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  • Need clarity, kindly explain! If L1is the line of intersection of the planes 2x−2y+3z−2=0, x−y+z+1=0 and L2 is the line of intersection of the planesx+2y−z−3=0, 3x−y+2z−1=0, then the distance of the origin fr

If L1 is the line of intersection of the planes 2x−2y+3z−2=0, x−y+z+1=0 and L2 is the line of intersection of the planes
x+2y−z−3=0, 3x−y+2z−1=0, then the distance of the origin from the plane, containing the lines L1 and L2, is :

  • Option 1)

    \frac{1}{\sqrt{2}}

  • Option 2)

    \frac{1}{4\sqrt{2}}

  • Option 3)

    \frac{1}{3\sqrt{2}}

  • Option 4)

    \frac{1}{2\sqrt{2}}

 

Answers (1)

best_answer

p_{1}+\lambda p_{2}=0

(2x-2y+3z-2)+\lambda (x-y+z+1)=0

(2+\lambda )x+(-2-\lambda )y+ (3+\lambda )z- (2-\lambda )=0

also L_{2}  has two planes 

\begin{vmatrix} (2+\lambda ) & -2-\lambda & 3+\lambda \\ 1&2 &-1 \\ 3 &-1 &2 \end{vmatrix}=0

\lambda =5

 

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

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

ax+by+cz+d= 0 and

a_{1}x+b_{1}y+c_{1}z+d_{1}= 0 is given by 

\left ( ax+by+cz+d \right )+\lambda \left ( a_{1}x+b_{1}y+c_{1}z+d _{1}\right )= 0

 

 

-

 

 


Option 1)

\frac{1}{\sqrt{2}}

This is incorrect

Option 2)

\frac{1}{4\sqrt{2}}

This is incorrect

Option 3)

\frac{1}{3\sqrt{2}}

This is correct

Option 4)

\frac{1}{2\sqrt{2}}

This is incorrect

Posted by

Himanshu

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