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  • Solve this problem - Three Dimensional Geometry - JEE Main-3

Equation of plane passing through line of intersection of x+y+2z-6=0 and 2x+y-z+4=0 and also parallel to line x=y=z will be

  • Option 1)

    3x+y+4z-14=0

  • Option 2)

    3x-y+4z+14=0

  • Option 3)

    3x-y-4z+14=0

  • Option 4)

    3x+y-4z+14=0

 

Answers (1)

best_answer

As we have learned

Equation of any plane passing through the line of intersection of two planes (vector form ) -

The equation of any plane passing through the line of intersection of two planes

\vec{r}\cdot \vec{n}= d\: and\: \vec{r}\cdot \vec{n_{1}}= d_{1} is given by \left ( \vec{r}\cdot \vec{n}-d \right )+\lambda \left ( \vec{r}\cdot \vec{n_{1}}-d_{1} \right )= 0

 

- wherein

 

 Equation of plane will be 

(x+y+2z-6)+\lambda(2x+y-z+4)=0

\Rightarrow (1+2\lambda)x+(1+\lambda)y+(2-\lambda)z+(4\lambda-6)=0

But above plane is parallel to line x=y=z so normal to plane  will also be perpendicular to line 

\therefore(1+2\lambda)1+(1+\lambda)1-(2-\lambda)1=0

\therefore 2\lambda=-4\Rightarrow \lambda=-2

\thereforePlane becomes 3x+y-4z+14=0


Option 1)

3x+y+4z-14=0

Option 2)

3x-y+4z+14=0

Option 3)

3x-y-4z+14=0

Option 4)

3x+y-4z+14=0

Posted by

Himanshu

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