Porisma

Worked solution

Use two plane normals and their common base point

Find the line of intersection of

Π1:r=(1,2,3)+α(1,2,1)+β(2,2,1)\Pi_1:\quad\mathbf r=(1,2,3)+\alpha(1,2,1)+\beta(2,2,1)

and

Π2:r=(1,2,3)+s(1,1,0)+t(4,6,7),\Pi_2:\quad\mathbf r=(1,2,3)+s(1,1,0)+t(4,6,7),

where all four parameters are real.

CJC · Lecture · Example 40

Answer

r=(1,2,3)+λ(12,14,7)\mathbf r=(1,2,3)+\lambda(12,14,7), λ∈R\lambda\in\mathbb R.

Compare your reasoning

The common base point is P=(1,2,3)P=(1,2,3). First find each plane normal from its two directions. For Π1\Pi_1,

(n1)x=2(1)−1(2)=0,(n1)y=1(2)−1(1)=1,(n1)z=1(2)−2(2)=−2.\begin{aligned}(n_1)_x&=2(1)-1(2)=0,\\(n_1)_y&=1(2)-1(1)=1,\\(n_1)_z&=1(2)-2(2)=-2.\end{aligned}

Hence n1=(1,2,1)×(2,2,1)=(0,1,−2)\mathbf n_1=(1,2,1)\times(2,2,1)=(0,1,-2). For Π2\Pi_2,

(n2)x=1(7)−0(6)=7,(n2)y=0(4)−1(7)=−7,(n2)z=1(6)−1(4)=2.\begin{aligned}(n_2)_x&=1(7)-0(6)=7,\\(n_2)_y&=0(4)-1(7)=-7,\\(n_2)_z&=1(6)-1(4)=2.\end{aligned}

Thus n2=(1,1,0)×(4,6,7)=(7,−7,2)\mathbf n_2=(1,1,0)\times(4,6,7)=(7,-7,2). A common-line direction is perpendicular to both normals:

mx=1(2)−(−2)(−7)=−12,my=(−2)7−0(2)=−14,mz=0(−7)−1(7)=−7.\begin{aligned}m_x&=1(2)-(-2)(-7)=-12,\\m_y&=(-2)7-0(2)=-14,\\m_z&=0(-7)-1(7)=-7.\end{aligned}

So n1×n2=(−12,−14,−7)=−(12,14,7)\mathbf n_1\times\mathbf n_2=(-12,-14,-7)=-(12,14,7). Reversing a nonzero direction does not change a line with all real parameters. Combine the printed direction (12,14,7)(12,14,7) with P to obtain the answer. The normals are nonparallel, so their cross product is nonzero and the intersection is a line. As a check, the direction has dot product 0 with both normals.

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