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The shortest distance from the plane `12 x+y+3z=327` to the sphere `x^2+y^2+z^2+4x-2y-6z=155` is a. `39` b. `26` c. `41-4/(13)` d. `13`

A

39

B

26

C

`41(4)/(13)`

D

13

Text Solution

Verified by Experts

The correct Answer is:
d
The given sphere is
`x^(2)+y^(2)+z^(2)+4x-2y-6z-155=0`
Its center is (-2,1,3) and radius `=sqrt(4+1+9+155)=sqrt169=13`

Therefore, distance of centre (-2,1,3) from the plane `12x+4y+3z=327` is
`|(12(-2)+4(1)+3(3)-327)|/(sqrt(144+16+9))`
Hence, the shortest distance is 13.
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CENGAGE-THREE-DIMENSIONAL GEOMETRY -Exercise (Single)
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  4. A tetrahedron has vertices O(0,0,0),A(1,2,1),B(2,1,3),a n dC(-1,1,2), ...

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  7. The point of intersection of the lines (x-5)/3=(y-7)/(-1)=(z+2)/1a ...

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  16. The projection of point P(vecp) on the plane vecr.vecn=q is (vecs), th...

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