Question
Let a plane pass through the point (1, –1, 2) and is
perpendicular to the vector 2i + j − 2k. Then its Cartesian equation is:Solution
We are given: • A point through which the plane passes: P(1, –1, 2) • A normal vector to the plane: n = 2i + j − 2k → this corresponds to vector n = (2, 1, –2) We are to find the Cartesian equation of the plane. General form of the plane equation: If a plane passes through point (x₀, y₀, z₀) and has normal vector (A, B, C), then the equation is: A(x – x₀) + B(y – y₀) + C(z – z₀) = 0 Substitute: • (x₀, y₀, z₀) = (1, –1, 2) • (A, B, C) = (2, 1, –2) So: 2(x – 1) + 1(y + 1) – 2(z – 2) = 0 Now simplify: 2x – 2 + y + 1 – 2z + 4 = 0 ⇒ 2x + y – 2z + 3 = 0
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