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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

The distance between the plane $x - 2 y + z - 6 = 0$ and the plane containing the sets of points $(1 + 2 λ, 2 + 3 λ, 3 + 4 λ)$ and $(2 + 3 μ, 3 + 4 μ, 4 + 5 μ)$ , where $λ, μ$ are parameters, is

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Solution:

Normal vector: $(2, 3, 4) \times (3, 4, 5) = - \hat{i} + 2 \hat{j} - \hat{k}$ . The plane is $- 1 (x - 1) + 2 (y - 2) - 1 (z - 3) = 0$ , i.e. $x - 2 y + z = 0$ . The required distance is $\frac{6}{1 + 4 + 1} = 6$ .

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

⏱ 2mℹ️ Source: QB

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