Finding shortest distance between an observer and a moving object

[MatinSAR]

Homework Statement

Find the shortest distance of an observer at the point (2, 1, 3) from a rocket in free
flight with velocity (1, 2, 3) m/s. The rocket was launched at time t = 0 from (1, 1, 1).
Lengths are in kilometers.

Relevant Equations

Please see below.

The rocket has constant velocity so we can write it's equation of motion as :
$$\vec r = \vec r_0+t \vec v_0 $$ We can write it for components along each axis :
$$x = x_0 + v_{0,x}t$$$$y = y_0 + v_{0,y}t$$$$z = z_0 + v_{0,z}t$$We put known values in above equations :$$x=1000+t$$$$y=1000+2t$$$$z=1000+3t$$We found position of rocket as a function of time. So we can use Distance formula in 3D to find distance of the observer from rocket as a function of time.$$D(t)=((t-1000)^2+(2t)^2+(3t-2000)^2)^{(\frac 1 2)}$$

We want to find "t" in which D(t) becomes minimum.
$$\frac {d}{dt} D(t) = 0$$
$$\frac {d}{dt} ((t-1000)^2+(2t)^2+(3t-2000)^2)^{(\frac 1 2)}=0$$$$2t-2000+8t+18t-12000=0$$$$28t-14000=0$$ $$t=500$$Minimum Distance between them :
$$D(t_{min})=D(t=500)=((500-1000)^2+(2*500)^2+(3*500-2000)^2)^{(\frac 1 2)}=1224.74 m$$

Is my answer correct?

 

[PeroK]

That's what I get, although I took a few shortcuts. Like moving the origin to the launch point of the rocket.

 

[MatinSAR]

Like moving the origin to the launch point of the rocket.

Good idea.

@PeroK Thank you for your help.

@erobz Thank you for your help.

 

[erobz]

Latex tip: I think that "\dfrac" must keep the exponent larger than it should be, I would just use "\frac{}{}".

Also, parenthesis that automatically size " \left( [your math] \right)"

I assume you know about "\sqrt{}",but don't prefer it.

 

[MatinSAR]

Latex tip: I think that "\dfrac" must keep the exponent larger than it should be, I would just use "\frac{}{}".

Agree. Thank you.

I assume you know about "\sqrt{}",but don't prefer it.

Yes. But now I see that here it is better to use "\sqrt{}".

 

[kuruman]

Another shortcut that avoids minimizing is to note that when the shortest distance occurs, the velocity of the rocket is perpendicular to the rocket's displacement from the observer.
Let $\vec X_O=~$ observer's position
Let $\vec X_R=~$ rocket's initial position
Let $\vec V_R=~$ rocket's velocity
The displacement vector is
$\vec D = \vec X_R+\vec V_R~t-\vec X_O$
and the time of closest approach is given by$$0=\vec V_R\cdot \vec D=\vec V_R\cdot (\vec X_R+\vec V_Rt-\vec X_O)=V_R^2~t+\vec V_R\cdot(\vec X_R-\vec X_O) \implies t = \frac{\vec V_R\cdot(\vec X_O-\vec X_R)}{V_R^2}. $$

 

[MatinSAR]

Another shortcut that avoids minimizing is to note that when the shortest distance occurs, the velocity of the rocket is perpendicular to the rocket's displacement from the observer.

I haven't known about it. Thanks a lot for your time.

 

[PeroK]

Note also that the velocity units are not relevant (as long as they are the same in every direction). Nor is the magnitude of the velocity relevant. So, you don't actually have to convert km to m or vice versa. Only the direction of the velocity matters.

 

[PeroK]

PS if you follow through with the @kuruman solution, then you should see the magnitude of velocity cancel out.

 

[kuruman]

PS if you follow through with the @kuruman solution, then you should see the magnitude of velocity cancel out.

And that makes eminent sense. The straight line of the rocket's trajectory depends only on the direction but not the magnitude of the rocket's velocity. To elaborate on post #8, one can always draw that line regardless of how fast the rocket is moving and then find the perpendicular to it from the location of the observer.

 

[MatinSAR]

Another shortcut that avoids minimizing is to note that when the shortest distance occurs, the velocity of the rocket is perpendicular to the rocket's displacement from the observer.
Let $\vec X_O=~$ observer's position
Let $\vec X_R=~$ rocket's initial position
Let $\vec V_R=~$ rocket's velocity
The displacement vector is
$\vec D = \vec X_R+\vec V_R~t-\vec X_O$
and the time of closest approach is given by$$0=\vec V_R\cdot \vec D=\vec V_R\cdot (\vec X_R+\vec V_Rt-\vec X_O)=V_R^2~t+\vec V_R\cdot(\vec X_R-\vec X_O) \implies t = \frac{\vec V_R\cdot(\vec X_O-\vec X_R)}{V_R^2}. $$

My method is probably unacceptable. Because this exercise is from mathematical methods for physicists(arfken ) and it was in dot product section. I guess I should solve the problem again but using this metod.

@PeroK @kuruman Thanks a lot for your help.

