Question

The plane $2x-y+3z+5=0$ is rotated through ${90}^{\circ }$ about its line of intersection with the plane $5x-4y-2z+1=0.$ Let the equation of the plane in the new position be $kx+my-26z-13=0$. Find $k+m$? (Note: $k$ and $m$ are integers in shortest form.)

Answer 1

Plane through the intersection of given planes is
$(2x-y+3z+5)+\lambda (5x-4y-2z+1)=0$
$(2+5\lambda )x-(1+4\lambda )y+(3-2\lambda )z+(5+\lambda )=0$ ...( 1 )
This plane makes an angle of ${90}^{\circ }$ with the plane
$2x-y+3z+5=0$
Applying the condition of perpendicularity
$\sum a_1{a}_2=0,$
$2(2+5\lambda )+1(1+4\lambda )+3(3-2\lambda )=0$
$8\lambda +14=0$

$\therefore \lambda =-\frac{7}{4}$
$\therefore$ required equation of plane from (1) on substituting $\lambda =-\frac{7}{4}$ is
$27x-24y-26z-13=0$
$\Rightarrow k+m=3$