Minimum Distance in a 3-Dimensional Space

Geometry Level pending

Let D D be the minimum distance between x = 2 y 2 + 3 z 2 x = 2y^{2}+3z^{2} and x + 8 y + 18 z + 36 = 0 x+8y+18z+36 = 0 . What is the value of 1 D 2 \frac{1}{D^{2}} ?

Proposed by Tommy \textit{Proposed by Tommy}

386 389 387 388

Alan Yan by Alan Yan , 20, USA

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1 solution

Alan Yan
Sep 3, 2015

Given a point ( x 0 , y 0 , z 0 ) (x_0,y_0,z_0) and a plane A x + B y + C z + D = 0 Ax+By+Cz+D=0 , the distance between the point and the plane is

∣ A x 0 + B y 0 + C z 0 + D ∣ A 2 + B 2 + C 2 \large \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}

Now, substituting x x in, we need to minimize

∣ ( 2 y 2 + 3 z 2 ) + 8 y + 18 z + 36 ∣ 389 = ∣ 2 ( y + 2 ) 2 + 3 ( z + 3 ) 2 + 1 ∣ 389 ≥ 1 389 = D \frac{|(2y^2+3z^2) + 8y + 18z + 36|}{\sqrt{389}} = \frac{|2(y+2)^2 +3(z+3)^2 +1|}{\sqrt{389}} \geq \frac{1}{\sqrt{389}} = D

Therefore, 1 D 2 = 389 \frac{1}{D^2} = \boxed{389}

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