Proper pencil

Property — Cartesian equation of a proper pencil

If $\pi_1\colon a_1 x+b_1 y+c_1 z+d_1=0$ and $\pi_2\colon a_2 x+b_2 y+c_2 z+d_2=0$ are two distinct planes intersecting along the line $r$, the proper pencil with axis $r$ has equation $\alpha(a_1 x+b_1 y+c_1 z+d_1) + \beta(a_2 x+b_2 y+c_2 z+d_2) = 0, \qquad (\alpha,\beta)\ne (0,0):$ every plane of the pencil is a linear combination of the two generators. Setting $\lambda=\beta/\alpha$ (when $\alpha\ne 0$) it is also written $(a_1+\lambda a_2)x+(b_1+\lambda b_2)y+(c_1+\lambda c_2)z+(d_1+\lambda d_2)=0,$ with a single parameter $\lambda\in\mathbb{R}$ (the value $\lambda=\infty$ corresponds to $\pi_2$ alone).

Example — Construction of a proper pencil

Let $\pi_1\colon x+y-z+1=0$ and $\pi_2\colon 2x-y+z=0$. The line $r=\pi_1\cap\pi_2$ is the axis of the pencil $(x+y-z+1)+\lambda(2x-y+z)=0 \iff (1+2\lambda)x+(1-\lambda)y+(-1+\lambda)z+1=0.$ For $\lambda=1$: $\;3x+1=0 \iff x=-\tfrac13$; for $\lambda=-1$: $\;-x+2y-2z+1=0$.

Example — Plane of the pencil passing through a point

Which is the plane of the previous pencil passing through $P(1;0;-1)$? Substituting $P$ into the equation of the pencil: $(1+0-(-1)+1)+\lambda(2-0+(-1))=0 \iff 3+\lambda=0 \iff \lambda=-3.$ The plane sought is $(x+y-z+1)-3(2x-y+z)=0 \iff -5x+4y-4z+1=0$.

Links

Topics: Analytic geometry in space
Concepts: Pencil of planes · Plane
Methods: Proper pencil of planes
Skills: Analytic geometry · Using formulae