Start Free Mock LoginAlready registered? Login here

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Mathematics3D Coordinate GeometryPerpendicular distance of a point from a planeMedium2 minQB

MathematicsMediumsingle choice

The shortest distance from the plane $12 x + 4 y + 3 z = 327$ to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

Options:

Answer:

Solution:

Centre of the sphere is (-2, 1, 3) and its radius is $4 + 1 + 9 + 155 = 13$ . Length of the perpendicular from the centre on the plane is $\frac{- 24 + 4 + 9 - 327}{144 + 16 + 9} = \frac{338}{13} = 26$ . So the plane is outside the sphere and the required distance is $26 - 13 = 13$ .

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

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...