I am trying to find the shortest distance from the point (3,0,-8) to the plane x+y+z = 8 and I keep getting the same incorrect solution. First, I found the equation fo the distance to be: $d=\sqrt{(x-3)^2 + (y^2) + (z+8)^2}$. Then I found the critical points of $d^2$. The critical point is $(\frac{14}{3}, \frac{5}{3}).$ The minimum distance must be at this critical point so I substituted these values into the equation for distance, with $z$ in terms of $x$ and $y$: $d=\sqrt{((\frac{14}{3})-3)^2 + (\frac{5}{3}^2) + (16-\frac{14}{3} – \frac{5}{3})^2}.$ This yields $3\sqrt{11}$, which is not the correct answer. Could someone help me understand what I am doing incorrectly?
[Math] Finding the Shortest Distance from Point to Plane
multivariable-calculusoptimizationpartial derivative
Related Solutions
I think that the method of Lagrange multipliers is the easiest way to solve my question, but how can I find the Lagrangian function?
As shown by other answers and in note 1 there are easier ways to find the shortest distance, but here is a detailed solution using the method of Lagrange multipliers. You need to find the minimum of the distance function
$$\begin{equation} d(x,y,z)=\sqrt{x^{2}+y^{2}+(z-1)^{2}} \tag{1a} \end{equation}$$
subject to the constraint given by the surface equation $z=f(x,y)=\frac32(x^2+y^2)$ $$\begin{equation} g(x,y,z)=z-\frac{3}{2}\left( x^{2}+y^{2}\right) =0. \tag{2} \end{equation}$$ Since $\sqrt{x^{2}+y^{2}+(z-1)^{2}}$ increases with $x^{2}+y^{2}+(z-1)^{2}$ you can simplify the computations if you find the minimum of $$\begin{equation} [d(x,y,z)]^2=x^{2}+y^{2}+(z-1)^{2} \tag{1b} \end{equation}$$ subject to the same constraint $(2)$. The Lagrangian function is then defined by $$\begin{eqnarray} L\left( x,y,z,\lambda \right) &=&[d(x,y,z)]^2+\lambda g(x,y,z) \\ L\left( x,y,z,\lambda \right) &=&x^{2}+y^{2}+(z-1)^{2}+\lambda \left( z- \frac{3}{2}\left( x^{2}+y^{2}\right) \right), \tag{3} \end{eqnarray}$$ where $\lambda $ is the Lagrange multiplier. By this method you need to solve the following system $$\begin{equation} \left\{ \frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial y}=0, \frac{\partial L}{\partial z}=0,\frac{\partial L}{\partial \lambda } =0,\right. \tag{4} \end{equation}$$ which results in
$$\begin{eqnarray} \left\{ \begin{array}{c} 2x+3\lambda x=0 \\ 2y+3\lambda y=0 \\ 2z-2-\lambda =0 \\ -z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 \end{array} \right. &\Leftrightarrow &\left\{ \begin{array}{c} x=0\vee 2+3\lambda =0 \\ y=0\vee 2+3\lambda =0 \\ 2z-2-\lambda =0 \\ -z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0 \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} x=0 \\ y=0 \\ \lambda =2 \\ z=0 \end{array} \right. \vee \left\{ \begin{array}{c} \lambda =-2/3 \\ z=2/3 \\ x^{2}+y^{2}=4/9 \end{array} \right. \tag{5} \end{eqnarray}$$
For $x=y=x=0$ we get $d(0,0,0)=1$. And for $x^2+y^2=4/9,z=2/3$ we get the minimum distance subject to the given conditions $$\begin{equation} \underset{g(x,y,z)=0}\min d(x,y,z)=\sqrt{\frac{4}{9}+(\frac{2}{3}-1)^{2}}=\frac{1}{3}\sqrt{5}. \tag{6} \end{equation}$$
It is attained on the intersection of the surface $z=\frac{3}{2}\left( x^{2}+y^{2}\right) $ with the vertical cylinder $x^{2}+y^{2}=\frac{4}{9}$ or equivalently with the horizontal plane $z=\frac{2}{3}$.
$$\text{Plane }z=\frac{2}{3} \text{(blue) and surface }z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $$
Notes.
- As the solution depends only on the sum $r^{2}=x^{2}+y^{2}$ we could just find $$\begin{equation} \min [d(r)]^2=r^{2}+(\frac{3}{2}r^{2}-1)^{2} \tag{7} \end{equation}$$ and then find $d(r)=\sqrt{[d(r)]^2}$ at the minimum.
- The surface $z=\frac{3}{2}\left( x^{2}+y^{2}\right) $ is a surface of revolution around the $z$ axis.
The flaw is in your diagram. For your method to work, the point $A$ must lie on the projection of the line onto the plane, otherwise the vector $\vec{AR}$ will be oblique to the plane.
While drawing the figure you have assumed that the point $A$ is lying on the projection of $L$ onto $\Pi$ while it's actually not.
Best Answer
You are trying to minimize $d=\sqrt{(x−3)^2+y^2+(z+8)^2}$ subject to $x+y+z = 8$. This is equivalent to minimizing $$ (x−3)^2+y^2+(z+8)^2 $$ s.t. $$ x+y+z=8 $$ because the new function is a strictly monotonic transformation of the original function. If you make the relevant substitution $z=8-x-y$ your FOC's would look like $$ 2(x-3) - 2(16-x-y) = 0 $$ and $$ 2y - 2(16-x-y) = 0 $$ implying that $$ x = y+3. $$ It should be straightforward from here onwards.
note: thanks Carl below for suggesting edit.