Question
What is the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line whose direction ratios are 3, 2, 1 ?
What is the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line whose direction ratios are 3, 2, 1 ?
B. 3x + 2y + z = 6
The direction ratios of the normal to the plane are proportional to the direction ratios of the line perpendicular to it. Thus, the normal vector has direction ratios 3, 2, 1 . The equation of a plane passing through a point (x_1, y_1, z_1) with normal direction ratios a, b, c is given by a(x - x_1) + b(y - y_1) + c(z - z_1) = 0. Substituting (x_1, y_1, z_1) = (1, 1, 1) and a, b, c = 3, 2, 1 , we get: 3(x - 1) + 2(y - 1) + 1(z - 1) = 0 3x - 3 + 2y - 2 + z - 1 = 0 3x + 2y + z = 6 Answer: 3x + 2y + z = 6
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