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Find the equation of the plane passing through the points A(1, 1, 2) and B(2, -2, 2) and perpendicular to the plane 6x – 2y + 2z = 9. |
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Answer» Plane passes through (1,1,2) and (2,- 2,2), A(x - 1) + B(y -1) + C(z - 2) = 0 (1) A(x - 2) + B(y + 2) + C(z - 2) = 0 (2) Subtracting (1) from (2), A(x - 2 - x + 1) + B(y + 2 - y-1) = 0 A - 3B = 0 (3) Now plane is perpendicular to 6x - 2y + 2z = 9 6A - 2B + 2C = 0 (4) Using (3) in (4) 18A - 2B + 2C = 0 16B + 2C = 0 C = - 8B Putting values in equation (1) 3B(x - 1) + B(y + 2) - 8B(z - 2) = 0 B(3x - 3 + y + 2 - 8z + 16) = 0 3x + y - 8z + 15 = 0 |
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