Question

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

Answer 1

Step $1$: Write the equation of plane parallel to given plane

Equation of plane parallel to given plane $x-2y+2z-3=0$ are of the form

$x-2y+2z+k=0$

Step $2$: Apply formula for perpendicular distance of point from a plane

The distance between plane $ax+by+cz+\lambda =0$ and a point $\left(x_1,y_1,z_1\right)$ is given as

Here $ax+by+cz+\lambda =0$$\equiv$$x-2y+2z+k=0$ and $\left(x_1,y_1,z_1\right)$$\equiv$$\left(1,2,3\right)$,$d=1$(as it is given unit distance)

$d=\left|\frac{a{x}_1+b{y}_1+c{z}_1+\lambda }{\sqrt{a^2+b^2+c^2}}\right|$

$\Rightarrow$ $1=\left|\frac{1\left(1\right)-2\left(2\right)+2\left(3\right)+k}{\sqrt{1^2+{\left(-2\right)}^2+2^2}}\right|$

$\Rightarrow$ $1=\left|\frac{k+3}{3}\right|$

$\Rightarrow$ $3=\left|k+3\right|$

$\Rightarrow$ $k=-6$ or $k=0$

Step $3$: Write the equations of parallel planes and find value of $K$

Hence, the equations of the parallel planes are

$x-2y+2z-6=0$ and $x-2y+2z=0$

comparing $x-2y+2z+6=0$ with $ax+by+cz+d=0$ we get,

$a=1,b=-2,c=2,d=-6$

Substituting these values in $\left(b-d\right)=K\left(c-a\right)$ we get

$-2+6=K\left(2-1\right)$

$\Rightarrow$ $K=4$

comparing $x-2y+2z=0$ with $ax+by+cz+d=0$we get,

$a=1,b=-2,c=2,d=0$

Substituting these values in $\left(b-d\right)=K\left(c-a\right)$ we get

$\left(-2\right)=K\left(2-1\right)$

$\Rightarrow$ $K=-2$

But we only require the positive value of $K$

$\Rightarrow$ $K=4$

Hence, the positive value of $K$ satisfying all given conditions is $4$.