 

[PeroK]

We should really tie all this together. Note that if we let $\vec D_0 = \vec X_R - \vec X_O$, then we have:
$$\vec D = \vec Vt + \vec D_0$$$$D^2 = \vec D \cdot \vec D$$$$\frac d {dt}(D^2) = 2\vec D \cdot \frac d {dt}(\vec D) = 2\vec V \cdot \vec D$$And we can see that $\dfrac d {dt}(D^2) = 0$ when $\vec V \cdot \vec D = 0$. Which must be the minimum.

Moreover:
$$\vec V \cdot \vec D = 0 \ \Rightarrow \ V^2t + \vec V \cdot \vec D_0 = 0 \ \Rightarrow \ t = -\frac{\vec V \cdot \vec D_0}{V^2}$$And, for the minimum distance we have:
$$\vec D = \vec Vt + \vec D_0 = \vec D_0 - \frac{\vec V \cdot \vec D_0}{V^2}\vec V = \vec D_0 - (\hat V \cdot \vec D_0)\hat V \ \ \ (\text{where} \ \hat V = \frac{\vec V}{V})$$$$D^2 = D_0^2 - 2(\vec D_0 \cdot \hat V)^2 + (\hat V \cdot \vec D_0)^2$$$$D^2 = D_0^2 - (\vec D_0 \cdot \hat V)^2$$Let's check that in this case:
$$\vec D_0 = (-1, 0, -2), \hat V = \frac 1 {\sqrt {14}}(1,2,3), \ D_0^2 = 5, \ \vec D_0 \cdot \hat V =- \frac 7 {\sqrt {14}} $$$$D^2 = \frac 3 2$$

 

[MatinSAR]

We should really tie all this together. Note that if we let $\vec D_0 = \vec X_R - \vec X_O$, then we have:
$$\vec D = \vec Vt + \vec D_0$$$$D^2 = \vec D \cdot \vec D$$$$\frac d {dt}(D^2) = 2\vec D \cdot \frac d {dt}(\vec D) = 2\vec V \cdot \vec D$$And we can see that $\dfrac d {dt}(D^2) = 0$ when $\vec V \cdot \vec D = 0$. Which must be the minimum.

Moreover:
$$\vec V \cdot \vec D = 0 \ \Rightarrow \ V^2t + \vec V \cdot \vec D_0 = 0 \ \Rightarrow \ t = -\frac{\vec V \cdot \vec D_0}{V^2}$$And, for the minimum distance we have:
$$\vec D = \vec Vt + \vec D_0 = \vec D_0 - \frac{\vec V \cdot \vec D_0}{V^2}\vec V = \vec D_0 - (\hat V \cdot \vec D_0)\hat V \ \ \ (\text{where} \ \hat V = \frac{\vec V}{V})$$$$D^2 = D_0^2 - 2(\vec D_0 \cdot \hat V)^2 + (\hat V \cdot \vec D_0)^2$$$$D^2 = D_0^2 - (\vec D_0 \cdot \hat V)^2$$Let's check that in this case:
$$\vec D_0 = (-1, 0, -2), \hat V = \frac 1 {\sqrt {14}}(1,2,3), \ D_0^2 = 5, \ \vec D_0 \cdot \hat V =- \frac 7 {\sqrt {14}} $$$$D^2 = \frac 3 2$$

Great!!
Thank you for the help .

 

[kuruman]

We should really tie all this together. Note that if we let $\vec D_0 = \vec X_R - \vec X_O$, then we have:
$$\vec D = \vec Vt + \vec D_0$$$$D^2 = \vec D \cdot \vec D$$$$\frac d {dt}(D^2) = 2\vec D \cdot \frac d {dt}(\vec D) = 2\vec V \cdot \vec D$$And we can see that $\dfrac d {dt}(D^2) = 0$ when $\vec V \cdot \vec D = 0$. Which must be the minimum.

Moreover:
$$\vec V \cdot \vec D = 0 \ \Rightarrow \ V^2t + \vec V \cdot \vec D_0 = 0 \ \Rightarrow \ t = -\frac{\vec V \cdot \vec D_0}{V^2}$$And, for the minimum distance we have:
$$\vec D = \vec Vt + \vec D_0 = \vec D_0 - \frac{\vec V \cdot \vec D_0}{V^2}\vec V = \vec D_0 - (\hat V \cdot \vec D_0)\hat V \ \ \ (\text{where} \ \hat V = \frac{\vec V}{V})$$$$D^2 = D_0^2 - 2(\vec D_0 \cdot \hat V)^2 + (\hat V \cdot \vec D_0)^2$$$$D^2 = D_0^2 - (\vec D_0 \cdot \hat V)^2$$Let's check that in this case:
$$\vec D_0 = (-1, 0, -2), \hat V = \frac 1 {\sqrt {14}}(1,2,3), \ D_0^2 = 5, \ \vec D_0 \cdot \hat V =- \frac 7 {\sqrt {14}} $$$$D^2 = \frac 3 2$$

That's a lot of work to get to $$D^2 = D_0^2 - (\vec D_0 \cdot \hat V)^2.$$
Note that the initial positions of rocket and observer, points A and O respectively, and the point P of closest approach form a right triangle shown below.

Distance AP is the projection of the initial displacement on the rocket's path, $\mathbf{D}_0\cdot \mathbf{\hat v}$. The Pythagorean theorem gives the distance of closest approach, $$D_{\text{min}}=\sqrt{D_0^2-(\mathbf{D}_0\cdot \mathbf{\hat v})^2}.$$