geometrymultivariable-calculus
I have tried doing some research on this and am looking for the easiest way to compute this distance. For example, Let $l$ be the line determined by $x=y=z$. Find the shortest distance from this line to the point $(a, b, c)$. What is the easiest way to approach this?
Edit: Form is not important, assume we know the direction vector $\vec v$ and a point on the line $P$.
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.
The shortest distance should intuitively be the length of the segment which connects our point to our line and forms a right angle. Consider that we can determine this first by determining a vector from a point on our line to $A$ followed by finding its projection onto $\vec{PQ}$; the orthogonal part is then just the difference between this vector and our original vector from our line to our point and its magnitude yields the length of said segment:$$d=\left\|\vec{AP}-\frac{\vec{AP}\cdot\vec{PQ}}{\|\vec{PQ}\|^2}\vec{PQ}\right\|$$
Best Answer
First, draw a picture. Let $Q$ be the point $(a,b,c)$ and let $\theta$ be the angle between the vectors $\vec{v}$ and $\vec{PQ}$. Then, the point $R$ on the line $\ell$ which is closest to $Q$ is the point such that $\angle QRP = 90^{\circ}$, i.e. $QR$ is perpendicular to $\ell$. Then, the shortest distance from $Q$ to $\ell$ is $\|\vec{QR}\| = \|\vec{PQ}\|\sin\theta$. Using the formula $\|\vec{PQ} \times v\| = \|\vec{PQ}\|\|\vec{v}\|\sin\theta$, we get $\|\vec{QR}\| = \|\vec{PQ}\|\sin\theta = \dfrac{\|\vec{PQ} \times v\| }{\|v\|}$.
Alternatively, Google "distance between a point and a line" and click on the Wikipedia article.