Question

The plane $4x+7y+4z+81=0$ is rotated through a right angle about its line of intersection with the plane $5x+3y+10z=25$. If the equation of plane in its new position is $x-4y+6z=k$, then the value of $k$ is equal to

• A. $73$
• B. $-89$
• C. $37$
• D. $106$

Answer 1

The correct option is D $106$

The equation of the plane through the line of intersection of the given planes is
$4x+7y+4z+81+\lambda (5x+3y+10z-25)=0$
$\Rightarrow (4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+(81-25\lambda )=0$
Since this plane is perpendicular to the plane $4x+7y+4z+81=0$
$\therefore 4(4+5\lambda )+7(7+3\lambda )+4(4+10\lambda )=0$
$\Rightarrow 16+20\lambda +49+21\lambda +16+40\lambda =0$
$\Rightarrow \lambda =-1$
$\therefore$ Required plane is $x-4y+6z=106$
Hence, the value of $k$ is $106.$