# Homework Help: 4D: mimimum distance between point and parametric line

1. Aug 13, 2010

### dustydude

1. The problem statement, all variables and given/known data
There is points:
P=(1,2,3,4)
Q=(4,3,2,1)
and line L passes through P and is parallel to A
A=(1,1,1,1).
X(t) is anypoint on line L.

1. Find the distance between X and Q as a fucntion of t.
2. Find the minimum distance between Q and the line.(which is 2(51/2))

2. Relevant equations
Parametric equations....
distance between two points.

3. The attempt at a solution

Im really not sure if my approach is right at all.... i just want to understand where im going wrong.

For 1.
I tried the parametric equation of the line which is
X(t)=P+At

to go from X(t) to Q its
X(t)-Q=(P-Q)+At=(-3,-1,1,3)+(1,1,1,1)t
X(t)-Q=(-3+t,-1+t,1+t,3+t)
the distance between X(t) and Q would be
d=||(X(t)-Q)|| or d=((X(t)-Q).(X(t)-Q))1/2

For 2.
then the distance squared as a function of t would be
d2=(X(t)-Q).(X(t)-Q)

this gives d2=20+4t2
the derivative is (d2)'=2((d2)'d(2
(d2)'=(2(8t)(20+4t2)
(d2)'=320t+64t3

set the derivative to 0 for minimum and i get
t=(-320/64)1/2

Its clearly not the right answer of 2(5)1/2

I dont know where im going wrong.

Last edited: Aug 13, 2010
2. Aug 13, 2010

### hunt_mat

You can also minimise d^2 to get the shortest distance between a point and a line. This is just the normal from the line, can you just use that?

Mat

3. Aug 13, 2010

### dustydude

I've minimised d^2 by taking the derivative and setting to 0

4. Aug 13, 2010

### vela

Staff Emeritus
You differentiated incorrectly. What Mat is suggesting is that you minimize f(t)=20+4t2. There's no need to use the chain rule.

5. Aug 13, 2010

### dustydude

How can you minimize f(t)=20+4t2?
Is it not only do when is a minus?

6. Aug 13, 2010

### vela

Staff Emeritus
You can do it by inspection. It's a parabola, right?

Or you can solve f'(t)=0. Note I just called it f(t) instead of d2 since the notation seemed to be leading you astray.

7. Aug 13, 2010

### HallsofIvy

y= 20+ 4t2 is a parabola opening upward. It's minimum value is at the vertex (which is very easy to find here- t2 is never negative).

You may be thinking that y= 20+ 4t2 has no maximum- and that y= 20- 4t2 has no minimum.

8. Aug 13, 2010

### dustydude

so t=5 is the time where the distance between the line is at a minimum?
.....visualisation helps sometimes

the substitue t into X(t)

and then find distance betweeb X(t) and Q?

9. Aug 13, 2010

### dustydude

sorry i mean t=20

10. Aug 13, 2010

### vela

Staff Emeritus
No. How did you get t=20?

11. Aug 13, 2010

### dustydude

I see now i get what u mean by minimize.
Find the value of d^2 which is the smallest.

The smallest value of d^2 is 20.
so d=2(5)^1/2

Last edited: Aug 13, 2010