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Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Mathematics3D Coordinate GeometryPerpendicular distance of a point from a planeMedium2 minQB

MathematicsMediumsingle choice

Let $A (1, 1, 1), B (2, 3, 5), C (- 1, 0, 2)$ be three points. The equation of a plane parallel to the plane ABC which is at a distance 2 from it is

Options:

Answer:

Solution:

$A B = ⟨ 1, 2, 4 ⟩$ and $A C = ⟨ - 2, - 1, 1 ⟩$ , so the normal to plane ABC has direction ratios \{2, -3, 1) and the plane is $2 x - 3 y + z = 0$ . A parallel plane $2 x - 3 y + z = k$ at distance 2 has $\frac{∣ k ∣}{4 + 9 + 1} = 2$ , so $k = \pm 214$ . The plane $2 x - 3 y + z + 214 = 0$ is among the options.

Stream:JEESubject:MathematicsTopic:3D Coordinate GeometrySubtopic:Perpendicular distance of a point from a plane

⏱ 2mℹ️ Source: QB

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