Question

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

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

Answer 1

The correct option is A $106$

Equation of plane in new orientation is given by,
$4x+7y+4z+81+\lambda (5x+3y+10z-25)=0$
$\Rightarrow (4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+(81-25\lambda )=0$
Now this plane is perpendicular to it's previous orientation $4x+7y+4z+81=0$
$\Rightarrow (4+5\lambda )4+(7+3\lambda )7+(4+10\lambda )4=0$
$\Rightarrow \lambda =-1$
Hence, our required plane is $x-4y+6z=106$.