First slide

Planes in 3D

Question

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

Difficult

A

39
B

26
C

$11 \frac{4}{13}$
D

13

Solution

Shortest distance =perpendicular distance between the plane and sphere
                                =distance of plane from centre of sphere –radius

                                $= | \frac{- 2 \times 12 + 4 \times 1 + 3 \times 3 - 327}{\sqrt{144 + 9 + 16}} | - \sqrt{4 + 1 + 9 + 155}$

                               $= 26 - 13 = 13$

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