Question

The plane $2x-y+3z+5=0$ is rotated through ${90}^{\circ }$ about its line of intersection with the plane $5x-4y-2z+1=0$. The equation of the plane in the new position is

• A. $6x-9y-29z-31=0$
• B. $27x-24y-26z-13=0$
• C. $43x-32y-2z+27=0$
• D. $26x-43y-151z-165=0$

Answer 1

The correct option is B $27x-24y-26z-13=0$

Equation of a plane passing through the line of intersection of the given planes is
$(2x-y+3z+5)+\lambda (5x-4y-2z+1)=0$
$(2+5\lambda )x-(1+4\lambda )y+(3-2\lambda )z+(5+\lambda )=0..$(i)
Now plane $2x-y+3z+5=0$ and (i) are perpendicular.
$\Rightarrow 2(2+5\lambda )+(1+4\lambda )+3(3-2\lambda )=0$ $\Rightarrow$ $\lambda =-7/4$
Thus required equation of the plane is
$4(2x-y+3z+5)-7(5x-4y-2z+1)=0$
$\Rightarrow$ $27x-24y-26z-13=0$