda \right) = 0\]
\[ \Rightarrow 6 - 6 + 2\lambda = 0\]
\[ \Rightarrow 2\lambda = 0\]
\[ \Rightarrow \lambda = 0\]
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