Answer
The point $\left( {\frac{1}{3}, - \frac{2}{3},\frac{2}{3}} \right)$ on the plane is closest to the point $P = \left( {1,0,0} \right)$.
Work Step by Step
We have the plane $z=x+y+1$ and a point $P = \left( {1,0,0} \right)$.
The square of the distance from $P$ to the plane is given by
$f\left( {x,y} \right) = {\left( {x - 1} \right)^2} + {y^2} + {\left( {x + y + 1} \right)^2}$
$ = 2{x^2} + 2{y^2} + 2xy + 2y + 2$
The partial derivatives are
${f_x} = 4x + 2y$, ${\ \ \ }$ ${f_y} = 4y + 2x + 2$
The point on the plane closest to $P$ is the point where $f\left( {x,y} \right)$ is minimal. Thus, we find the critical points of $f$ by solving the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 4x + 2y = 0$, ${\ \ }$ ${f_y} = 4y + 2x + 2 = 0$
From the first equation we get $y=-2x$. Substituting it in the second equation gives $x = \frac{1}{3}$.
So, the critical point is $\left( {\frac{1}{3}, - \frac{2}{3}} \right)$.
The second partial derivatives are
${f_{xx}} = 4$, ${\ \ \ }$ ${f_{yy}} = 4$, ${\ \ \ }$ ${f_{xy}} = 2$
The discriminant is $D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = 12 > 0$. By Theorem 2, $f\left( {\frac{1}{3}, - \frac{2}{3}} \right) = \frac{4}{3}$ is a local minimum.
Using the equation $z=x+y+1$ we obtain the point $\left( {\frac{1}{3}, - \frac{2}{3},\frac{2}{3}} \right)$ on the plane which is closest to the point $P = \left( {1,0,0} \right)$.
