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Find the point on the plane 3x + 2y + z = 6 that is nearest the origin.

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Answered 1 year ago
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Let O(0,0,0)O(0,0,0) and A(x,y,z)A(x,y,z), then we get the following:

d(OA)=(x0)2+(y0)2+(z0)2=x2+y2+z2d(OA)=\sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=\sqrt{x^2+y^2+z^2}

So, we will minimize d(OA)=x2+y2+z2d(OA)=\sqrt{x^2+y^2+z^2} if we minimize D(OA)=x2+y2+z2D(OA)=x^2+y^2+z^2. We are given that the plane is 3x+2y+z=6z=63x2y3x+2y+z=6 \Rightarrow z=6-3x-2y. Now, we get the following:

D(x,y)=x2+y2+(63x2y)2Dx=2x6(63x2y)=0,Dy=2y6(63x2y)=0D(x,y)=x^2+y^2+(6-3x-2y)^2 \Rightarrow D_x=2x-6(6-3x-2y)=0, D_y=2y-6(6-3x-2y)=0

x=97,y=67,z=37\Rightarrow x=\dfrac{9}{7}, y=\dfrac{6}{7}, z=\dfrac{3}{7}

So, we got that the critical point is (97,67)\left(\dfrac{9}{7}, \dfrac{6}{7} \right), then we get

Dxx=20>0,Dyy=10,Dxy=12DxxDyy(Dyy)2=56>0D_{xx}=20>0, D_{yy}=10, D_{xy}=12 \Rightarrow D_{xx}D_{yy}-(D_{yy})^2=56>0

The conclusion is that the point (97,67,37)\left(\dfrac{9}{7}, \dfrac{6}{7}, \dfrac{3}{7} \right) is the point of a local minimum of DD, so, that is the point which is on the given plane and it is the nearest to the origin.

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