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Question

Find the equation of the plane passing through the point (1,-2,1) and perpendicular to the line joining the points A(3, 2,1) and B(1, 4, 2).

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Solution

The equation of a plane passing through the point 1, -2, 1 is,ax - 1 + by + 2 + cz - 1 = 0 ......1Here, a, b, c are the direction ratios of the normal to the plane.Equation of the line joining the points A3, 2, 1 and B1, 4, 2 is, x - 31 - 3 = y - 24 - 2 = z - 12 - 1x - 3-2 = y - 22 = z - 11 .......2Now, direction ratios of line AB are :a1 , b1, c1 = -2, 2, 1Since, the plane 1 is to the line AB, thena, b, c a1 , b1, c1 = -2, 2, 1 Putting the values of a, b and c in 1, we get-2x - 1 + 2y + 2 + 1z - 1 = 0-2x + 2 + 2y + 4 + z - 1 = 0-2x + 2y + z + 5 = 02x - 2y - z - 5 = 0

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