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Write the equation of the plane passing through the origin and parallel to the plane 5x - 3y + 7z + 11 = 0. |
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Answer» Let the equation of plane be A1x + B1y + C1z + D1 = 0 Direction ratios of parallel planes are related to each other as \(\frac{A_1}{A_2}\) = \(\frac{B_1}{B_2}\) = \(\frac{C_1}{C_2}\) = k (constant) Putting the values from the equation of a given parallel plane, \(\frac{A_1}{5}\) = \(\frac{B_1}{-3}\) = \(\frac{C_1}{7}\) = k A1 = 5k, B1 = - 3k, C1 = 7k Putting in equation plane 5kx - 3ky + 7kz + D1 = 0 As the plane is passing through (0,0,0), it must satisfy the plane, 5k.0 - 3k.0 + 7k.0 + D1 = 0 D1 = 0 5kx - 3ky + 7kz = 0 5x - 3y + 7z = 0 So, required equation of plane is 5x - 3y + 7z = 0. |
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