Question

A particle is traveling along the line (x-3)/2=(y+1)/(-2)=z-1. Write the equation of its path in the form $\mathbf{r}\mathbf{=}{\mathbf{r}}_{\mathbf{0}}\mathbf{+}\mathbf{At}$. Find the distance of closest approach of the particle to the origin (that is, the distance from the origin to the line). If t represents time, show that the time of closest approach is $\mathbf{t}\mathbf{=}\mathbf{-}\left(r_0\times A\right)\mathbf{/}{\left|A\right|}^{\mathbf{2}}$. Use this value to check your answer for the distance of closest approach. Hint: See Figure 5.3. If P is the point of closest approach, what is $\mathbf{A}\mathbf{\times }{\mathbf{r}}^{\mathbf{2}}$?

Short answer

The value of $\mathrm{A}\times {\mathrm{r}}^2\mathrm{is}\sqrt{2}$.

Step-by-step solution

Given information:

A particle is traveling along the line $\left(x-3\right)/2=\left(y+1\right)/\left(-2\right)=z-1$.

Concept used

The symmetric form $\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_0}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_0}{\mathrm{c}}$implies that $\left(x_0,y_0,z_0\right)$is on the line.

Given that a vector v=ai+bj+ck is parallel to the line described by the equation

$\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_0}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_0}{\mathrm{c}}$

The vector along the line is A=2i-2j+k .

So, the equation of the particles path is then $\mathrm{r}-{\mathrm{r}}_0+\mathrm{At}=\left(3,-1,1\right)+\left(2,-2,1\right)\mathrm{t}$.

Letbe the distance from the origin to the line. Ifis a vector in the direction of the line, then$u-\frac{\mathrm{A}}{\left|\mathrm{A}\right|}$is a unit vector in that direction.

$$\begin{gathered} \mathrm{A}-\left|\mathrm{A}\right|=\sqrt{{\left(2\right)}^2+{\left(-2\right)}^2+{\left(1\right)}^2}-\sqrt{9}-3 \\ \mathrm{u}=\frac{1}{3}\left(2\mathrm{i}-2\mathrm{j}+\mathrm{k}\right) \end{gathered}$$

Distance between the points

The point (3,-1, 1) is on the line, so the vector $\mathrm{B}=\left(3,-1,1\right)-\left(0,0,0\right)$is a vector from the origin to the point $\left(3,-1,1\right),\mathrm{B}=3\mathrm{i}-\mathrm{j}+\mathrm{k}$.

Hence,

$$\begin{gathered} \mathrm{d}-\left|\mathrm{B}\times \mathrm{u}\right|=\frac{1}{3}\left|\left(3\mathrm{i}-\mathrm{j}+\mathrm{k}\right)\times \left(2\mathrm{i}-2\mathrm{j}+\mathrm{k}\right)\right| \\ =\left(\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 3& -1& 1\\ 2& -2& 1\end{array}\right)=-\mathrm{i}-\mathrm{j}-4\mathrm{k} \\ \mathrm{d}-\left|\mathrm{B}\times \mathrm{u}\right|=\frac{1}{3}\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2+{\left(-4\right)}^2}=\frac{\sqrt{18}}{2}=\sqrt{2} \end{gathered}$$

The distance of the closest point

Now consider the equation of the form$\mathrm{r}={\mathrm{r}}_0+\mathrm{At}$.

The distance squared from the origin at any time is given by $\left|{\mathrm{r}}^2\right|=r\times r$. Thus

$$\begin{gathered} \mathrm{r}\times \mathrm{r}=\left({\mathrm{r}}_0+\mathrm{At}\right)\times \left({\mathrm{r}}_0+\mathrm{At}\right) \\ ={\mathrm{r}}_0\times {\mathrm{r}}_0+\mathrm{A}\times {\mathrm{At}}^2+2{\mathrm{r}}_0\times \mathrm{At} \end{gathered}$$

But $\mathrm{A}\times \mathrm{A}$is simply ${\left|\mathrm{A}\right|}^2$, and${\mathrm{r}}_0={\left|{\mathrm{r}}_0\right|}^2,$

${\left|\mathrm{r}\right|}^2={\left|{\mathrm{r}}_0\right|}^2+{\left|\mathrm{A}\right|}^2{\mathrm{t}}^2+2{\mathrm{r}}_0\times \mathrm{At}$

Now we need to minimize the distance${\left|\mathrm{r}\right|}^2$, so we need to find${\mathrm{t}}_0$at which the derivate of the square vanishes (that is$\frac{\mathrm{d}{\left|\mathrm{r}\right|}^2}{\mathrm{dt}}=0)$)

$$\begin{gathered} \frac{\mathrm{d}{\left|\mathrm{r}\right|}^2}{\mathrm{dt}}=2{\left|A\right|}^2{\mathrm{t}}_0+2{\mathrm{r}}_0\times \mathrm{A}=0 \\ {\mathrm{t}}_0=\frac{\mathrm{A}\times {\mathrm{r}}_0}{{\left|\mathrm{A}\right|}^2} \end{gathered}$$

Calculating for the given and

$$\begin{gathered} {\mathrm{t}}_0=\frac{\left(2\right)\left(3\right)+\left(-2\right)\left(-1\right)+\left(1\right)\left(1\right)}{9} \\ =\frac{9}{9} \\ =-1 \\ \mathrm{r}={\mathrm{r}}_0-\mathrm{A} \\ =\left(3.-1,1\right)-\left(2,-2,1\right) \\ =\left(1,1,0\right) \\ \mathrm{d}=\left|\mathrm{r}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2} \\ =\sqrt{2} \end{gathered}$$