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  • How to solve this problem- - Three Dimensional Geometry - JEE Main-5

If the plane 2x-y+2z+3=0 has the distances \frac{1}{3} and \frac{2}{3} units

from the planes 4x-2y+4z+\lambda=0 and 2x-y+2z+\mu=0,

respectively, then the maximum value of \lambda+\mu is equal to : 

  • Option 1)

    9

  • Option 2)

    15

  • Option 3)

    5

  • Option 4)

    13

 

Answers (1)

best_answer

Given plane 2x-y+2z+3=0

 \frac{1}{3} unit distance from the plane 4x-2y+4z+\lambda=0

|\frac{\lambda-6}{\sqrt{16+4+16}}|=\frac{1}{3}

|\lambda-6|=2

=> \lambda=8,4

Now ,  \frac{2}{3} unit distance from the plane 2x-y+2z+\mu=0

|\frac{\mu-3}{\sqrt{4+1+4}}|=\frac{2}{3}

|\mu-3|=2

=> \mu=5,1

\therefore maximum value of \lambda+\mu = 8 + 5 = 13

So, correct option is (4)


Option 1)

9

Option 2)

15

Option 3)

5

Option 4)

13

Posted by

Plabita

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