Question 449 of 521MediumMCQ
The shortest distance from the plane 12x+4y+3z=327 to the sphere x2+y2+z2+4x−2y−6z=155 is Solutions · 1
Official solution
Shortest distance = perpendicular distance between the plane and sphere = distance of plane from center of sphere - radius
\[ = \left| {{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327} \over {\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155} \]
=26−13=13 Answer the question to unlock