The question is "The shortest squared distance, $x^2 + y^2 + z^2$, from the origin $(0, 0, 0)$ to a point $(x, y, z)$ on the plane $x + y + z = 1$ is"
I don't seem to understand where to start so an in-depth answer would be really helpful. Thank you.
(edit: The answer to this problem is 1/3)
On
If you want to use the Lagrange multiplier method the Lagrange function is $\mathcal{L}(x,y,z,\lambda) = f(x,y,z) - \lambda g(x,y,z)$. In you the specific case you minimize
$$\mathcal{L}(x,y,\lambda) = x^2+y^2+z^2 - \lambda \cdot (1-x-y-z)$$
The partial derivatives are
$\frac{\partial \mathcal L}{\partial x}=2x-\lambda=0 \Rightarrow 2x=\lambda \quad (1)$
$\frac{\partial \mathcal L}{\partial y }=2y-\lambda=0 \Rightarrow 2y=\lambda \quad (2)$
$\frac{\partial \mathcal L}{\partial z}=2z-\lambda=0 \Rightarrow 2z=\lambda\quad (3)$
$\frac{\partial \mathcal L}{\partial \lambda }=1-x-y-z=0 \quad (4)$
Dividing $(1)$ by $(2)$ gives $\frac{x}y=1\Rightarrow x=y$, assuming $y,\lambda\neq 0$.
Dividing $(2)$ by $(3)$ results in $\frac{y}z=1\Rightarrow y=z$, assuming $z,\lambda\neq 0$.
Plugging $y$ for $x$ and $z$ into $(4)$: $1-y-y-y=0\Rightarrow 1=3y\Rightarrow y=\frac13$
On
OK, then, here's another (geometric) way:
The shadow of any vector from the origin to a point in the plane on the vector $(1,1,1)$ has length $1/\sqrt 3,$ since for each such vector $\bf{v},$ we have that $|(1,1,1)| × |\bf{v}| \cos\phi=1,$ with $\phi$ being the angle between these vectors and the directing vector.
It follows that the smallest distance is the length of this shadow, and it's square is $1/3.$
You know how the radius of a circle is perpendicular to the tangent?
Likewise, the shortest segment from the origin to the plane $\color{blue}1x+\color{blue}1y+\color{blue}1z=1$
is perpendicular to the plane, i.e., in the direction $\color{blue}{(1,1,1)}$ , i.e., $(x,y,z)=(x,x,x)$.
If $x+x+x=1,$ then $x=\frac13,$ so $x^2+y^2+z^2=\frac13^2+\frac13^2+\frac13^2=\frac13.$