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प्रश्न
Show that the following planes are at right angles.
x − 2y + 4z = 10 and 18x + 17y + 4z = 49
उत्तर
`\text{ We know that the planes} a_1 x + b_1 y + c_1 z + d_1 = 0 and a_2 x + b_2 y + c_2 z + d_2 = 0 \text{ are perperndicular to each other only if }`
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 . \]
\[\text{ The given planes are } x - 2y + 4z = 10 \text{ and } 18x + 17y + 4z = 49 . \]
\[ \Rightarrow a_1 = 1; b_1 = - 2; c_1 = 4; a_2 = 18; b_2 = 17; c_2 = 4\]
\[\text{ Now} ,\]
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = \left( 1 \right) \left( 18 \right) + \left( - 2 \right) \left( 17 \right) + \left( 4 \right) \left( 4 \right) = 18 - 34 + 16 = 0\]
\[\text{ So, the given planes are perpendicular} .\]
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