Parametric equations distance

[natashajane]

Calculate the distance between the 2 lines and use this distance to prove that the are not going to intersect.


x(t) = 2 + t
y(t) = -1 –t
z(t) = t

x(t) = 3 – s
y(t) = 1
z(t) = 1 + s

I have no idea where to start with this question! please help!

 

[Schrodinger's Dog]

Are you unable to make an equation given those?

 

[natashajane]

I suppose so if that's what I have to do

 

[natashajane]

For some reason the question is relating this to two balls traveling and has given the diameter of the two balls as well being 0.3 for the first one and 0.1 for the second.

 

[HallsofIvy]

The "distance between two (skew) lines" is always measured along the unique line perpendicular to both. The line x= 2+ t, y= -1-t, z= t has "direction vector" i- j+ k and so that is perpendicular to any plane perpendicular to the line can be written x- y+ z= C for some number C. The line x= 3- s, y= 1, z= 1+ s intersects that plane when 3-s-1+ 1+s= 2= C. That is, x= 3- x, y= 1, z= 1+ s lies in the plane x- y+ z= 2 perpendicular to the first line. That line, x= 2+ t, y= -1- t, z= t, intersects the plane when 2+ t- 1- t+ t= t+ 1= 2 or t= 1. That gives x= 2+1= 3, y= -1-1= -2, z= 1.

Now, what is the distance from the point (3, -2, 1) to the line x= 3- s, y= 1, z= 1+ s?

 

[natashajane]

Ive worked it out a different way... does this make sense:

x(t) = 2 + t
y(t) = -1 –t
z(t) = t

x(t) = 3 – s
y(t) = 1
z(t) = 1 + s

The directional vectors u and v, of the two lines are taken from the coefficients of the respective parameters.
u = <1, -1, 1>
v = <-1, 0, 1>

The minimum distance between them is perpendicular to both directional vectors. Take the cross product.
n = u x v = <1, -1, 1> x <1, 0, 1> = <-1, 2, -1>

Calculate a point on each line by setting the parameters (t) equal to zero.
P=(2,-1,0) Q=(3,1,1)

The two skew lines can be contained in parallel planes that have the normal vector n. The distance from any point on one plane to the other plane will be the same. Calculate the parallel plane that contains the first line. Use point P and the normal vector. For a vector in the plane use the arbitrary point R(x, y, z). The dot product of orthogonal vectors is zero.

n • PR = 0
<-1, 2, -1> • <x - 2, y + 1, z - 0> = 0
(x - 2) - 2(y + 1) + -1(z - 0) = 0
x - 2 - 2y - 2 -z = 0
x - 2y –z – 4 = 0

The distance from point Q in the second plane to the first plane is the minimum distance between the skew lines.
D= (|ax_1+ by_1+ cz_1+ d|)/√(a^2+ b^2+ c^2 )
D= (|3 -2 -1 -4|)/√(1^2+ 〖(-2)〗^2+ 〖(-1)〗^2 )

D= (-4)/√(1+4+1)= (-4)/√6= -1.632