4D: mimimum distance between point and parametric line

[dustydude]

Homework Statement

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 function of t.
2. Find the minimum distance between Q and the line.(which is 2(5^1/2))

Homework Equations

Parametric equations...
distance between two points.

The Attempt at a Solution

Im really not sure if my approach is right at all... i just want to understand where I am 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))^1/2For 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 don't know where I am going wrong.

 

[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

 

[dustydude]

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

 

[vela]

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

 

[dustydude]

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

 

[vela]

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.

 

[HallsofIvy]

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

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+ 4t^{2 has no maximum- and that y= 20- 4t2 has no minimum.}

 

[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?

 

[dustydude]

sorry i mean t=20

 

[vela]

No. How did you get t=20?

 

[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