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

  • Option 1)

    11\frac{3}{4}\;

  • Option 2)

    \; 13\;

  • Option 3)

    \; 39\;

  • Option 4)

    \; 26

 

Answers (1)

As we learnt in

Distance of a point from plane (Cartesian form) -

The length of perpendicular from P(x_{1},y_{1},z_{1}) to the plane

ax+by+cz+d= 0 is given by  \frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}

 

-

 

 Center of sphere = C(-2,1,3) and r= 13

Distance of center from  plane = \frac{\left | 12(-2) +4(1) + 3(3)- 327 \right |}{13}

=\frac{338}{13}= 26

Hence shortest distance is 26-13= 13 units.

 


Option 1)

11\frac{3}{4}\;

Incorrect option

Option 2)

\; 13\;

Correct option

Option 3)

\; 39\;

Incorrect option

Option 4)

\; 26

Incorrect option

Posted by

Vakul

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