Consider the dashed line through the (0,0,0) and (0,1,1), and the dotted line through (0, 1, 0) and (1, 0, 0) as in the figure.

Question

Consider the dashed line through the (0,0,0) and (0,1,1), and the dotted line through (0, 1, 0) and (1, 0, 0) as in the figure.

(a) Let f(t) be the point that moves along the dashed line, with constant speed, such that f(0) = (0, 0, 0) and f(1) = (0, 1, 1). Let g(t) be the point that moves along the dotted line, with constant speed, such that g(0) = (0, 1, 0) and g(1) = (1, 0, 0). Calculate f(t) − g(t).

(b) Calculate the smallest value of ||f(t) − g(t)||. (Hint: minimize the square ||f(t) − g(t)||2.)

(c) Is the distance in part (2b) the smallest distance between any pair of points where one is in the dashed line and one in the dotted line? (Hint: It could help to calculate the cross product of the vectors parallel to the two lines.)

Answer 1

A SULUTION IN A NUTSHELL

Firstly lets put some order to this chaos of points and lines and.. planes...

Let A(0,0,0) and B(0,1,1) be the points of the "dashed line" which from now on we will call it line L_1,

and C(0,1,0) and D( 1,0,0) be the points of the " dotted line" which from now on we will call it line L_2.

The directing vector u of line L₁ is u = <0,1,1> and a vector valued function of the line L₁

is f (t) = < 0,t,t > , t ∈ ℜ

The directing vector v of line L₂ is v = <1, -1 ,0 > and a vector valued function of the line L₂

is g (s) = < 1+s, -s ,0 > , s ∈ ℜ

Any arbitrarily chosen point K on the line L₁ is of the form K ( 0,t,t) and any arbitrarily chosen point Ω

on line L₂ must be of the form Ω ( 1+s, -s ,0 ).

The distance d of points K and Ω is a real function of two variables t and s.

That is d (s, t ) = [ (1 +s)² + (t +s)² + t² ]^1/2

To minimize d suffices to minimize ζ (s, t ) = (1 +s)² + (t +s)² + t²

∂ ζ / ∂s = 4s +2t +2

∂ ζ / ∂t = 2s +4t

Solving the system 4s +2t +2=0 and 2s +4t =0 we obtain the critical point (s, t ) = (-2/3 , 1/3 )

Using the Second Derivative test we can safely conclude that at (-2/3 , 1/3 ) the function ζ has a minimum.

Therefore the function d (s, t ) has a minimum at ( -2/3, 1/3 ).

d( -2/3, 1/3 )= [ 1/9 + 1/9 +1/9 ] ^1/2 = √3 /3

In addition we can be pinpoint the " points" K and Ω as being K ( 0, 1/3, 1/3 ) and Ω ( 1/3, 2/3, 0 ) that make

the distance between L₁ and L₂ the smallest .

Additionally calculating the vector KΩ = < 1/3 , 1/3, - 1/3 > easily can be verified that

KΩ is orthogonal to both u and v.

Answer 2

t=0 to 1

f(t) =(0,t,t),,, g(t)=(t,-t+1.0)

correction

f(t)-g(t)=(-t,2t-1,t)

|f(t)-g(t)|=√[(t²)+(2t-1)²+(-t)²]=√[6t²-4t+1]

taking the derivative and setting to zero

derivative=(12t-4)*(1/2)*(6t²-4t+1)^(-1/2)

critical point t=1/3

the derivative is negative t<1/3 and poistive t>1/3 so it is minimum

distance for the motion and paths

end correction

but, as part c). indicates it may not be the mimimum distance if

the two lines have independent parameters

so the minimum distance is not bound to movement

let line1(L1) be (0,t,t), let line2(L2) be (s,-s+1,0)

want direcition of L2-L1 perpendicular to direction L2 (dot product=0)

direction L2=(1,-1,0)

2t+s-1=0

want direcition of L1-L2 perpendicular to direction L1 (dot product=0)

direction L1=(0,1,1)

t+2s-1=0

solving, t=1/3 s=1/3 ,,,& at those values

|L2-L1|=√3/3

check Is vector L2-L1 at t=s=1/3 perpendicular to direction of L1

dot product (1/3-0,2/3/-1/3.0-1/3) and (0,1,1)=0

so vectors perpendicular so distance is minimum