Equations of Planes — Question 9

Question 9 – Solution

Strategy Use the point-to-plane distance formula with absolute values. Equal normal lengths simplify the two sign cases.

Step 1: Distance equation The normal lengths are $$\sqrt{1^2+(-2)^2+2^2}=3,\qquad \sqrt{2^2+2^2+(-1)^2}=3.$$ Thus equality of point-to-plane distances becomes $$|x-2y+2z-3|=|2x+2y-z+6|.$$

Step 2: Resolve the absolute values Let the two expressions be $F_1$ and $F_2$. The equation $|F_1|=|F_2|$ gives two cases.

If $F_1=F_2$, then $$x-2y+2z-3=2x+2y-z+6\Longrightarrow x+4y-3z+9=0.$$ If $F_1=-F_2$, then $$x-2y+2z-3=-2x-2y+z-6\Longrightarrow 3x+z+3=0.$$ Therefore the locus consists of $$\boxed{x+4y-3z+9=0},\qquad \boxed{3x+z+3=0}.$$

Interpretation On either plane, the signed perpendicular displacements from $\Pi_1$ and $\Pi_2$ have equal magnitude. These are precisely the internal and external angle-bisector planes, and both contain the original planes’ intersection line.

Perpendicularity Normals may be chosen as $\left\langle 1,4,-3\right\rangle$ and $\left\langle 3,0,1\right\rangle$. Their dot product is $3-3=0$, so the bisector planes are perpendicular.

Verification Substituting any point from either boxed plane into the absolute-value equation recovers equal distances.