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Question

Find the minimum distance from the point (2, -1, 1) to the plane x + y - z = 2.

Solution

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Let A(2,−1,1)A(2,-1,1) and B(x,y,z)B(x,y,z), then we get the following:

d(AB)=(x−2)2+(y+1)2+(z−1)2=(x−2)2+(y+1)2+(z−1)2d(AB)=\sqrt{(x-2)^2+(y+1)^2+(z-1)^2}=\sqrt{(x-2)^2+(y+1)^2+(z-1)^2}

So, we will minimize d(AB)=(x−2)2+(y+1)2+(z−1)2d(AB)=\sqrt{(x-2)^2+(y+1)^2+(z-1)^2} if we minimize D(AB)=(x−2)2+(y+1)2+(z−1)2D(AB)=(x-2)^2+(y+1)^2+(z-1)^2. We are given that the plane is x+y−z=2⇒z=x+y−2x+y-z=2 \Rightarrow z=x+y-2. Now, we get the following:

D(x,y)=(x−2)2+(y+1)2+(x+y−3)2⇒Dx=2(x−2)+2(x+y−3)=0,Dy=2(y+1)+2(x+y−3)=0D(x,y)=(x-2)^2+(y+1)^2+(x+y-3)^2 \Rightarrow D_x=2(x-2)+2(x+y-3)=0, D_y=2(y+1)+2(x+y-3)=0

⇒x=83,y=−13,z=13\Rightarrow x=\dfrac{8}{3}, y=-\dfrac{1}{3}, z=\dfrac{1}{3}

So, we got that the critical point is (83,−13)\left(\dfrac{8}{3}, -\dfrac{1}{3} \right), then we get

Dxx=4>0,Dyy=4,Dxy=2⇒DxxDyy−(Dyy)2=12>0D_{xx}=4>0, D_{yy}=4, D_{xy}=2 \Rightarrow D_{xx}D_{yy}-(D_{yy})^2=12>0

The conclusion is that the point (83,−13,13)\left(\dfrac{8}{3}, -\dfrac{1}{3}, \dfrac{1}{3} \right) is the point of a local minimum of DD, which means that is required point.

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