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

Solution

VerifiedLet $O(0,0,0)$ and $A(x,y,z)$, then we get the following:

$d(OA)=\sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=\sqrt{x^2+y^2+z^2}$

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

$D(x,y)=x^2+y^2+(6-3x-2y)^2 \Rightarrow D_x=2x-6(6-3x-2y)=0, D_y=2y-6(6-3x-2y)=0$

$\Rightarrow x=\dfrac{9}{7}, y=\dfrac{6}{7}, z=\dfrac{3}{7}$

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

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

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

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