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Find the equations of the plane passing through the origin and parallel to the plane 2x – 3y + 7z + 13 = 0. |
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Answer» Formula : Plane = r . (n) = d Where r = any random point n = normal vector of plane d = distance of plane from origin If two planes are parallel , then their normal vectors are same. Therefore , Parallel Plane 2x – 3y + 7z + 13 = 0 Normal vector = (2i - 3j + 7k) ∴ Normal vector of required plane = (2i - 3j + 7k) Equation of required plane r . (2i - 3j + 7k) = d In cartesian form 2x - 3y + 7z = d Plane passes through point (0,0,0) therefore it will satisfy it. 2(0) - (0) + 3(0) = d d = 0 Equation of required plane 2x - 3y + 7z = 0 |
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