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  • I need help with - Three Dimensional Geometry - JEE Main-2

If (2, 3, 5) is one end of a diameter of the sphere x^{2}+y^{2}+z^{2}-6x-12y-2z+20=0, then the coordinates of the other end of the diameter are

  • Option 1)

    (4, 3, 5)

  • Option 2)

    (4, 3, –3)

  • Option 3)

    (4, 9, –3)

  • Option 4)

    (4, –3, 3)

 

Answers (1)

best_answer

As we learnt in 

Section Formula -

1)    Internal Division

\left ( \frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}, \frac{mz_{2}+nz_{1}}{m+n} \right )

2)    External Division

\left ( \frac{mx_{2}-nx_{1}}{m-n}, \frac{my_{2}-ny_{1}}{m-n}, \frac{mz_{2}-nz_{1}}{m-n} \right )

3)    Mid Point Formula

\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2} \right )

- wherein

 

 Coordinate of centre of sphere= (3,6,1)

Let other end be (x,y,z)

\frac{x+2}{2}= 3,\frac{y+3}{2}= 6,\frac{z+5}{2}= 1

x= 4, y= 9, z=-3


Option 1)

(4, 3, 5)

Incorrect Option

 

Option 2)

(4, 3, –3)

Incorrect Option

 

Option 3)

(4, 9, –3)

Correct Option

 

Option 4)

(4, –3, 3)

Incorrect Option

 

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