Class 12

Math

3D Geometry

Three Dimensional Geometry

Find the Cartesian and vector equations of the planes through the line of intersection of the planes $r⋅(i^−j^ )+6=0$ and $r⋅(3i^+3j^ −4k^)=0$, which are at a unit distance from the origin.

The equations of the given planes are

$(xi^+yj^ +zk^)⋅(i^−j^ )+6=0$ and $(xi^+yj^ +zk^)⋅(3i^+3j^ −4k^)=0$

$⇒x−y+6=0$ and $3x+3y−4z=0$

Any plane through their intersection is

$(x−y+6)+λ(3x+3y−4z)=0$

$⇒(1+3λ)x+(3λ−1)y−4λz+6=0$ ..........(i)

which is at a unit distance from the origin. .... (given)

$∴(1+3λ)_{2}+(3λ−1)_{2}+(−4λ)_{2} 6 =1$

$⇒34λ_{2}+2=36$

$⇒λ_{2}=1$

$⇒λ=±1$

So, the required planes are

$2x+y−2z+3=0$ and $x+2y−2z−3=0$

In vector form, they are

$r⋅(2i^+j^ −2k^)+3=0$ and $r⋅(i^+2j^ −2k^)−3=0$.