Equations of Planes — Question 8

Question 8 – Solution

Strategy Every combination $F_1+\lambda F_2=0$ vanishes wherever both original plane expressions vanish. Choose $\lambda$ using normal-vector perpendicularity.

Step 1: Build the pencil Move all terms to the left: $$F_1=x+y+z-1,\qquad F_2=x-y+2z-3.$$ At every point on the common line, $F_1=F_2=0$. Therefore every equation $$\boxed{F_1+\lambda F_2=0\ (\lambda\in\mathbb R),\quad\text{together with }F_2=0}$$ contains that line. The member $F_2=0$ is also needed for the full pencil; its normal has dot product $-3$ with $\left\langle 1,2,-1\right\rangle$, so it is not the requested perpendicular member. Collecting coefficients shows its normal is $$\vec n_\lambda=\left\langle 1+\lambda,1-\lambda,1+2\lambda\right\rangle.$$

Step 2: Impose perpendicularity The normal of $\Sigma$ is $\vec m=\left\langle 1,2,-1\right\rangle$. Perpendicular planes have perpendicular normals, so $$\vec n_\lambda\cdot\left\langle 1,2,-1\right\rangle=2-3\lambda=0,$$ because $(1+\lambda)+2(1-\lambda)-(1+2\lambda)=2-3\lambda$. Thus $\lambda=2/3$. Multiplying by $3$ gives $$3F_1+2F_2=0,$$ and expanding yields $$\boxed{5x+y+7z=9}.$$

Verification Any point common to $\Pi_1$ and $\Pi_2$ has $F_1=F_2=0$, so it satisfies $3F_1+2F_2=0$. Its normal $\left\langle 5,1,7\right\rangle$ dots with $\left\langle 1,2,-1\right\rangle$ to $0$.