Question
Let a plane pass through the point (1, β1, 2) and is
perpendicular to the vector 2i + j β 2k. Then its Cartesian equation is:Solution
We are given: β’ A point through which the plane passes: P(1, β1, 2) β’ A normal vector to the plane: n = 2i + j β 2k β this corresponds to vector n = (2, 1, β2) We are to find the Cartesian equation of the plane. General form of the plane equation: If a plane passes through point (xβ, yβ, zβ) and has normal vector (A, B, C), then the equation is: A(x β xβ) + B(y β yβ) + C(z β zβ) = 0 Substitute: β’ (xβ, yβ, zβ) = (1, β1, 2) β’ (A, B, C) = (2, 1, β2) So: 2(x β 1) + 1(y + 1) β 2(z β 2) = 0 Now simplify: 2x β 2 + y + 1 β 2z + 4 = 0 β 2x + y β 2z + 3 = 0
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