Determine the distance between the following pair of parallel lines.?

Question

r = (1,2,2) + t(2,3,-1)
r = (-1,1,3) + t(2,3,-1)

Showing work would be great.

Answer 1

Hi,

The point you know on the second line is (-1,1,3). Any point on the first line has to have the coordinates of the form
(1 + 2T, 2 + 3T, 2 - T). To find the distance between them, we can use the distance formula,

Distance = sqrt((X2 - X1)^2 + ( Y2 - Y1)^2 + (Z2 - Z1)^2) which for these points is:

Dist = sqrt[(1 + 2T -(-1))^2 + (2 + 3T - 1)^2 + ( 2 - T - 3)^2 ]

Dist = sqrt[ (2T + 2)^2 + (3T + 1)^2 + ( -T - 1)^2 ]

Dist = sqrt[ 4T^2 + 8T + 4 + 9T^2 + 6T + 1 + T^2 + 2T + 1 ]

Dist = sqrt[ 14T^2 + 16T + 6]

Since this ends up being a quadratic equation inside the radical, we know it has a minimum value at its vertex. This would be the value of T when the minimum distance would occur. You could plug this T value back into the first line's point, (1 + 2T, 2 + 3T, 2 - T), to find the point on line 1 closest to the point at (-1,1,3) on line 2. Then we could use the distance formula again to find the actual distance between the 2 points with all numerical values.

For the quadratic 14T^2 + 16T + 6, its low point is its vertex, which is found from the formula x = -b/(2a). In this case b = 16 and a = 14, so -b/(2a) = -16/28 = -4/7. The time when these points are closest is when T = -4/7.

(1 + 2T, 2 + 3T, 2 - T) becomes
(1 + 2(-4/7), 2 + 3(-4/7), 2 - (-4/7)) = ( -1/7, 2/7, 18/7)

To find the distance between the lines use the distance formula to find the distance between ( -1/7, 2/7, 18/7) and
(-1,1,3).

Distance = sqrt((X2 - X1)^2 + ( Y2 - Y1)^2 + (Z2 - Z1)^2) which for these points is:

Distance = sqrt((-1 - (-1/7))^2 + ( 1 - (2/7))^2 + (3 - (18/7))^2)

= sqrt[( -6/7)^2 + (5/7)^2 + (3/7)^2 ]

= sqrt[ 36/49 + 25/49 + 9/49 ] = sqrt[ 70/49 ] = 1.195

I'd say that's the distance between these 2 parallel lines.



I hope that helps you!! :-)

Answer 2

Given two parallel lines r1 and r2, find the distance between them.

r1 = <1, 2, 2> + s<2, 3, -1>
r2 = <-1, 1, 3> + t<2, 3, -1>

The directional vector v, of the lines is <2, 3, -1>.
The magnitude of v is √[2² + 3² + (-1)²] = √(4 + 9 + 1) = √14.

To find a point on r1, let s = 0 and we have point P(1, 2, 2).
To find a point on r2, let t = 0 and we have point Q(-1, 1, 3).

Create the vector PQ.

PQ = <1+1, 2-1, 2-3> = <2, 1, -1>

The distance d, between the two lines is given by:

d = || PQ X v || / || v ||
d = || <2, 1, -1> X <2, 3, -1> || / √14
d = || <2, 0, 4> || / √14
d = √(2² + 0² + 4²) / √14
d = √(20/14) = √(10/7) ≈ 1.1952286

Answer 3

The 2 lines are parallel so the length of the vector that is between them is equal to the distance between the lines

Using the position vectors A(1,2,2) and B(-1,1,3) from the equation of the lines, you can find the vector AB.

The vector AB will be the vector lying between them. (Try drawing it to get the visual)

The vector AB = ((-1-1),(1-2),(3-2)) = (-2,-1,1)
The length of AB = sqrt ((-2)^2 + (-1)^2 + (1)^2)
=sqrt (4+1+1)
= sqrt (6)