Class 12

Math

3D Geometry

Three Dimensional Geometry

Find the equation of the plane passing through the intresection of the planes $x−2y+z=1$ and $2x+y+z=8$ and parallel to the line with direction ratio proportional to $1,2,1,$ find also the perpendicular distance of $(1,1,1)$ from this plane.

The equation of the plane passing through the intersection of the given planes is

$(x−2y+z−1)+λ(2x+y+z−8)=0$

$⇒(1+2λ)x+(−2+λ)y+(1+λ)z−1−8λ=0$ ......$(1)$

This plane is parallel to the line whose direction ratios are proportional to $1,2,1$

So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to $1,2,1$

$⇒(1+2λ)1+(−2+λ)2+(1+λ)1−1−8λ=0$

$⇒5λ−2=0$

$⇒λ=52 $

Substituting this in $(1)$, we get

$⇒(1+2×52 )x+(−2+52 )y+(1+52 )z−1−852 =0$

$⇒9x−8y+7z−21=0$ .......$(2)$ which is the required equation of the plane.

Perpendicular distance of plane $(2)$ from $(1,1,1)$

$=9_{2}+(−8)_{2}+7_{2} ∣9×1−8×1+7×1−21∣ $

$=194 ∣−13∣ =19413 $units.