The equation of the plane parallel to the plane x-2y+2z-3=0 which is at a unit distance fr…

Question

The equation of the plane parallel to the plane $x-2y+2z-3=0$ which is at a unit distance from the point $(1,2,3)$ is $ax+by+cz+d=0$. If $(b-d)=K(c-a)$, then the positive value of $K$ is:

Answer: 4

Accepted Answer

$$x-2y+2z+\lambda=0.$$
Distance from $(1,2,3)$: $\dfrac{|1-4+6+\lambda|}{3}=1 \implies |\lambda+3|=3 \implies \lambda=0,-6.$

Taking $\lambda=-6$: $x-2y+2z-6=0$, so $a=1,b=-2,c=2,d=-6$.
$$b-d=-2-(-6)=4,\qquad c-a=2-1=1 \implies K=4.$$

Solutions · 1