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  • I need help with - Three Dimensional Geometry - JEE Main-6

Equation of plane parallel to 2x-y+2z-5=0  at a distance of 1 unit from it is

  • Option 1)

    2x-y-2z+8=0

  • Option 2)

    2x-y-2z+2=0

  • Option 3)

    2x-y+2z+8=0

  • Option 4)

    2x-y+2z-2=0

 

Answers (1)

best_answer

As we have learned

Distance between parallel planes (Cartesian form ) -

Distance between two planes ax+by+cz+d= 0 and

ax+by+cz+d_{1}= 0 is given by

\frac{\left [ d_{1}-d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}

-

 

 Let equation of plane is 2x-y+2z+k=0 distance between them is \frac{|k+5|}{3}=1\Rightarrow k+5=3,-3

\therefore k=-8,-2

\therefore Planes are 2x-y+2z-8=0 and 2x-y+2z-2=0

 


Option 1)

2x-y-2z+8=0

Option 2)

2x-y-2z+2=0

Option 3)

2x-y+2z+8=0

Option 4)

2x-y+2z-2=0

Posted by

Himanshu

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