Equations of Planes — Question 6

Question 6 – Solution

Strategy The shortest segment from a point to a plane follows the plane’s normal. Use Cauchy–Schwarz for a global proof.

Step 1: Follow the normal The normal is $\vec n=\left\langle 1,2,2\right\rangle$ and $$\|\vec n\|=\sqrt{1^2+2^2+2^2}=3.$$ A point on the normal line through the origin has form $(t,2t,2t)$. Imposing the plane equation gives $$t+2(2t)+2(2t)=9\Longrightarrow 9t=9\Longrightarrow t=1.$$ Hence $$\boxed{H=(1,2,2)},\qquad \boxed{d(O,\Pi)=\|H\|=3}.$$

Step 2: Global proof For any $X=(x,y,z)$ on $\Pi$, $\vec n\cdot X=9$. Cauchy–Schwarz gives $$9=|\vec n\cdot X|\le\|\vec n\|\|X\|=3\|X\|,$$ so $\|X\|\ge 3$. Equality occurs precisely when $X$ is parallel to $\vec n$, as $H$ is.

Step 3: Tangent sphere A sphere centered at the origin and reaching $H$ has radius $3$, so its equation is $$\boxed{x^2+y^2+z^2=9}.$$ Its radius to $H$ is normal to $\Pi$, so the plane is tangent there.