Equations of Planes — Question 7

Question 7 – Solution

Strategy Put the equations over the same normal. Their constant difference divided by the normal length is the separation.

Step 1: Distance Both planes have normal $\vec n=\left\langle 2,-2,1\right\rangle$, whose length is $$\|\vec n\|=\sqrt{2^2+(-2)^2+1^2}=3.$$ Because their left sides are identical, the distance is the absolute difference of constants divided by $\|\vec n\|$: $$\boxed{d(\Pi_1,\Pi_2)=\frac{|7-(-5)|}{3}=4}.$$

Step 2: Halfway plane A parallel plane has form $2x-2y+z=c$. Equal distance from $c$ to $7$ and $-5$ requires $$7-c=c-(-5)\Longrightarrow 2c=2\Longrightarrow c=1.$$ Thus $$\boxed{2x-2y+z=1}.$$

Step 3: Shortest connector Choose $A=(3,0,1)$; indeed $2(3)-2(0)+1=7$. The unit normal is $\vec n/3$, so moving four units toward $\Pi_2$ gives $$B=A-\frac 43\left\langle 2,-2,1\right\rangle=\boxed{\left(\tfrac 13,\tfrac 83,-\tfrac 13\right)},$$ Substitution gives $2(1/3)-2(8/3)-1/3=-5$, so $B\in\Pi_2$. Its midpoint with $A$ is $$\boxed{M=\left(\tfrac 53,\tfrac 43,\tfrac 13\right)}.$$

Verification $2(5/3)-2(4/3)+1/3=1$. Also $B-A$ is parallel to the common normal, so $AB$ is perpendicular to both planes and is therefore a shortest connector.