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Find the equation of the plane passing through the point (1, 4, - 2) and parallel to the plane 2x – y + 3z + 7 = 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 – y + 3z + 7 = 0 Normal vector = (2i - j + 3k) ∴ Normal vector of required plane = (2i - j + 3k) Equation of required plane r . (2i - j + 3k) = d In cartesian form 2x - y + 3z = d Plane passes through point (1,4, - 2) therefore it will satisfy it. 2(1) - (4) + 3( - 2) = d d = 2 – 4 – 6 = - 8 Equation of required plane 2x - y + 3z = - 8 2x - y + 3z + 8 = 0 |
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