Equations of Planes — Question 5

Question 5 – Solution

Strategy Move from $P$ along the normal $\vec n=\left\langle 2,-1,2\right\rangle$. The signed equation value determines the required multiple.

Step 1: Signed displacement Write the plane as $F(x,y,z)=2x-y+2z-3=0$. At $P$, $$2(4)-(-1)+2(5)-3=16,\qquad \|\vec n\|^2=9.$$

Step 2: Projection A point on the perpendicular through $P$ is $P-s\vec n$. To reach the plane, $$F(P-s\vec n)=F(P)-s\|\vec n\|^2=16-9s=0,$$ so $s=16/9$. Hence $$H=P-\frac{16}{9}\vec n =\left(4-\tfrac{32}{9},-1+\tfrac{16}{9},5-\tfrac{32}{9}\right) =\boxed{\left(\tfrac 49,\tfrac 79,\tfrac{13}{9}\right)}.$$

Step 3: Distance Since $P-H=(16/9)\vec n$, $$d(P,\Pi)=\frac{16}{9}\|\vec n\|=\frac{16}{9}(3),\qquad \boxed{d(P,\Pi)=\frac{16}{3}}.$$

Step 4: Reflection Since $H$ is the midpoint of $PP'$, $$P'=2H-P=P-\frac{32}{9}\vec n=\boxed{\left(-\tfrac{28}{9},\tfrac{23}{9},-\tfrac{19}{9}\right)}.$$

Verification Substituting $H$ into the plane gives $3$. Also $P-H=(16/9)\vec n$, so $PH$ is perpendicular to the plane, and direct averaging of $P$ and $P'$ gives $H$.