Question

The plane denoted by $P_1:4x+7y+4z+81=0$ is rotated through a right angle about its line of intersection with the plane $P_2:5x+3y+10z=25.$ If the plane in its new position be denoted by $P,$ and the distance of this plane from the origin is $d,$ then find the value of $\left[k/2\right]$ (where $[\cdot ]$ represents greatest integer less than or equal to $k$ ).

Answer 1

$4x+7y+4z+81=0$ $\left(i\right)$

$5x+3y+10z=25$ $\left(ii\right)$

Equation of plane passing through their line of intersection is

$(4x+7y+4z+81)+\lambda (5x+3y+10z-25)=0$

or $(4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+81-25\lambda =0$ $\left(iii\right)$

plane (iii) $\perp$ to (i), so

$4(4+5\lambda )+7(7+3\lambda )+4(4+10\lambda )=0$

$\therefore \lambda =-1$

From (iii), equation of plane is $-x+4y-6z+106=0$ $\left(iv\right)$

Distance of (iv) from $(0,0,0)d=\frac{106}{\sqrt{1+16+36}}=\frac{106}{\sqrt{53}}$

$\frac{d}{2}=7.28$

$\left[\frac{d}{2}\right]=7$